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3 years ago

Which sets of ratios are a proportion? (Choose all that apply) *

Mathematics

1 answer:

vladimir1956 [14]3 years ago

5 0

Answer:

<h2>Sets of Ratios that are Proportion</h2>

3/11 and 17/6

60/55 and 7/25

7/6 and 52/48

<h3>Solution:</h3>

> 12/7 and 36/21

$\frac{12}{7} = \frac{36}{21} \\ 12(21) = 36(7) \\ 252 = 252$

> 4/10 and 8/20

$\frac{4}{10} = \frac{8}{20} \\ 4(20) = 8(10) \\ 80 = 80$

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ivanzaharov [21]

Answer:

It is C

Step-by-step explanation:

5 0

3 years ago

What is the volume of a rectangular prism that is 16 meters by 25 meters by 37 meters?

ElenaW [278]

Answer:

14,800

Step-by-step explanation:

To find the volume of an object, simply multiply its hight, length, and width together.

16*25*37

16*25=400

400*37=14,800

The rectangular prism has a volume of 14,800.

Hope this helps!

4 0

3 years ago

Find two consecutive odd integers if twice the larger, increased by the smaller equals 37,

Alexxx [7]

Answer:

11 and 13

Step-by-step explanation:

the first (smaller) odd integer can be called x

the second (larger) odd integer can be called x+2

2(x+2)+x = 37

<em>[simplify . . .]</em>

2x+4+x = 37

3x+4 = 37

<em>[subtract 4 from each side]</em>

3x = 33

<em>[divide each side by 3; find smaller odd integer]</em>

x = 11

<em>[add 2 to find larger odd integer]</em>

x+2 = 13

6 0

3 years ago

A rectangular box with a square base is to be constructed from material that costs $9 per ft2 for the bottom, $5 per ft2 for the

kramer

Answer:

$18.73ft^3$

Step-by-step explanation:

Let

Side of square base=x

Height of rectangular box=y

Area of square base=Area of top= $(side)^2=x^2$

Area of one side face= $l\times b=xy$

Cost of bottom=$9 per square ft

Cost of top=$5 square ft

Cost of sides=$4 per square ft

Total cost=$204

Volume of rectangular box= $V=lbh=x^2y$

Total cost= $9(x^2)+5x^2+4(4xy)=14x^2+16xy$

$204=14x^2+16xy$

$204-14x^2=16xy$

$y=\frac{204-14x^2}{16x}=\frac{102-7x^2}{8x}$

Substitute the values of y

$V(x)=x^2\times \frac{102-7x^2}{8x}=\frac{1}{8}(102x-7x^3)$

Differentiate w.r.t x

$V'(x)=\frac{1}{8}(102-21x^2)=0$

$V'(x)=0$

$\frac{1}{8}(102-21x^2)=0$

$102-21x^2=0$

$102=21x^2$

$x^2=\frac{102}{21}=4.85$

$x=\sqrt{4.85}=2.2$

It takes positive because side length cannot be negative.

Again differentiate w.r. t x

$V''(x)=\frac{1}{8}(-42x)$

Substitute the value

$V''(2.2)=-\frac{42}{8}(2.2)=-11.55$

Hence, the volume of box is maximum at x=2.2 ft

Substitute the value of x

$y=\frac{102-7(2.2)^2}{8(2.2)}=3.87 ft$

Greatest volume of box= $x^2y=(2.2)^2\times 3.87=18.73 ft^3$

5 0

3 years ago

When BA=10 ft, find the area of the region that is NOT shaded. Round to the nearest whole number. (reply so i can draw it out, p

nexus9112 [7]

The area of the region that is not shaded in the diagram is determined through the following solution:

<span>A = <span>(1 − (<span><span>60°/</span><span>360°</span></span>)*</span>π) * (10 ft ) ^<span>2</span></span> <span>A = <span>56*</span>π*100 [<span>ft2</span>]</span> <span><span>A = <span>250/3 </span>π [<span>ft2</span>]
A = 262 [ft2]</span></span>

5 0

3 years ago