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3 years ago

In △Rst, the measure of

Mathematics

2 answers:

Scilla [17]3 years ago

7 0

Answer:

∠ R ≈ 53°

Step-by-step explanation:

Using the sine ratio in the right triangle

sin R = $\frac{opposite}{hypotenuse}$ = $\frac{ST}{SR}$ = $\frac{7.7}{9.7}$ , thus

∠ R = $sin^{-1}$ ( $\frac{7.7}{9.7}$ ) ≈ 53° ( to the nearest degree )

forsale [732]3 years ago

5 0

Answer:

Step-by-step explanation:

You might just duh me but the answer is obviously less than 7.7 but greater that anything above one so what you do is process of elimination hope this way of solving helps

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A rectangular box with a volume of 272ft^3 is to be constructed with a square base and top. The cost per square foot for the bot

ASHA 777 [7]

Answer:

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

The length of one side of the base of the given box is 3 ft.

The height of the box is 30.22 ft.

Step-by-step explanation:

Given that, a rectangular box with volume of 272 cubic ft.

Assume height of the box be h and the length of one side of the square base of the box is x.

Area of the base is = $(x\times x)$

$=x^2$

The volume of the box is = area of the base × height

$=x^2h$

Therefore,

$x^2h=272$

$\Rightarrow h=\frac{272}{x^2}$

The cost per square foot for bottom is 20 cent.

The cost to construct of the bottom of the box is

=area of the bottom ×20

$=20x^2$ cents

The cost per square foot for top is 10 cent.

The cost to construct of the top of the box is

=area of the top ×10

$=10x^2$ cents

The cost per square foot for side is 1.5 cent.

The cost to construct of the sides of the box is

=area of the side ×1.5

$=4xh\times 1.5$ cents

$=6xh$ cents

Total cost = $(20x^2+10x^2+6xh)$

$=30x^2+6xh$

Let

C $=30x^2+6xh$

Putting the value of h

$C=30x^2+6x\times \frac{272}{x^2}$

$\Rightarrow C=30x^2+\frac{1632}{x}$

Differentiating with respect to x

$C'=60x-\frac{1632}{x^2}$

Again differentiating with respect to x

$C''=60+\frac{3264}{x^3}$

Now set C'=0

$60x-\frac{1632}{x^2}=0$

$\Rightarrow 60x=\frac{1632}{x^2}$

$\Rightarrow x^3=\frac{1632}{60}$

$\Rightarrow x\approx 3$

Now $C''|_{x=3}=60+\frac{3264}{3^3}>0$

Since at x=3 , C''>0. So at x=3, C has a minimum value.

The length of one side of the base of the box is 3 ft.

The height of the box is $=\frac{272}{3^2}$

=30.22 ft.

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

7 0

3 years ago

Estimate Width of a room in cm, m, or km

bearhunter [10]

The width of a room can be from around 15 cm to about 40 cm.
meaning 1 m
and about less than 1/6 of a km
Hope it helped!

7 0

3 years ago

The legs of a right triangle are 12 cm and 16 cm. What is the length of the hypotenuse? 15 cm 17 cm 20 cm 28 cm

GalinKa [24]

Answer: 20 cm

To solve this, use the Pythagorean Theorem, a² + b² = c², where a and b are the legs of the right triangle, and c is the hypotenuse of the right triangle.
**The Pythagorean Theorem only works for right triangles.**

Substitute in the values.
a² + b² = c²
12² + 16² = c²

Solve for c, the hypotenuse.
12² + 16² = c²
144 + 256 = c²
400 = c²
c = √400
c = 20

7 0

3 years ago

Read 2 more answers

The line with x-intercept of 10 and y-<br> intercept of -2.

Yuri [45]

Answer:

y = $\frac{1}{5}$ x - 2