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3 years ago

What is the value of the expression shown below? 2 over 3 to the power of 2 + 5 × 2 − 4

Mathematics

2 answers:

trasher [3.6K]3 years ago

3 0

The answer would be 6 and 4 over 9

Alisiya [41]3 years ago

3 0

Answer: <em> 6 and 4 over 9 </em>

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Find the area of the triangle.

s344n2d4d5 [400]

Answer:

10.5 cm^2

Step-by-step explanation:

Since we have two sides and the angle between those sides, we can use the alternative area formula:

$A=\frac{1}{2}ab\sin(C)$

a and b are the two sides while C is the angle in between the two sides.

Plug in the numbers:

$A=\frac{1}{2}(7)(6)\sin(150)$

Recall the unit circle. Sin(150) is 1/2.

$A=21(\frac{1}{2})$

$A=21/2=10.5cm^2$

7 0

3 years ago

If the sum of a number and six is tripled, the result is eight less than twice the number. Find the number.

PSYCHO15rus [73]

3(x+6)=2x-8

x= -26

the answer is -26

7 0

3 years ago

If a + 15 = 19, then a = ?

Softa [21]

Answer:

a=4

Step-by-step explanation:

4+15=19 it makes sense right?

hope it helps

7 0

3 years ago

Read 2 more answers

Help!!!!!!<br> Please and thank you

fenix001 [56]

Answer:

1. A

2. B

3. A

Step-by-step explanation:

Hope this helps!

5 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