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3 years ago

Which student evaluated the power correctly?

Mathematics

2 answers:

Levart [38]3 years ago

7 0

Answer:

Aida

Step-by-step explanation:

5 to the power of 4 is equivalent to 5 x 5 x 5 x 5. Hector, Camila, and Juan all thought of the exponents in wrong ways.

Reika [66]3 years ago

6 0

Answer:

The first one

Step-by-step explanation:

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Find the distance between lines 8x−15y+5=0 and 16x−30y−12=0.

AlladinOne [14]

Answer:

The distance between the two parallel lines is 11/17 units

Step-by-step explanation:

we have

8x-15y+5=0 -----> equation A

16x-30y-12=0 ----> equation B

Divide by 2 both sides equation B

8x-15y-6=0 ----> equation C

Compare equation A and equation C

Line A and Line C are parallel lines with different y-intercept

step 1

Find the slope of the parallel lines (The slope of two parallel lines is the same)

8x-15y+5=0

15y=8x+5

$y=\frac{8}{15}x+\frac{1}{3}$

the slope is

$m=\frac{8}{15}$

step 2

Find the slope of a line perpendicular to the given lines

Remember that

If two lines are perpendicular then their slopes are opposite reciprocal (the product of their slopes is -1)

$m1*m2=-1$

we have

$m1=\frac{8}{15}$

therefore

$m2=-\frac{15}{8}$

step 3

Find the equation of the line perpendicular to the given lines

assume any point that lie on line A

$y=\frac{8}{15}x+\frac{1}{3}$

For $x=0$

$y=\frac{1}{3}$

To find the equation of the line we have

$point\ (0,1/3)$ ---> is the y-intercept

$m=-\frac{15}{8}$

The equation in slope intercept form is

$y=-\frac{15}{8}x+\frac{1}{3}$ -----> equation D

step 4

Find the intersection point of the perpendicular line with the Line C

we have the system of equations

$y=-\frac{15}{8}x+\frac{1}{3}$ ----> equation D

$8x-15y-6=0$ ----> $y=\frac{8}{15}x-\frac{2}{5}$ ----> equation E

equate equation D and equation E and solve for x

$\frac{8}{15}x-\frac{2}{5}=-\frac{15}{8}x+\frac{1}{3}$

$\frac{8}{15}x+\frac{15}{8}x=\frac{1}{3}+\frac{2}{5}$

Multiply by 120 both sides to remove fractions

$64x+225x=40+48$

$289x=88$

$x=88/289$

Find the value of y

$y=-\frac{15}{8}(88/289)+\frac{1}{3}$

$y=-\frac{206}{867}$

the intersection point is $(\frac{88}{289},-\frac{206}{867})$

step 5

Find the distance between the points $(0,\frac{1}{3})$ and $(\frac{88}{289},-\frac{206}{867})$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

substitute the values

$d=\sqrt{(-\frac{206}{867}-\frac{1}{3})^{2}+(\frac{88}{289}-0)^{2}}$

$d=\sqrt{(-\frac{495}{867})^{2}+(\frac{88}{289})^{2}}$

$d=\sqrt{(\frac{245,025}{751,689})+(\frac{7,744}{83,521})}$

$d=\sqrt{\frac{314,721}{751,689}}$

$d=\frac{561}{867}\ units$

Simplify

$d=\frac{11}{17}\ units$

therefore

The distance between the two parallel lines is 11/17 units

see the attached figure to better understand the problem

4 0

3 years ago