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2 years ago

Three apple pies, all the following shape, are left on a counter. How much total pie has been left?

Mathematics

1 answer:

faltersainse [42]2 years ago

6 0

1 pie is the total pieeeeeeeeeee

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What is the surface area of the triangular prism

Harrizon [31]

Answer:

$surface \: a = 9 \times 5 \\ = 45ft \times 2 \\ = {90}^{2} ft\\ surface \: b = \frac{1}{2} bh \\ = \frac{1}{2 } \times 6 \times 4 \\ = 3 \times 4 \\ = 12 \times 2 \\ = {144}^{2} ft \\ = 90 + 144 \\ = {234}^{2} ft$

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2 years ago

Θ is an acute angle. Find the value of θ in radians.

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2 years ago

A town's yearly snowfall in inches over a 10-year period is recorded in this table. What is the mean of the snowfall amounts?

lawyer [7]

So to find the mean you add them all together and divide by the number of numbers there is so 11+15+19+17+23+20+17+18+21=174 and there are 10 numbers so it would be 174/10 which is 17.4 so you answer is 17.4

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2 years ago

(1/1+sintheta)=sec^2theta-secthetatantheta pls help me verify this

Xelga [282]

Answer:

See Below.

Step-by-step explanation:

We want to verify the equation:

$\displaystyle \frac{1}{1+\sin\theta} = \sec^2\theta - \sec\theta \tan\theta$

To start, we can multiply the fraction by (1 - sin(θ)). This yields:

$\displaystyle \frac{1}{1+\sin\theta}\left(\frac{1-\sin\theta}{1-\sin\theta}\right) = \sec^2\theta - \sec\theta \tan\theta$

Simplify. The denominator uses the difference of two squares pattern:

$\displaystyle \frac{1-\sin\theta}{\underbrace{1-\sin^2\theta}_{(a+b)(a-b)=a^2-b^2}} = \sec^2\theta - \sec\theta \tan\theta$

Recall that sin²(θ) + cos²(θ) = 1. Hence, cos²(θ) = 1 - sin²(θ). Substitute:

$\displaystyle \displaystyle \frac{1-\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta \tan\theta$

Split into two separate fractions:

$\displaystyle \frac{1}{\cos^2\theta} -\frac{\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta\tan\theta$

Rewrite the two fractions:

$\displaystyle \left(\frac{1}{\cos\theta}\right)^2-\frac{\sin\theta}{\cos\theta}\cdot \frac{1}{\cos\theta}=\sec^2\theta - \sec\theta \tan\theta$

By definition, 1 / cos(θ) = sec(θ) and sin(θ)/cos(θ) = tan(θ). Hence:

$\displaystyle \sec^2\theta - \sec\theta\tan\theta \stackrel{\checkmark}{=} \sec^2\theta - \sec\theta\tan\theta$

Hence verified.

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2 years ago

What is the area of a rectangle with vertices at (0, −4) , (−1, −3) , (2, 0) , and (3, −1) ? Enter your answer in the box

Bad White [126]

We are given vertices of a rectangle (0, −4) , (−1, −3) , (2, 0) , and (3, −1).

Length is the distance between (0, −4) and (−1, −3) points.

Width is distance between (−1, −3) and (2, 0) points.

<u>Computing length:</u>

$\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$