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3 years ago

Pleaase help!Find h. 11 in. h = ✓ [?] in. 14 in Enter

Mathematics

1 answer:

Aleksandr [31]3 years ago

3 0

Answer:

$h=8.48\ \text{inches}$

Step-by-step explanation:

Given that,

Radius of cone, r = 7 inches

Slant height of the cone, l = 11 inches

We need to find the height of the cone. Let the height be h. Using trigonometry to find it as follows :

$h^2+r^2=l^2\\\\h=\sqrt{l^2-r^2}\\\\h=\sqrt{11^2-7^2}\\\\h=8.48\ \text{inches}$

So, the height of the cone is equal to $8.48\ \text{inches}$ .

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4. Tom rides his bike a total of 6 14 miles to work over the course of 5 days.

irina [24]

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3 years ago

The age of Sita is five years more than Gautam's age. 5 years ago, the ratio of their ages was 3:2. Find their present ages.

vampirchik [111]

<h2>Question:-</h2>

The age of Sita is five years more than Gautam's age. 5 years ago, the ratio of their ages was 3:2. Find their present ages.

<h2>Solution:-</h2>

The age of Sita = (x + 5) years.

And, the age of Gautam = x years.

Now,

5 years ago, the ratio of their ages was 3 : 2

$\therefore$ $\bf \frac{3}{2} \: = \frac{x \: + \: 5}{x}$

Now by cross multiplying, we get,

3x = 2(x + 5)

3x = 2x + 10

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Now,

The present age of Sita = 15 + 5 = 20 years. (Answer)

And, the present age of Gautam = 10 + 5 = 15 years. (Answer)

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3 years ago

State if each pair of ratios forms a proportion 4/3 and 8/6 yes or no??

Reptile [31]

We have: $\frac{4}{3}=\frac{4.2}{3.2}= \frac{8}{6}$

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Ok done. Thank to me :>

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What is the average value of y=cosx/x^2+x+2 on the closed interval "[-1, 3]"?

soldi70 [24.7K]

We have
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Note that [cos(x)/(x² + x + 2)] does not have an elementary anti-derivative.
By approximating techniques:
1/4 ∫ cos(x)/(x^2 + x + 2) dx (from x=-1 to 3) ≈ 0.182951

the answer is
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3 years ago

A pair of shoes is on sale for 80% of the original price. The sale price is $60. What was the original price?

Paraphin [41]

Answer:

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4 years ago

Read 2 more answers

Pleaase help!Find h. 11 in. h = ✓ [?] in. 14 in Enter​

Pleaase help!Find h. 11 in. h = ✓ [?] in. 14 in Enter