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3 years ago

Find the missing number. 30,______ 19, 13 1/2.

Mathematics

1 answer:

Andru [333]3 years ago

5 0

The answer is 24.5 because $19-13 \frac{1}{2} = 5 \frac{1}{2}$ so you add 5.5 from 19 to get the answer

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Is AABC AXYZ? If so, identify the similarity postulate or theorem that<br> applies

zysi [14]

Answer:

Similar-AA

Step-by-step explanation:

If you know 2 angles of the triangle you automatically know the third. If all the corresponding angles of two triangles are the same, they are similar.

(p.s. This sign ~ means similar)

7 0

3 years ago

If 5 pounds of broccoli costs $12.00 then how much will 7 pounds of broccoli cost

kondaur [170]

Each pound costs 2.40,
to check this 2.40 * 5 = 12
so 2.40 * 7 = 16.80

Answer: 16.80

8 0

3 years ago

The bar graph below displays students’ responses to the question "What caffeinated drinks do you consume?”

tamaranim1 [39]

Answer:

Yes, because the data are grouped into categories.

Step-by-step explanation:

7 0

3 years ago

A helicopter is at 110 feet above landing pad, angle of elevation is 29. How far the landing agent is from the landing pad? 60.9

Lorico [155]

You have to use soh coa toa.. the equation is set up to use sine

7 0

3 years ago

How can you prove that csc^2(θ)tan^2(θ)-1=tan^2(θ)

Oxana [17]

Answer:

Make use of the fact that as long as $\sin(\theta) \ne 0$ and $\cos(\theta) \ne 0$ :

$\displaystyle \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ .

$\displaystyle \csc(\theta) = \frac{1}{\sin(\theta)}$ .

$\sin^{2}(\theta) + \cos^{2}(\theta) = 1$ .

Step-by-step explanation:

Assume that $\sin(\theta) \ne 0$ and $\cos(\theta) \ne 0$ .

Make use of the fact that $\tan(\theta) = (\sin(\theta)) / (\cos(\theta))$ and $\csc(\theta) = (1) / (\sin(\theta))$ to rewrite the given expression as a combination of $\sin(\theta)$ and $\cos(\theta)$ .

$\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \left(\frac{1}{\sin(\theta)}\right)^{2} \, \left(\frac{\sin(\theta)}{\cos(\theta)}\right)^{2} - 1 \\ =\; & \frac{\sin^{2}(\theta)}{\sin^{2}(\theta)\, \cos^{2}(\theta)} - 1\\ =\; & \frac{1}{\cos^{2}(\theta)} - 1\end{aligned}$ .

Since $\cos(\theta) \ne 0$ :

$\displaystyle 1 = \frac{\cos^{2}(\theta)}{\cos^{2}(\theta)}$ .

Substitute this equality into the expression:

$\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots\\ =\; & \frac{1}{\cos^{2}(\theta)} - 1 \\ =\; & \frac{1}{\cos^{2}(\theta)} - \frac{\cos^{2}(\theta)}{\cos^{2}(\theta)} \\ =\; & \frac{1 - \cos^{2}(\theta)}{\cos^{2}(\theta)}\end{aligned}$ .

By the Pythagorean identity, $\sin^{2}(\theta) + \cos^{2}(\theta) = 1$ . Rearrange this identity to obtain:

$\sin^{2}(\theta) = 1 - \cos^{2}(\theta)$ .

Substitute this equality into the expression:

$\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots \\ =\; & \frac{1 - \cos^{2}(\theta)}{\cos^{2}(\theta)} \\ =\; & \frac{\sin^{2}(\theta)}{\cos^{2}(\theta)}\end{aligned}$ .

Again, make use of the fact that $\tan(\theta) = (\sin(\theta)) / (\cos(\theta))$ to obtain the desired result:

$\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots \\ =\; & \frac{\sin^{2}(\theta)}{\cos^{2}(\theta)}\\ =\; & \left(\frac{\sin(\theta)}{\cos(\theta)}\right)^{2} \\ =\; & \tan^{2}(\theta)\end{aligned}$ .

5 0

2 years ago