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without solving is the solution of -2x=-15 greater than or less than -15? explain

without solving is the solution of -2x=-15 greater than or less than -15? explain.

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Is the solution of -2x = -15 greater than or less than -15 ...

Is the solution of -2x = -15 greater than or less than -15 ... Is the solution of -2x = -15 greater than or less than -15? Explain ... Dante is showing his steps to ...

Equations and Inequalities - Inequalities - In Depth

Solving an inequality is very ... This gives "x is less than –6," not "x is greater than ... If the solution had been "x is less than or equal to –6 ...

Without solving determine whether the solution of 1/3x = 21 ...

Without solving determine whether the solution of 1/3x = 21 is greater than or less than 21. Explain ... Without solving, the solution is greater than 21.

Solving Inequalities - Maths Resources

Solving inequalities is very like solving equations ... 3y/3 < 15/3. y < 5. And that is our solution: ... Less Than or Greater Than Inequalities Solving Inequality ...

Inequalities - A complete course in algebra

How to solve inequalities. ... not be a number that is less than 1 and greater than 5. There is no such ... also be greater than −5. Therefore, 1) has the solution ...

Solve Linear Inequalities - Tutorial

Solve Linear Inequalities - Tutorial. Solve linear inequalities: ... ( less than or equal) ... ( greater than or equal ), ...

5%2D1 Solving Inequalities by Addition and Subtraction

... a number less than 6, and a number greater ... and solve each problem. Check your solution. ... Shelter 3 h 15 min

5%2D4 Solving Compound Inequalities

Then check your solution. Eight less than a number is no more than ... Let n = 15. So, the solution ... substitute a number greater than 0 and less than or equal to ...

Solving Linear Inequalities: Introduction and Formatting

Demonstrates how to solve linear inequalities ... Solving linear inequalities is ... Since –3 is not included in the solution (this is a "less than", ...

3 Ways to Graph Inequalities - wikiHow

How to Graph Inequalities. ... since an inequality shows a set of values greater than or less than, ... since the solution includes values greater than 1.

How to solve this solution set: 2x>-26

yes you solve it just like an =. the only difference is that if you multiply or divide by a negative then you flip the sign in this case you divide both sides by 2, so you don't flip the sign 2x  > -26 x > -13

without solving is the solution of -2x=-15 greater than or less than -15? explain

dum quesshun!!!! -2x=-15-->2x=15 x=7.5

Prove, by contradiction, that if w,z∈C.... (elementary algebra)

Given the validity of the comment attached to the question, and it's resolution, we can take a look at the implication of the question. To be true for w and z generally, we can take a specific case to demonstrate the argument. Let w=0.8+0.6i, which satisfies the requirement |w|=1, a special case of |w|<1. For any value of n, the binomial expansion of (a+b)^n=1 only when a+b=1, i.e., b=1-a. The binomial expansion is (a+b)^n=a^n+nba^(n-1)+(n(n-1)/2)b^2a^(n-2)+...+b^n=1. The general term is (n!/(r!(n-r)!))a^(n-r)b^r, where 01/2. Let w=0.03+0.04i, |w|=0.05, z=1-(0.03+0.04i)=0.97-0.04i, |z|=0.9708 approx>1/2. As |w| gets smaller, |z| gets larger, so moving further away from 1/2 and towards 1. What is w+z? In all of the above cases w+z=1, by definition, and so the binomial expansion also =1. Just seen your comment. Let w=0.96+0.28i, |w|=1, z=0.04-0.28i, |z|=0.2828 approx, which is <1/2. So the proposition would not be true for the binomial expansion of (w+z)^n=1 for all n. That means we require a different expression than w+z. However, we do know that (...)^n must be 1 for all n and we need to find an alternative expression similar to the one given. Let w=a+isqrt(p^2-a^2), |w|=sqrt(a^2+p^2-a^2)=p<1. This is the general form for w. In the above, z=1-a-isqrt(p^2-a^2) and |z|=sqrt((1-a)^2+p^2-a^2)=sqrt(1-2a+a^2+p^2-a^2)=sqrt(1-2a+p^2). When p=1, this is sqrt(2(1-a)) which can be less than 1/2 (7/81/2) and wz/(w+z)=1 so ((w+z)/wz)^n=((1/z)+(1/w))^n=(wz/(w+z))^n=1. ((1/z)+(1/w))^n=(w+z)^n=1. Therefore, w and z are particular complex numbers for which the proposition is true, and the binomial expansion applies.

Shawn took in \$69.15 in two hours work on Saturday selling Belts for \$8.05 and earrings for \$4.50. How many of each did Shawn sell?

Shawn took in \$69.15 in two hours work on Saturday selling Belts for \$8.05 and earrings for \$4.50. How many of each did Shawn sell? I'm not too sure how ​you are expected to solve this problem. You will end up with a Diophantine Equation (one involving integer coefficients and integer solutions only) If you have never heard of Diophantus, then the precise mathematical solution later on should be ignored. Instead ...   The (Diophatine) equation is 161B + 90E = 1363 You should be able to show that Shawn can only sell a maximum of 8 belts (i.e. if he sells no earrings), and a maximum of 15 earrings (i.e. if he sells no belts). So start with the smaller number (8). Shawn sells up to 8 belts. So put B = 1 in the (Diophantine ) equation and see if B is an integer. If not, then B=1 cannot be a solution, So now try again with B = 2, and so on ..     My precise mathematical solution now follows. Let B be the number of belts sold @ \$8.05 per belt. Let E be the number of earrings sold @ \$4.50 per pair. Total money made is P = \$69.15 Total money made is: B*\$8.05 + E*\$4.50 Equating the two expressions, 8.05B + 4.50E = 69.15 Making the coefficients as integers we end up with a Diophantine equation. 161B + 90E = 1383 We now manipulate the coefficients 161 and 90. We should end up with a relation between the two that should produce the value 1. This manipulation is in two parts. In the first line, we are writing the larger coefft, 161, as a multiple of the smaller coefft plus a remainder. 1st Part 161 = 1x90 + 71  90 = 1x71 + 19     now we write the 1st multiple as a multiple of the 1st remainder + remainder 71 = 3x19 + 14    we write the 2nd multiple as a multiple of the 2nd remainder + remainder 19 = 1x14 + 5 14 = 2x5 + 4 5 = 1x4 + 1 ============ 2nd Part We rewrite that last equation so that 1 is on the left hand side 1 = 5 – 1x4 We now use the remainder from the next equation up. 1 = 5 – 1x(14 – 2x5) 1 = 3x5 – 1x14 And again we use the remainder in the next equation up 1 = 3(19 – 1x14) – 1x14 1 = 3x19 – 4x14 1 = 3x19 – 4(71 – 3x19) 1 = 15x19 – 4x71 1 = 15(90 – 1x71) – 4x71 1 = 15x90 – 19x71 1 = 15x90 – 19(161 – 1x90) 1 = 34x90 – 19x161 ================ Therefore, 1383 = (-19x1383).161 + (34x1383).90 1383 = -26,277*161 + 57,022*90 Which implies B = -26,277 E = 47,022 which is obviously incorrect! But all is not lost. These two values for B and E are simply two values that happen to satisfy the original Diophantine equation. What we need is the general solution, which now follows. In general, B = -26,277 + 90k    (using the coefft of E) E = 47,022 – 161k    (using the coefft of B) If you substitute these expressions into the original Diophantine equation, the k-terms will cancel out, leaving you with the original eqn. We know that B and E must be smallish numbers and that neither can be negative. For example, if no earrings were sold, Shawn would need to sell over 8 belts to clear \$69, and if no belts were sold, then Shawn would need to sell over 15 earrings. So Shawn needs to sell between 0 and 8 belts and between 0 and 15 earrings. Setting k = 292, B = -26,277 + 90*292 = 3 E = 47,022 – 161*292 = 10 B = 3, E = 10 If k is greater than or less than 292 by 1, or more, then either B will be negative or E will be negative. So this is the answer. Check 3*8.05 + 10*4.50 = 25.15 + 45.00 = 69.15 – Correct! Answer: Shawn sells 3 belts and 10 earrings

What is the probability that the difference of the two (2) numbers drawn is 10 or less?

After one number, a, is drawn, 49 remain. Let's suppose a=1, then the numbers 2-11 (10 tickets) would satisfy the criterion that b, the second number, satisfies |b-a|≤10. So that's 10/49. If a=2, then if b=1 or 3≤b≤12 |b-a|≤10. That's 11 tickets, 11/49. If a=25, 15≤b≤24 or 26≤b≤35, that's 20 tickets, 20/49. If a=30, then 20≤b≤29 or 31≤b≤40, 20 tickets again, 20/49. The rule is then that for 11≤a≤40 the probability for b is 20/49. The probability of picking a in range for this probability for b is 30/50. The combined probability is 30/50*20/49=12/49.  At the other end of the scale for 50≥a≥41 we have the same probabilities as for 1≤a≤10; when a=50, b can be 40-49 (10 tickets); when a=49, b can be 50 or 39-48 (11 tickets), and so on. We can put together a 50X50 table, but we can also summarise it: ("Y" means "yes, the values of 'a' (row number) and 'b' (column number) represented by this little square satisfy |a-b|≤10".) The small square with the red diagonal represents the whole of the large table on the left, which is only shown up to its 40th row. The red diagonal, containing X's for no-go, or disallowed combinations, excludes a=b. One way of calculating probability is to work out the fraction (number of "successful " events) ÷ (total possible number of events). This is the same as dividing the "area" occupied by Y's in the table by the "area" of the table itself, excluding the red diagonal. This is better represented by the small square with the red diagonal. This approach of using areas is the fastest way to arrive at the probability and it's easy to calculate: Area of table=50^2-50=2450=50*49 (there are 50 red squares in the red diagonal to be removed). Area of a trapezoidal corner piece=19+18+...+10=145. Total area of both trapezoids=290. (The right-angled trapezoids are formed by the Y's in rows a=1 to 10 and b=1 to 20, and a=41 to 50 and b=31 to 50. The trapezoid consists of a 10X10 square=100, less the diagonal of 10, which makes 90; and a right-angled triangle containing 55 Y's. Total 145. Not the usual way of calculating the area of a triangle!) Area of central "parallelogram"=30*20=600. Total area of Y's (central band)=890. So the probability of picking two numbered tickets, the difference of which is 10 or less, is 890/2450=89/245. In case you're not convinced that this simple solution is correct, let's go back to the calculations at the beginning. For any particular value of a there's always a 1/50 probability of picking it, but, depending on the value we have varying probabilities for b. Take a=1: the combined probability of |a-b|≤10 is 1/50*10/49; for a=2 it's 1/50*11/49, giving us the progression: 1/50*10/49+1/50*11/49+1/50*12/49+...+1/50*19/49. This also applies to a=50, a=49, ..., a=41. The total probability for the "ends" is (2/2450)(10+11+12+...+19)=2*145/2450. In the "middle section" we have constant 20/49 for b in the range 11≤a≤40=30/50*20/49=600=600/2450. The total is (290+600)/2450=890/2450=89/245. So the figures match the "areas".

find all solutions 2 cos^theta-cos theta+3=0

2cos^2(theta)-cos(theta)+3=0 appears to be the intended question. 2cos^2(theta)-cos(theta)+3=(2cos(theta)-3)(cos(theta)-1)=0. cos(theta) cannot be greater than 1 or less than -1, so cos(theta)=3/2 can be ruled out, leaving cos(theta)=1 as the only solution. However, values of theta are countless and are given by theta=2πn where n is an integer, or 360n in terms of degrees: 0, 360, 720, etc.

Solve the inequality -2/3 > z/5

We can cross-multiply without upsetting the inequality provided we use positive numbers: -10>3z. We can divide both sides by 3: -10/3>z. We can switch the inequality around: z<-10/3. This means that z can be any value strictly less than -3 1/3. To prove we have the right answer, let's put z=-5, which is less (more negative) than -3 1/3: -2/3>-1. This is true because -2/3 is not as negative as -1, so it's bigger. Put z=0 which is bigger than -3 1/3 and we get -2/3>0, which is false and is the answer we would expect because we picked an invalid value for z.

sinx=e^(-x^2) where x is between -pi and pi

x=-3.141644, 0.680598 and 3.141541 (-3.142, 0.681 and 3.142 to nearest thousandth). The negative solution is slightly less than -(pi) but the larger positive solution is less than (pi). So there are two solutions in range. When x becomes large (positive or negative), the exponential becomes small so that it approaches zero, and sin(x) approaches zero, so x approaches n(pi) where n is an integer, as can be seen by the solutions close to +(pi). For 0 Read More: ...

Solve compound inequality 4v+4 less than or equal to 12 or 3v-3 less than -12. Write the solution in interval notation if there is no solution write no solution.

Solve compound inequality 4v+4 less than or equal to 12 or 3v-3 less than -12. Write the solution in interval notation if there is no solution write no solution. Our inequalities are, 4v + 4 <= 12 || 3v - 3 < -12   (divide the 1st inequality by 4 and the 2nd one by -3, to give) v + 1 <= 3  ||  v - 1 < -4        (rearrange both inequalities) v <= 2  || v <-3 We have a compound inequality here, v <= 2 or v < -3, which is satisfied by the single inequality, v <= 2. Solution: v <= 2  (since any number less than or equal  2 will be either less than or equal to 2 or less than -3, or both)

What is the solution set of {x | x < -3} ∩ {x | x > 5}?

∩ means "intersect."  It means where the thing before it and the thing after it are true at the same time. No number is less than -3 and greater than 5 at the same time, so the answer is 3) (the empty set) The empty set is sort of the answer when there is no answer. For example, if we want to solve the system of equations x + y = 7, 2x + 2y = 8 we get: 2x + 2y = 8 2(x+y) = 8 x+y = 4 but we also have x+y=7, which leads to: 7 = 4 This can never happen, so there's no combination of values for x and y that make x + y = 7, 2x + 2y = 8 true at the same time.  The answer is the empty set.