All categories

3 years ago

Help^^^^^^^^^^^^^^************

Mathematics

2 answers:

Paha777 [63]3 years ago

7 0

Answer and Step-by-step explanation:

The answer you chose is correct:

D. 2(5w + 5)

This is equivalent to 10w + 10.

This is because the 2 is a factor of 10, so when distributed out of the 10's, we get 5.

#teamtrees #PAW (Plant And Water)

shutvik [7]3 years ago

6 0

I’m pretty sure that the one you marked is correct

You might be interested in

Write each of these in scientific notation 1) 0.00000402 2) 1,900,000

Mkey [24]

Answer:

See below in bold.

Step-by-step explanation:

1. 0.00000402

= 4.02 * 10^-6 (Counting the digits after the decimal point until we get to the 4 gives us the -6).

2. 1,900,000

= 1.9 * 10^6 ( counting the number of digits after the 1 gives us 6).

3 0

3 years ago

Solve -vp+40<95 for v

Leya [2.2K]

-vp+40<95

subtract 40 from each side

-vp < 55

divide by -p (which flips the inequality) assuming p>0

v > -55/p

5 0

3 years ago

HELP PLEASE ASAP IM SO CONFUSED!!!!!!!!

liubo4ka [24]

The slope-intercept form:

$y=mx+b$

m - slope

b - y-intercept → (0, b)

We have the points (-4, -6) and (2, 6). Substitute:

$m=\dfrac{6-(-6)}{2-(-4)}=\dfrac{12}{6}=2\\\\y=2x+b$

Put the coordinates of the point (2, 6) to the equation of a line:

$6=2(2)+b\\\\6=4+b\qquad|-4\\\\2=b\to b=2$

<h3>Answer: y = 2x + 2</h3>

8 0

3 years ago

Which equation has the solutions x=1+or-square root of 5?

stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

case a) $x^{2}+2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=-4+1$

$x^{2}+2x+1=-3$

Rewrite as perfect squares

$(x+1)^{2}=-3$

$(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i$

therefore

case a) is not the solution of the problem

case b) $x^{2}-2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=-4+1$

$x^{2}-2x+1=-3$

Rewrite as perfect squares

$(x-1)^{2}=-3$

$(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i$

therefore

case b) is not the solution of the problem

case c) $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=4+1$

$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

case d) $x^{2}-2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=4+1$

$x^{2}-2x+1=5$

Rewrite as perfect squares

$(x-1)^{2}=5$

$(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}$

therefore

case d) is the solution of the problem

therefore

the answer is

$x^{2}-2x-4=0$

5 0

3 years ago

Read 2 more answers

MATH HELP PLEASE I WILL MARK BRAINLIEST !!!! WHY IS THIS NOT FACTORABLE ?

Leya [2.2K]

I think that -25 should be +25.

so it would be (7x-5)^2

8 0

3 years ago