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Can i get that answer? 8+3x = -7

8+3x= 7      this ecuation

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3x+7=8 - Math homework help -

Is that your homework? Bulls eye!!! ... Normally we charge $1.00 for this answer Today, you can get it FOR FREE if you ... 3x+7=8. 3x+7=8.
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How do you solve -3x+1= 5x-7? | Socratic

... (blue)(step 1):- 1=5x+3x-7 color(blue)(step 2:- 1+7=5x+3x color(blue)(step 3:- 8=8x color(blue)(step 4:- x ... You can reuse this answer Creative Commons ...
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Checking Answers Using Algebra Calculator - MathPapa

Checking Answers Using Algebra Calculator. ... Example Answer x=6 ... For system of equations x+y=8 and y=x+2, check ...
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How can x+8=-x+7 be set up as a system of equatio... - OpenStudy

How can x+8=-x+7 be set up as a system of equations A. Y=x+8 Y=-x+7 B. y=x+8 Y=-3x+7 C. Y=x+5 Y=-2x+11 D. Y=8 Y=-3x+7This equation ... Get our expert's. answer on ...
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3x-7=1-x - Algebra homework help

You can ask homework questions and get assistance. ... Normally we charge $1.00 for this answer Today, you can get it FOR FREE if you ... (8 words) 3x-7=1-x 3x+x-7=1 ...
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How do you solve 2x-y=7 and 3x+y=8? | Socratic

Substitute this expression for y in the second equation and get 3x + (2x-7) = 8 5x-7=8 5x ... How do you solve #2x-y=7# and #3x+y=8#? ... Start with a one sentence ...
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What is the solution to the compound inequality 5x + 7 > −8 ...

What is the solution to the compound inequality 5x + 7 > −8 and 3x + 7 ≤ 19? - 2064399. 1. ... Not sure about the answer?
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Which best describes the solution set for the inequality ...

Which best describes the solution set for the inequality below? 3x + 7 ≤ 4x – 8 or –2x + 3 ≥ 1 - 2939643 ... Answer quality is ensured by our experts.
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Can You Solve: 7(5y-2)>6(6y-1) -3x+8 |

Answer to Can you solve: 7(5y-2)>6(6y-1) -3x+8... Textbook Solutions Expert Q&A ... Get this answer with Chegg Study View this answer. OR.
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solve for X, show work 5x + 7 + 3x = -8 + 3x - OpenStudy

solve for X, show work 5x + 7 + 3x = -8 + 3xadd the x's on either side of the equation and tell me what you get. ... We 've verified this expert answer for you, ...
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Suggested Questions And Answer :

statistic probability

i. All the possible answers in Quiz 1 by chance can be represented by the terms in the expansion of (0.25+0.75)^8, where 0.25 is the probability of a correct choice of answer for each individual question and 0.75 is the probability of an incorrect choice of answer. The binomial expression has a value of 1, which is a probability of 1, i.e., certainty, made up of the sum of all the terms. Let p=0.25 and 1-p is 0.75. The expansion is p^8+... + 8Cr*p^r(1-p)^(8-r), where r indicates the rth term between r=0 and 8 and 8Cr is the combination function nCr for n=8. The value of 8Cr is given by (8*7*...*(8-r+1))/(1*2*...*r). The coefficients are, in order: 1, 8, 28, 56, 70, 56, 28, 8, 1, the 8th row of Pascal's triangle. The binomial terms start with the probability of 8 correct answers, then 7 correct 1 incorrect, 6 correct 2 incorrect, 5 correct 3 incorrect, and so on. The probability of answering incorrectly 3 question is (3/4)^3=27/64, so the probability of answering at least one question in 3 correctly by chance is 1-27/64=37/64, in other words, this is the probability that at most 3 questions are attempted before obtaining a correct answer. This is better than an evens chance (50%), so is more likely than unlikely. ii. In Quiz 2 p=1/5 or 0.20 and n=15. The probability of less than 4 correct answers is the sum of the last four terms of the binomial series for 15, which give all incorrect, 1, 2, 3 correct. The coefficients are respectively 1, 15, 105, 455. The p terms alone are 0.03518, 008796, 0.002199 and 0.0005498. When we multiply by the coefficients and add the terms together we get 0.6482 or 64.82% probability. iii. The probability of 6 or more correct answers in Quiz 1 is 1-(probability of up to 2 incorrect answers)=1-(0.75^8+8*0.75^7*0.25+28*0.75^6*0.25^2)=1-0.6785=0.3215 or 32.15%. This is less than 1/3, so the odds are 2:1 against Siti getting 6 or more correct answers by chance alone, which makes it reasonably unlikely. iv and v. The problem says "before the 8th question", so there need to be 5 correct answers within the first 7 questions. Using p=0.2, the probability of getting 5 correct answers out of 5 is 0.2^5=0.00032; the probability of getting 5 correct out of 6 is 6*0.2^5*0.8=0.001536; and the probability of 5 out of 7 is 21*0.2^5*0.8^2=0.0043008. Therefore the probability of getting 5 correct before the 8th question is the sum of these=0.0061568 or 0.62%, a most unlikely occurrence, less than 1 in 162.
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times student will get 3 out of 10 questions right randomly guessing 4 choice questions

This is a binomial distribution, where p=1/4=0.25, the chance of getting the correct answer, and 1-p=0.75, the chance of getting the wrong answer. The distribution's last 4 terms are: (1-p)^10, 10(1-p)^9p, 45(1-p)^8p^2, 120(1-p)^7p ^3. These represent the chances of 0, 1, 2, 3 correct answers in 10. The coefficients are the number of ways of 0, 1, 2, 3 correct answers can be mixed with wrong answers. We need the last in this list which evaluates to 120*0.75^7*0.25^3=0.25 approx. and is the probability of getting exactly 3 right out of 10.
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What is 5/3 divide by 1/3?

If the question had been: what's 5 apples divided by one apple you would have said 5, wouldn't you? The question has thirds instead of apples, but the answer is still 5. Don't forget that to divide by a fraction you invert the fraction and multiply: 5/3 times 3 is 5, because 3 cancels out the denominator in 5/3. I'm sorry you didn't get your answer in 10 minutes as you wished, but I've only just seen it! There are thousands of questions coming in and questions cannot always be answered just as they come in, and you have to remember that the website is available all over the world, which is split into many time zones. Add to that the fact that the questions are answered by users, like me, who have a life outside answering and asking questions! There are meals to eat, shopping to be done, going to the office or factory, going to school, you know the sort of thing. And, of course, while your question is coming in, a user may be answering another question, perhaps a harder one. I'm sure everyone does their best to help people who have questions to answer. But, be assured that there are many users who will be only too pleased to help you. It's just a question of when. Here's a clue that may help you. Below your question you will usually find a list of similar questions, many of which will probably have been answered (shown in green with the number of answers), so if you look at them, you may well find out how to do your own question. Now that will save everyone's time!  
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In a mathematics emathematics, each correct answergets 2 marks and a wrong answer attract -1, ola got a total of 19 in aquiz with 20 questions. How many correct answers did he get?

If we call the number of right answers R and the number of wrong answers W, 2R-W=19. Also we know W+R=20. Add the two equations: 3R=39 so R=13, 13 correct answers. 13 correct answers gains 26 points; 7 incorrect answers loses 7 points and 26-7=19.
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The first die is 3 or the sum is 8

The answer posted above is correct. However the same answer can be arrived at in another way also. Tho get a total of 8 for 2 dies none of the dice must have a '1'. The probability that the first dice will not have a '1', or that it will have 2, 3, 4, 5, or 6, are: Probability of first dice not getting 1 = 5/6 This is because the probability of getting any number from 1 to 6 are equal. For each occurrence of the number 2 to 6 for dice the second dice must get a number given by: Required number if dice 2 = 8 - n Where: n = number of dice 1 As the probability of getting any one number is equal to 1/6, the probability that the second dice will have exactly the required number is: Probability of second dice getting the number (8 - n) = 1/6 The probability of getting sum of 8: Probability that sum is 8 = (Probability of first dice not getting 1)*(Probability of second dice getting the number (8 - n)) = (5/6)*(1/6) = 5/36 Answer: Probability of sum of two dice as 8 = 5/36
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Divide 2x4 - 5x3 - 8x2 - 17x - 4 by x + 4

I'm assuming the question reads: divide 2x^4-5x^3-8x^2-17x-4 by x+4, where the caret (^) means "to the power of". Long division in algebra is similar to long division in arithmetic. We lay out the dividend, which is the long expression, as we would lay out a number we're going to divide into. To the left of it we write the divisor, which is x+4. How many x's go into 2x^4? The answer is 2x^3 because 2x^3*x=2x^4. OK, the first term in the quotient (the answer) is 2x^3. We multiply x+4 by 2x^3, which gives us 2x^4+8x^3, and we write this under the dividend so that the powers of x line up. We should have 2x^4 under 2x^4 in the long expression and 8x^3 under 5x^3 in the long expression. We now have to work out 2x^4-5x^3-(2x^4+8x^3). Always beware of the + and - signs, it's so easy to make a mistake. The subtraction gives us -13x^3. We now look at this result and the next term in the long expression which is -8x^2. How many times does x go into -13x^3? The answer is -13x^2 times. That's the next term in the quotient following 2x^3. Multiply x+4 by -13x^2 and we get -13x^3-52x^2, subtract this from -13x^3-8x^2 and we get 44x^2. We continue the process bringing in successive terms from the dividend until we get to the end, when we'll perhaps be left with a remainder. The answer is the quotient=2x^3-13x^2+44x-193 remainder 768. Is that the answer you get?
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what is the answer to 18-4t/0.5=20 ?

Perhaps it was the fraction that put you off, or was it the decimal in the denominator. I guess you want to find the value of t. Let's get rid of the fraction by multiplying both sides of the equation by 0.5. It's not clear in the question whether the problem is 18-(4t/0.5)=20 or (18-4t)/0.5=20. The brackets tell us what has to be worked out first. I'm going to provide solutions whichever one of these is the problem. First, 18-(4t/0.5)=20 Multiplying both sides by 0.5 we get 18*0.5-4t=20*0.5 giving us 9-4t=10. The decimals have been swallowed up in the multiplication. 0.5 is the same as 1/2, but if you didn't know that, multiply by 5 and move the decimal point back one place. So, for example, 5*18=90. The decimal point invisibly sits after 90, so moving it back a place makes it 9.0 which is just 9. So 9-4t=10. We need to get the numbers on one side of the equation and the unknown on the other side. At this stage we don't want to separate 4 from 4t. OK, let's keep the unknown on the left, so we get -4t=10-9, giving us -4t=1. When we move things from one side of an equation to the other plus becomes minus, and minus becomes plus, multiply becomes divide, and divide becomes multiply. So this is the same as 4t=-1, and t=-1/4. Second, (18-4t)/0.5=20 Multiply both sides by 0.5 and we get 18-4t=20*0.5, in other words 18-4t=10. Let's separate the unknown from the numbers, but this time we'll take the unknown over to the right (it doesn't matter which side really). So we get 18-10=4t. Therefore 8=4t, the same as 4t=8, so t=2. That looks like the simpler of the two answers, so my guess is that it's what the question really was, don't you think? It's important that questions are written correctly because, as you can see, we came up with two quite different answers.
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10 vedic maths rules for class IX

2 instead of 5: 34/5 can be calculated by multiplying 34 by 2 instead of dividing by 5. 34*2=68. We just move the decimal point one place to the left: 34/5=6.8. 124/5=24.8 because 124*2=248. Move the decimal point: 248 becomes 24.8. 34*5 is the same as 34/2=17 but we add a zero to make 17 into 170. 73*5 is the same as 73/2=36.5 then move the decimal point one place to the right (or add zero): 36.5 becomes 365=73*5. So we only need to know how to multiply and divide by 2 to divide and multiply by 5. We just move the decimal point. Divisibility by 9 or remainder after dividing by 9. All multiples of 9 contain digits which added together give 9. As we add the digits together, each time the result goes over 9 we add the digits of the result together and use that result and continue in this way up to the last digit. Is 12345 divisible by 9? Add the digits together 1+2+3=6. When we add 4 we get 10, so we add 1 and zero=1 then we add 5 to get 6. The number is not exactly divisible by 9, but the remainder is 6. We can also ignore any 9's in the number. Now try 67959. We can ignore the two 9's. 6+7=13, and 1+3=4; 4+5=9, so 67959 is divisible by 9. Multiplying by 11. Example: 132435*11. We write down the first and last digits 1 ... 5. Now we add the digits in pairs from the left a digit step at a time. So 1+3=4; 3+2=5: 2+4=6; 4+3=7; 3+5=8. Write these new digits between 1 and 5 and we get 1456785=132435*11. But we had no carryovers here. Now try 864753*11. Write down the first and last digits: 8 ... 3. 8+6=14, so we cross out the 8 and replace it with 8+1=9, giving us 94 ... 3. Next pair: 6+4=10. Again we go over 10 so we cross out 4 and make it 5. Now we have 950 ... 3. 4+7=11, so we have 9511 ... 3. 7+5=12, giving us 95122 ... 3; 5+3=8, giving us the final result 9512283.  Divisibility by 11. We add alternate digits and then we add the digits we missed. Subtract one sum from the other and if the result is zero the original number was divisible by 11. Example: 1456785. 1 5 7 5 make up one set of alternate digits and the other set is 4 6 8. 1+5+7=13. We drop the ten and keep 3 in mind to add to 5 to give us 8. Now 4 6 8: 4+6=10, drop the ten and add 0 to 8 to give us 8 (or ignore the zero). 8-8=0 so 11 divides into 1456785. Now 9512283: set 1 is 9 1 2 3 and set 2 is 5 2 8; 9+1=0 (when we drop the ten); 2+3=5; set 1 result is 5; 5+2+8=5 after dropping the ten, and 5-5=0 so 9512283 is divisible by 11. Nines remainder for checking arithmetic. We can check the result of addition, subtraction, multiplication and (carefully) division. Using Method 2 above we can reduce operands to a single digit. Take the following piece of arithmetic: 17*56-19*45+27*84. We'll assume we have carried out this sum and arrived at an answer 2365. We reduce each number to a single digit using Method 2: 8*2-1*9+9*3. 9's have no effect so we can replace 9's by 0's: 8*2 is all that remains. 8*2=16 and 1+6=7. This tells us that the result must reduce to 7 when we apply Method 2: 2+3+6=11; 1+1=2 and 2+5=7. So, although we can't be sure we have the right answer we certainly don't have the wrong answer because we arrived at the number 7 for the operands and the result. For division we simply use the fact that a/b=c+r where c is the quotient and r is the remainder. We can write this as a=b*c+r and then apply Method 2, as long as we have an actual remainder and not a decimal or fraction. Divisibility by 3. This is similar to Method 2. We reduce a number to a single digit. If this digit is 3, 6 or 9 (in other words, divisible by 3) then the whole number is divisible by 3. Divisibility by 6. This is similar to Method 6 but we also need the last digit of the original number to be even (0, 2, 4, 6 or 8). Divisibility by 4. If 4 divides into the last two digits of a number then the whole number is divisible by 4. Using 4 or 2 times 2 instead of 25 for multiplication and division. 469/25=469*4/100=1876/100=18.76. 538*25=538*100/4=134.5*100=13450. We could also double twice: 469*2=938, 938*2=1876, then divide by 100 (shift the decimal point two places to the left). And we can divide by 2 twice: 538/2=269, 269/2=134.5 then multiply by 100 (shift the decimal point two places left or add zeroes). Divisibility by 8. If 8 divides into the last three digits of a number then the whole number is divisible by 8. Using 8 or 2 times 2 times 2 instead of 125 for multiplication and division. Similar to Method 9, using 125=1000/8. Using addition instead of subtraction. 457-178. Complement 178: 821 and add: 457+821=1278, now reduce the thousands digit by 1 and add it to the units: 278+1=279; 457-178=279. Example: 1792-897. First match the length of 897 to 1792 be prefixing a zero: 0897; complement this: 9102. 1792+9102=1894. Reduce the thousands digit by 1 and add to the result: 894+1=895. Example: 14703-2849. 2849 becomes 02849, then complements to 97150. 14703+97150=111853; reduce the ten-thousands digit by 1 and and add to the result: 11854. Squaring numbers ending in 5. Example: 75^2. Start by writing the last two digits, which are always 25. Take the 7 and multiply by 1 more than 7, which is 8, so we get 56. Place this before the 25: 5625 is the square of 75. The square of 25 is ...25, preceded by 2*3=6, so we get 625. All numbers ending in 0 or 5 are exactly divisible by 5 (see also Method 1). All numbers ending in zero are exactly divisible by 10. All numbers ending in 00, 25, 50 or 75 are divisible by 25. Divisibility by 7. Example: is 2401 divisible by 7? Starting from the left with a pair of digits we multiply the first digit by 3 and add the second to it: 24: 3*2+4=10; now we repeat the process because we have 2 digits: 3*1+0=3. We take this single digit and the one following 24, which is a zero: 3*3+0=9. When we get a single digit 7, 8 or 9 we simply subtract 7 from it: in this case we had 9 so 9-7=2 and the single digit is now 2. Finally in this example we bring in the last digit: 3*2+1=7, but 7 is reduced to 0. This tells us the remainder after dividing 2401 by 7 is zero, so 2401 is divisible by 7. Another example: 1378. 3*1+3=6; 3*6=18 before adding the next digit, 7 (we can reduce this to a single digit first): 3*1+8=3*1+1=4; now add the 7: 4+7=4+0=4;  3*4=12; 3*1+2+8=5+1=6, so 6 is the remainder after dividing 1378 by 7.  See also my solution to:
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8/9 - 1/5=

8/9 - 1/5= how do i get the answer please explain dont just give answer please thanks When adding or subtracting fractions, you must always use a common denominator. The easiest way to determine that denominator is simply to multiply them: 9 * 5 = 45. Multiply each fraction by a fraction that is equivalent to 1, but will result in the same denominator. For this problem, multiply the first fraction by 5/5, and multiply the second fraction by 9/9. Both 5/5 and 9/9 are equal to 1, so you are NOT changing the VALUE of the fractions, merely their representation. (8/9) * (5/5) - (1/5) * (9/9) When multiplying fractions, multiply the numerators to get the new numerator, and multiply the denominators to get the new denominator. (8/9) * (5/5) - (1/5) * (9/9) (8 * 5) / (9 * 5)  -  (1 * 9) / (5 * 9) 40/45 - 9/45 Now, simply subtract. 40/45 - 9/45 31/45
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how can I divide a sextillion by eight million?

I may need stronger glasses; however, isn't the number you put up there 1 septillion (as in million, billion, trillion, quadrillion, quintillion, sextillion, septillion, octillion, nonillion, decillion, undecillion, duodecillion, tredecillion and my all time favorite--Quattrodecillion and etc, etc)?  In any event, I would just do it old school and start striking out 3 zero's at a time.  Thus, after getting rid of 2 sets of 3 zeros from 8 million leaves you with 8.  Getting rid of the equivalent 2 sets of 3 zeros from the other number leave you with 6 sets of 3 zeros with the '1' in front of them which should be 1 quintillion (or 1 quadrillion if you really wanted to start with 1 sextillion).  Still too many freakin zeros so then I guess I would start taking away 3 sets of zeros with the first set giving you 0.008 and either 1 quadrillion or 1 trillion--still too many zeros so I'm going to live dangerously and take away 2 more sets (i.e., 6 total zeros) just to get this over with.  Now your down to 0.000000008 and either just 1 billion or 1 million.  Now your down to something you can at least input into your calculator, but if it doesn't have scientific notation you still won't have your answer and buy a better calculator.  I was able to get an answer on my IPhone if I held it hortizonally and because it will answer using scientific notation.  So, if it is the number you typed in (1 septillion), the answer I got was 1.25 x 1017 (that 17 is supposed to be a superscript) or 125,000,000,000,000,000 a/k/a 125 quadrillion.
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