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Angelina_Jolie [31]
3 years ago
10

Question 4 pls help in a timer

Mathematics
2 answers:
OleMash [197]3 years ago
7 0
The first one is > the the second one is =
Verdich [7]3 years ago
4 0
Number 1 = >
1foot = 12inches
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The diameter of a circle if five times the length of a recatngle. The length of a rectangle is three times the width of the rect
Leno4ka [110]

Answer:

706.5 ft²

Step-by-step explanation:

A. Width of the rectangle

The formula for the area of a rectangle is

A = lw

But, l = 3w, so,

A = (3w)w

= 3w² = 12

      w² = 4         Divided each side by 3

      w   = 2 ft     Took the square root of each side

B. Length of rectangle

l = 3w = 6 ft

C. Diameter of circle

d = 5l = 5 × 6 = 30 ft

D. Radius of circle

d = 2r

r = d/2     Divided each side by 2

r = 30/2 = 15 ft

E. Area of circle

A = πr² = 3.14 × 15² = 3.14 × 225 = 706.5 ft²

The area of the circle is 706.5 ft².

8 0
3 years ago
Find the volume of each figure. Round your answers to the nearest tenth, if necessary​
olga nikolaevna [1]

Answer:

1.92 m³

hopefully this answer can help you to answer the next question.

5 0
3 years ago
A student took a 20 question test. 16 of the questions were correct. What percent of the questions did the student get correct?
elena-14-01-66 [18.8K]

Answer:

80%

Step-by-step explanation:

16/20 = 0.80

80%

5 0
3 years ago
Read 2 more answers
Is 22+ 32 = 42 a true statement? Explain.
ivolga24 [154]
No, this is not true, 22+32 =54 this is not a true statement
4 0
3 years ago
The equation a=1/2(b^1+b^2)h can be determined the area, a, of a trapezoid with height, h, and base lengths, b^1 and b^2 Which a
Evgesh-ka [11]

The complete question is as follows.

The equation a = \frac{1}{2}(b_1 + b_2 )h can be used to determine the area , <em>a</em>, of a trapezoid with height , h, and base lengths, b_1 and b_2. Which are equivalent equations?

(a) \frac{2a}{h} - b_2 = b_1

(b) \frac{a}{2h} - b_2 = b_1

(c) \frac{2a - b_2}{h} = b_1

(d) \frac{2a}{b_1 + b_2} = h

(e) \frac{a}{2(b_1 + b_2)} = h

Answer: (a) \frac{2a}{h} - b_2 = b_1; (d) \frac{2a}{b_1 + b_2} = h;

Step-by-step explanation: To determine b_1:

a = \frac{1}{2}(b_1 + b_2 )h

2a = (b_1 + b_2)h

\frac{2a}{h} = b_1 + b_2

\frac{2a}{h} - b_2 = b_1

To determine h:

a = \frac{1}{2}(b_1 + b_2 )h

2a = (b_1 + b_2)h

\frac{2a}{(b_1 + b_2)} = h

To determine b_2

a = \frac{1}{2}(b_1 + b_2 )h

2a = (b_1 + b_2)h

\frac{2a}{h} = (b_1 + b_2)

\frac{2a}{h} - b_1 = b_2

Checking the alternatives, you have that \frac{2a}{h} - b_2 = b_1 and \frac{2a}{(b_1 + b_2)} = h, so alternatives <u>A</u> and <u>D</u> are correct.

4 0
3 years ago
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