MATHEMATICS HIGH SCHOOL

Find the sum using fraction bars or a paper and pencil. Write your answer in simplest form.1/2 + 7/8

Answers

Answer 1

Answer: First find the LCD. That is 8.

1/2 = 4/8

7/8 + 4/8 = 11/8 or 1 3/8

Answer 2

Answer:

Answer:1 3/8

Step-by-step explanation:

First find the LCD. That is 8.

1/2 = 4/8

7/8 + 4/8 = 11/8 or 1 3/8

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How do you solve 2x^2-32=0 using square roots?

Please help!

Answers

$2x^2-32=0 \ \ / :2 \n \n x^2-16=0 \n \nx^2-4^2 = 0 \n \n (x-4)(x+4)=0 \n \n x-4 = 0 \ \ or \ \ x+4 = 0\n \n x=4\ \ or \ \ x=-4$

$2x^2-32=0\n\n2(x^2-16)=0\ /:x^2-16=0\n\nx^2-4^2=0\n\n(x-4)(x+4)=0\n\nx-4=0\ \ \ \ or\ \ \ \ \ x+4=0\n\nx=4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=-4$

Which is the best estimate for the quotient of 312.21 59.113?a. 5,000
b. 50
c. 5
d. 0.5

Answers

Answer=C

The first step in estimating is to round the numbers into simpler numbers so that we can easily do the math in our head...

312.21 becomes 300

59.113 becomes 60

300÷60=5

Answer=C

Answer=C

312.21 would be 300

59.113 would be 60

300÷60=5
Im pretty sure that is right IDK im in Middle School

Which two numbers add up to 6 and muliply to -126

Answers

xy=-126
x+y=6
minus x

y=6-x
sub
x(6-x)=-126
6x-x^2=-126
minus (6x-x^2) both sides
0=x^2-6x-126

in form
ax^2+bx+c=0
x= $\frac{ -b+/-\sqrt{b^(2)-4ac} }{2a}$

a=1
b=-6
c=-126

x= $\frac{ -(-6)+/-\sqrt{(-6)^(2)-4(1)(-126)} }{2(1)}$
x= $( 6+/-√(36+504) )/(2)$
x= $( 6+/-√(540) )/(2)$
x= $( 6+/-6√(150) )/(2)$
x= $3+/-3√(150)$

the numbers are
$3+3√(150)$ and $3-3√(150)$

Square EFGH stretches vertically by a factor of 2.5 to create square E′F′G′H′. The square stretches with respect to the x-axis. If point H is located at (-2, 0), what are the coordinates of H?A.(-5, 0)
B.(-2, 0)
C.(0.5, 0)
D.(-2, 2.5)

Answers

Square EFGH stretches vertically by a factor of 2.5 to create square E′F′G′H′. The square stretches with respect to the x-axis. If point H is located at (-2, 0), the coordinates of H will be at point "D.(-2, 2.5)." Since it stretches vertically, it assumed that it stretches all parallel with the y axis.

Answer with explanation:

Pre -image= Square E F G H

Image=Square E′F′G′H′

Vertical Stretch Factor= 2.5

Coordinates of Point H= (-2,0)

If, (x,y), is the coordinate of any point, and , m is vertical stretch factor,then coordinate of Image =(x, my)

→Here, Vertical stretch, factor is 2.5,and coordinates of H=(-2,0) then Coordinates of H' will be =(-2, 0×2.5)=(-2,0)

Option B: (-2,0)

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