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2 years ago

A rectangle has an area of 4y^2 – 49 square units. What are the dimensions of the rectangle?

Mathematics

1 answer:

klio [65]2 years ago

8 0

Answer:

Step-by-step explanation:hi

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What is the area of this trapezoid? 175 in² 140 in² 129 in² 85 in²

aksik [14]

Area of trapezoid = (B1 + B2)h/2

B1 = 15 in. + 4 in. + 6 in. = 25 in.
B2 = 15 in.
h = 7 in.

Area = (25 in. + 15 in.)(7 in.)/2 = 140 in.^2

3 0

2 years ago

Read 2 more answers

The volume is 120.9 in. What is the missing dimension (X)

Elina [12.6K]

Answer:

X = 2.5 in

Step-by-step explanation:

volume = L x W x H

120.9 = 9.3 x 5.2 x X

X = 2.5 in

5 0

2 years ago

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Show by substitution whether the number r is a solution of the corresponding quadratic equation.

Stels [109]

Answer:

It's choice B.

Step-by-step explanation:

When r= -11,

x^2 + 8x - 33 = (-11)^2 - 88 - 33

= 121 - 88 - 33

= 0

- so x = -11 gives a zero value and is a solution to the equation.

3 0

1 year ago

Substitute a = 4, b = 1 and c = 5 into the expressions: a + b - c

Soloha48 [4]

Answer:

a+b-c

4+1-5

5-5=0

THE ANSWER IS 0

3 0

2 years ago

three bags of sweets weigh 27/4 kg. two of them have the same weight and the third bag is heavier than each of the bags of equal

Nana76 [90]

I'm here buddy,

so, let's take the value of the two bags with equal weight as x.

= x + x + (x + $\frac{6}{5}$ ) = $\frac{27}{4}$

= 3x + $\frac{6}{5}$ = $\frac{27}{4}$

= 3x = $\frac{27}{4}$ - $\frac{6}{5}$

( let's take the LCM of 4 and 5 = 20

= 3x = $\frac{135}{20}$ - $\frac{24}{20}$

= 3x = $\frac{111}{20}$

= x = $\frac{111}{20}$ ÷ $\frac{3}{1}$ = $\frac{111}{20}$ × $\frac{1}{3}$ = $\frac{37}{20}$

So, the weight of the equal bags are $\frac{37}{20}$ and the weight of the third bag ( heavy one ) is $\frac{37}{20}$ + $\frac{6}{5}$ = $\frac{37}{20}$ + $\frac{24}{20}$ = $\frac{61}{20}$

1st bag = $\frac{37}{20}$ kg

2nd bag = $\frac{37}{20}$ kg

3rd bag = $\frac{61}{20}$ kg

Hope it helps...

7 0

2 years ago