Guide :

# how would i find the answer to 94. -70.735 .. step by step

the answer on a calculator  is 23.265 i just dont know how to do it on paper

## Research, Knowledge and Information :

### How to Divide: 6 Steps (with Pictures) - Instructables

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_ HOW TO FIGURE PERCENT _ ... Find 4% of \$70.00 ... Answer: \$707.20 STEP 1: Find 4% of \$680 STEP 2: Add the raise to \$680 \$680.

### 56. AB Calculus – Step-by-Step Name

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## Suggested Questions And Answer :

### zeros of 2x^6+5x^4-x²

All the powers of 2 are even so this 6-degree expression can be reduced to a cubic if we put y=x^2: 2y^3+5y^2-y. Putting this equal to zero to find the y zeroes: 2y^3+5y^2-y=0. This factorises: y(2y^2+5y-1)=0. Take the quadratic: 2y^2+5y-1=0 and rewrite it as: y^2+5y/2=1/2 by dividing through by 2 and moving the constant across. Now we can complete the square by dividing the y term by 2 and squaring its coefficient: y^2+5y/2+25/16=1/2+25/16, and you can see we've added 25/16 to both sides of the equation to make it balance. We now have a perfect square on the left: (y+5/4)^2=1/2+25/16=(8+25)/16=33/16, so we can take the square root of each side:  (y+5/4)=+sqrt(33)/4. Therefore, y=-5/4+sqrt(33)/4. This is the same result as we would have found using the formula for solving a quadratic equation. We need to calculate what these solutions are: 0.18614 and -2.68614. But we want x, not y, so we need to find the square root of these values of y. Assuming we don't want complex solutions (involving the imaginary square root of -1), we can only use the positive solution 0.18614 and take the square root of that, which is +0.43144 approx. so the real solution is x=0.43144 or -0.43144. (The complex solution is +1.63894i, where i=sqrt(-1).) But we're not finished yet, because y=0 was also a solution, and that means x=0 is a solution, so we have three possible real zeroes for x: 0, 0.43144 and -0.43144 (and two complex zeroes).

### 10+2(3+2x)=0

10 + 2 (3 + 2x) = 0 PEMDAS dictates you solve what is in parentheses first, but you cannot add what is in parentheses since they are not the same "language;" one is an integer and the other is a variable being multiplied by 2 (a coefficient).  No exponents here, so we start with multiplying/dividing:  Here our multiplication is in the form of distributive property: 2(3 + 2x) becomes 2*3 + 2*2x which is 6+4x.  Your equation is now 10 + 6 + 4x = 0. There is nothing more to multiply or divide since we don't know what x is yet., so we can add/subtract same "language" numbers:  10 + 6 is 16. Your equation is now 16 + 4x = 0, a simple two-step linear equation.  We want to find what x equals, so we need to get rid of anything on the side of the equation with x. The first step here is to get rid of the integer by doing the inverse of its operation.  Our integer is 16 and it's positive so it's being added.  The inverse (opposite) would be to subtract:  16 (- 16) + 4x = 0 (-16).  You must subtract on both sides of the equation or it will no longer be equal.  The 16's cancel out to zero on one side of the equation while 0 - 16 becomes -16 on the other side and leaves us with 4x = -16.  We need to change our 4x to 1x by using the inverse operation.  4x is 4 times x and the inverse of multiplication is division, so we will divide by 4 on both sides of the equal mark:  4x/4 = -16/4.  4x/4 is now 1x or just x (a coefficient of 1 is understood and not written next to the variable) and -16/4 is -4 (a negative divided by a positive is a negative), so x = -4.

### k3-k2-14k-16=0

First find the factor pairs of 16: (1,16), (2,8), (4,4) are the only ones. Also note that the coefficient of k is 1, meaning in the simplification that all coefficients of k will be 1. Also it's a cubic equation and if it is to factorise, we can expect a term like k+c where c is a number. We can also expect something like k^2+ak+b as another factor where a and b are constants. (k+c)(k^2+ak+b)=k^3+ak^2+bk+ck^2+ack+bc=k^3-k^2-14k-16=0 Now we look at the coefficients and constant term. For the constant bc=-16 which takes us back to the factor pairs. It's got to be just one of them. But we do need to take note of the minus sign. We don't need to think about the k^3 term because it will cancel out. k^2 term: a+c=-1, so c=-1-a or -(1+a) k term: b+ac=-14 What possibilities do we have for b and c? One must be positive and one negative so we have: (-1,16), (1,-16), (-2,8), (2,-8), (-4,4), (4,-4). We've already been given the answer so let's see what happens when we use (-8,2). a+2=-1 so a=-3, and we already have c=2 and b=-8 so we have all the numbers. b+ac=-8-6=-14 which is true. We have consistency. (k+c)(k^2+ak+b)=(k+2)(k^2-3k-8). Now, I shortened this because we should have tried all the factor pairs as if we didn't already know the answer. But we would have found only one that fitted, because the other answers would have been inconsistent. Nevertheless, the method shows how we compare coefficients.

### How to Find Square Root

98=49*2, so sqrt(98)=sqrt(49)*sqrt(2)=7sqrt(2)=7*1.4142=9.8994 approx. There's another way using the binomial theorem. 98=100-2=100(1-0.02). sqrt(100)=10 so sqrt(98)=10(1-0.02)^(1/2) because square root is the same as power 1/2. (1+x)^n expands to 1+nx+(n(n-1)/1*2)x^2+(n(n-1)(n-2)/1*2*3)x^3+... Putting n=1/2 and  x=-0.02, we get sqrt(98)=10(1-0.02)^(1/2)=10[1-(1/2)0.02+((1/2)(-1/2)/2)0.0004+...]. This gives us: 10(1-0.01-0.00005+...)=10*0.98995=9.8995. A third method is to use an iterative process, which means you keep repeating the same action over and over again. Look at this: x=10-(2/(10+x)). If we solve for x we get x=sqrt(98); but we're going to find x in an iterative way. Start with x=0 and work out the right hand side: 10-2/10=9.8. This gives us a new value for x, 9.8, which we feed back into the right hand side: 10-(2/(10+9.8))=10-2/19.8=9.8989..., giving us another value for x, 9.8989... which we feed back into the right hand side: 10-(2/(10+9.8989...))=9.89949..., giving us yet another value for x and so on. Very quickly we build up accuracy with each x. You can do this on a calculator, a basic one that doesn't even have square roots, using the memory to hold values for you. Here's a very simple program, where STO means store in memory (if your calculator doesn't have STO use MC (memory clear) followed by M+ (add to memory)); MR means read memory (the steps show what calculator keys to press in order; / may be ÷ on your calculator): 0= +10=STO 10-2/MR= GO TO STEP 2 OR STOP (display should show the answer for sqrt(98)) Note: In STEP 3 the division must be carried out before subtracting from 10, otherwise you get the wrong answer. If your calculator doesn't do this you need to replace STEP 3 with: 0-2=/MR=+10= You should only have to go round the loop a few times before you get a really accurate result. To find the square root of 2 directly the iteration equation is x=1+1/(1+x) and the program is: 0= +1=STO 1/MR+1= GO TO STEP 2 OR STOP STEP 3 should work on all calculators.

### how do you find the order and magnitude of symmetry for a star

A star is basically a ball of very hot gas in constant motion. From a distance it is a point of light, with the exception of the sun, our nearest star. The sun has a generally uniform appearance so its rotational symmetry could be considered of infinite order. But if its granular appearance is considered then there would be no symmetry since there is so much randomness. Its magnitude (brightness) is -27, while the full moon's magnitude is -13. The brighter an object the more negative is its magnitude. So faint stars have positive magnitude. A step of 5 in the magnitude between two stars is a change in brightness by a factor of 100, so a step of 1 is 100^(1/5). A piano keyboard can be used as an analogous system. The lower notes correspond to brighter stars while the higher notes correspond to dimmer stars. A piano keyboard is logarithmic as is the magnitude system. To use the magnitude system we need to define a reference magnitude. Middle C on a piano could represent magnitude 0. The note below would be a star with magnitude -1, or roughly 2.5 times brighter than a star with magnitude 0. The note above would be about 2.5 times dimmer. A star with magnitude -5 is a hundred times brighter than a magnitude 0 star, magnitude 5 100 times dimmer. The magnitude is not restricted to integers. The formula apparent magnitude, M=M0-2.5log[10](I[m]/I[m0]) applies, where I denotes measurable intensity, M0 is the reference magnitude (star with magnitude 0). Absolute magnitude requires knowledge of a star's distance. To determine the apparent magnitude of a star you would need to measure the brightness of a reference object (e.g., the moon) using, for example, power per unit area, then use the same units to measure the star's brightness. Knowing the magnitude of the moon's brightness, M0=-13, plug the values into the equation to find M.

### How to find c and y so that each quadrilateral is a parallelogram

The opposite internal angles of a parallelogram are equal, and adjacent angles are supplementary, but which of the given angles are opposite and which are adjacent? We know that all angles must be positive, so 7x-11>0 and 5y-9>0 so x>11/7 or 1.57 and y>1.8. We also know that all the angles of a parallelogram can be determined if just one is known, because of the relationships. This means that if we take any pair of angles we know that they are (a) equal or (b) supplementary. Take the first pair: 5x+29 and 5y-9: if (a), 5x+29=5y-9, so 5(y-x)=38 and y=(38+5x)/5; or if (b), 5(y+x)=160, or y+x=32 and y=32-x. Also, the remaining pair 3y+15 and 7x-11: if (a), 7x-3y=26, y=(7x-26)/3; or if (b), 7x+3y=176 and y=(176-7x)/3. We can see that if (a) is applied to the first pair at least one of x or y will contain a fraction. If (a) is applied to the second pair, which can be written 2x-8+(x-2)/3, x must be x=5, 8, 11, ..., 3n+2 for y to be an integer (where n is a positive integer) and y is 3, 10, 17, ..., and if (b), which can be written 58-2x-(x-2)/3, x must be 3n+2 for y to be an integer (where n is an integer 0 Read More: ...

### How to find equation of the circle passing through (9,1), (-3,-1) and (4,5)

The equation of a circle is more specific: (x-h)^2+(x-k)^2=a^2, where (h,k) is the centre and a the radius. Plug in the points: (9-h)^2+(1-k)^2=(-3-h)^2+(-1-k)^2=(4-h)^2+(5-k)^2=a^2 This can be written (9-h)^2+(1-k)^2=(3+h)^2+(1+k)^2=(4-h)^2+(5-k)^2=a^2 Leaving a out of it for the moment we can use pairs of equations, using difference of squares: 12(6-2h)+2(-2k)=0, 36-12h-2k=0, 18-6h-k=0, so k=18-6h. 7(-1-2h)+6(-4-2k)=0, -7-14h-24-12k=0, -31-14h-12k=0 or 31+14h+12k=0, 31+14h+12(18-6h)=0, 31+14h+216-72h=0. 247-58h=0 so h=247/58 This looks suspiciously ungainly. So rather than continuing, I'm going to call the three points: (Q,R), (S,T), (U,V) to provide a general answer to all questions of this sort. (Q-h)^2+(R-k)^2=(S-h)^2+(T-k)^2=(U-h)^2+(V-k)^2=a^2 [(equation 1)=(equation 2)=(equation 3)=a^2] Take the equations in pairs and temporarily ignore a^2. Equations 1 and 2: (Q-S)(Q+S-2h)+(R-T)(R+T-2k)=0 Q^2-S^2-2h(Q-S)+R^2-T^2-2k(R-T)=0 2k(R-T)=Q^2+R^2-(S^2+T^2)-2h(Q-S), so k=(Q^2+R^2-(S^2+T^2)-2h(Q-S))/(2(R-T)) Equations 2 and 3: S^2-U^2-2h(S-U)+T^2-V^2-2k(T-V)=0, so k=(S^2+T^2-(U^2+V^2)-2h(S-U))/(2(T-V)) At this point, we can substitute for k and end up with an equation involving the unknown h only: (Q^2+R^2-(S^2+T^2)-2h(Q-S))/(2(R-T))=(S^2+T^2-(U^2+V^2)-2h(S-U))/(2(T-V)) (T-V)(Q^2+R^2-(S^2+T^2)-2h(Q-S))=(R-T)(S^2+T^2-(U^2+V^2)-2h(S-U)) (T-V)(Q^2+R^2-(S^2+T^2))-2h(T-V)(Q-S)=(R-T)(S^2+T^2-(U^2+V^2))-2h(R-T)(S-U) 2h((R-T)(S-U)-(T-V)(Q-S))=(R-T)(S^2+T^2-(U^2+V^2))-(T-V)(Q^2+R^2-(S^2+T^2)) h=((R-T)(S^2+T^2-(U^2+V^2))-(T-V)(Q^2+R^2-(S^2+T^2)))/((R-T)(S-U)-(T-V)(Q-S)). Once h is found we can calculate k, and then we can substitute the values for h and k in any equation to find a^2 which is equal to each of the three equations. When we use Q=9, R=1, S=-3, T=-1, U=4, V=5 or -5, we appear to get very ungainly solutions. One way to find which values need to be changed may be to plot the values and work out where the circle should fit and where its centre has less complex values. If V=6 or -6, there is a simple solution: (x-3)^2+y^2=37 (centre at (3,0), a^2=37=6^2+1^2. (-3-3)^2=(9-3)^2=6^2; (-6)^2=6^2; (2-3)^2=(4-3)^2=1; (-1)^2=1^2 shows the combination of points that would lie on the circle: (9,1), (9,-1), (-3,1), (-3,-1), (4,6), (4,-6), (2,6), (2,-6) and this includes points A and B, but not C. Note that the equation k=18-6h is valid for h=3, k=0.