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3 years ago

8

What is the value of x if 5^X+2=5^9?

2 answers:

scoray [572]3 years ago

8 0

Answer:

I believe the answer is 7

Step-by-step explanation:

Dafna11 [192]3 years ago

6 0

What is the value of x if 5^X+2=5^9?

<u>7.</u>

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Find the solution to m^2+9m=0

sladkih [1.3K]

<h3>Answers: m = 0 or m = -9</h3>

Work Shown:

m^2+9m=0

m(m+9)=0

m = 0 or m+9 = 0

m = 0 or m = -9

For more information, check out the topics about factoring and the zero product property.

5 0

2 years ago

Read 2 more answers

Original chess tournament evolve 64 participants the number Of participants decreased by 1/2 after each round of play. Write the

marishachu [46]

Answer:

The first three terms is 64, 32, 16

Step-by-step explanation:

Given

Number of participants = 64

Decrement = 1/2

Required

Find the first three terms

To calculate the first three terms, we make use of the following formula;

Current Term = Previous Term - 1/2 * Previous Term

The first term = Initial number of participants

First Term = 64

Second Term = First Term - 1/2 * First Term

Second Term = 64 - 1/2 * 64

Second Term = 64 - 32

Second Term = 32

Third Term = Second Term - 1/2 * Second Term

Third Term = 32 - 1/2 * 32

Third Term = 32 - 16

Third Term = 16

Hence, the first three terms is 64, 32, 16

4 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

An electrical engineer wishes to compare the mean lifetimes of two types of transistors in an application involving high-tempera

Verizon [17]

Answer:

add the decimals thats all

Step-by-step explanation:

7 0

4 years ago

PLEASE HELP ME OUT ON MATH TO ALL THE MATH NERDS OUT THERE

finlep [7]

Answer:

4 847 575, 389362513

Step-by-step explanation:

is the answer

hope i helped

6 0

3 years ago

Read 2 more answers