3 years ago

How do i find the missing base of a trapezoid?

1 answer:

7 0

The tops and bottoms of a trapezoid are parallel, and a polygon's volume and angles are always 360.As long as you know the shape and about three of the lengths, you can find the base.

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HELP ASAP

Mamont248 [21]

Answer:

m = 0

Step-by-step explanation:

distribute

7 + 35m + 2m = 7 +2m

add 35m + 2m

7 + 37m = 7 + 2m

subtract 2 from 37m

7 + 35m = 7

subtract 7 from 7

35m = 0

0 ÷ 35 = 0

so technically the answer is 0

I'm not 100% sure but I do believe this is right

3 0

3 years ago

Write the ratio as a fraction in simplest form 30 to 18

Westkost [7]

30/18 = 10/6 = 5/3 hope this helps!

6 0

3 years ago

Read 2 more answers

Hi, does someone know what this question is looking for? There is no other instructions.

slamgirl [31]

Answer: 3y⁴(y² + 5)(y - 1)(y + 1)

Step-by-step explanation:

3y⁸ + 12y⁶ - 15y⁴

= 3y⁴(y⁴ + 4y² - 5)

= 3y⁴(y² + 5)(y² - 1)

= 3y⁴(y² + 5)(y - 1)(y + 1)

7 0

3 years ago

Benny has saved three thousand nine hundred cents over four days from selling lemonade.How many dollars does Benny have?

saw5 [17]

30 Dollars.

Hope this helps please mark this as Brainliest answer,

6 0

3 years ago

Sebastian solved the radical equation y + 1 = but did not check his solution. (y + 1)2 = y2 + 2y + 1 = –2y – 3 y2 + 4y + 4 = 0 (

Softa [21]

Answer:

There are no true solutions to the equation.

Step-by-step explanation:

The correct equation is

$y+1=\sqrt{-2y-3}$

Solve for y

squared both sides

$(y+1)^2=(-2y-3)$

$(y^2+2y+1)=(-2y-3)$

$y^2+2y+1+2y+3=0$

$y^2+4y+4=0$

$(y+2)(y+2)=0$

$y=-2$

Verify

substitute the value of y in the original expression

$-2+1=\sqrt{-2(-2)-3}$

$-1=1$ ----> is not true

therefore

There are no true solutions to the equation.

4 0

3 years ago