All categories

3 years ago

Which of the following problems can be solved using the equation below? x+15=2x

Mathematics

1 answer:

pshichka [43]3 years ago

6 0

Answer:

Answer is below

Step-by-step explanation:

Subtract 2x from both sides.

$x+15-2x=2x-2x\\-x+15=0$

Subtract 15 from both sides.

$-x+15-15=0-15\\-x=-15$

Divide both sides by -1.

$\frac{-x}{-1}=\frac{-15}{-1} \\x=15$

So therefore your answer is 15

You might be interested in

Anybody want to help with math? Pls hurry, I’ll give brainiest and everything.

Vsevolod [243]

Answer:

Step-by-step explanation:

You are going to add 9 to 27 because you are counting by nines on the bottom.

3 0

2 years ago

Craig earns extra money as a lifeguard. He earns $37 for 4 hours and $64.75 for 7 hours. Explain how you know whether the relati

nordsb [41]

Answer:No it is not proportional

Step-by-step explanation:

4 hours=$37

7 hours=$64.75

I t is not proportional because if 4 hours =$37 then 6 hours is supposed to be around $70 but 7 hours=$64.75

3 0

3 years ago

Ma<br>Find the co-ordinate the image of a <br>point 6,4). Rotate it than<br>270<br>de acee

telo118 [61]

Answer:

(4, - 6 )

Step-by-step explanation:

Under a counterclockwise rotation about the origin of 270°

a point (x, y ) → (y, - x ) , thus

(6, 4 ) → (4, - 6 )

6 0

3 years ago

5 wholes 2/3 gallon in 2 hours

Firdavs [7]

What do you mean? I don't think it's enough information

4 0

3 years ago

100 POINTS AND BRAINLIEST

zysi [14]

Answer:

8 square units and $\frac{40}{3}$ square units

Step-by-step explanation:

The area of the triangle ABC is 24 square units.

1. Triangles ABC and FBG are similar with scale factor $\frac{1}{3},$ then

$\dfrac{A_{\triangle FBG}}{A_{\triangle ABC}}=\dfrac{1}{9}\Rightarrow A_{\triangle FBG}=\dfrac{1}{9}\cdot 24=\dfrac{8}{3}\ un^2.$

2. Triangles ABC and DBE are similar with scale factor $\frac{2}{3},$ then

$\dfrac{A_{\triangle DBE}}{A_{\triangle ABC}}=\dfrac{4}{9}\Rightarrow A_{\triangle DBE}=\dfrac{4}{9}\cdot 24=\dfrac{32}{3}\ un^2.$

3. Thus, the area of the quadrilateral DFGE is

$A_{DFGE}=A_{\triangle DBE}-A_{\triangle FBG}=\dfrac{32}{3}-\dfrac{8}{3}=8\ un^2.$

and the area of the quadrilateral ADEC is

$A_{ADEC}=A_{\triangle ABC}-A_{\triangle DBE}=24-\dfrac{32}{3}=\dfrac{40}{3}\ un^2.$

4 0

3 years ago

Read 2 more answers