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3 years ago

Solve for x.

Mathematics

2 answers:

astra-53 [7]3 years ago

5 0

Answer:

Option: C is correct.

$x=\ln\sqrt{\dfrac{t+1}{1-t}}$

Step-by-step explanation:

We are asked to solve for 'x' such that we are given a equation as:

$\dfrac{e^x-e^{-x}}{e^x+e^{-x}}=t$

This equation could also be written as:

$e^x-e^{-x}=t(e^x+e^{-x})\\\\e^x-e^{-x}=te^x+te^{-x}\\\\e^x-te^x=te^{-x}+e^{-x}\\\\(1-t)e^x=(t+1)e^{-x}\\\\e^{2x}=\dfrac{t+1}{1-t}$

on taking logarithmic function on both the sides we have:

$2x=\ln (\dfrac{t+1}{1-t})$

(since $\ln (e^x)=x$ )

$x=\dfrac{1}{2}\times \ln (\dfrac{t+1}{1-t})$

as we know $\dfrac{1}{2}\times \ln b = \ln (b^\frac{1}{2})$

Hence, we have: $x=\ln\sqrt{\dfrac{t+1}{1-t}}$

Hence, option C is correct.

Butoxors [25]3 years ago

5 0

Answer:

OPTION C

Step-by-step explanation:

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Find cosx if sinx cotx =0.5

Arada [10]

Answer:

cos x = 0.5

Step-by-step explanation:

Using the trigonometric identity

cot x = $\frac{cosx}{sinx}$

Given

sin x cot x = 0.5, then

sin x × $\frac{cosx}{sinx}$ = 0.5 ← cancel sin x on numerator/ denominator

cos x = 0.5

5 0

4 years ago

You represent a chemical company that is being sued for paint damage to automobiles. You want to support the claim that the mean

denpristay [2]

Correct question is;

You represent a chemical company that is being sued for paint damage to automobiles. You want to support the claim that the mean repair cost per automobile is more than $650. How would you write the null and alternative hypotheses?

A) State the null and alternative hypotheses in words (context of the problem)

B) Write the null and alternative hypotheses in appropriate symbols (population parameters μ, σ or p)

C) Describe in words Type I error (the consequence of rejecting a true null hypothesis.)

D) Describe in words Type II error (the consequence of failing to reject a false null hypothesis.)

Answer:

A)In words;

Null hypothesis is that the cost is $650

Alternative hypothesis is that the cost is not $650

B) Null hypothesis; μ = 650

Alternative hypothesis; μ ≠ 650

C) Type I Error will be to make a claim that the repair cost is not $650, when it is actually $650.

D) Type II Error will be to make a claim that the repair cost is not $650, but fail to reject the claim that it is $650.

Step-by-step explanation:

We are told that You want to support the claim that the mean repair cost per automobile is more than $650.

Thus;

A)In words;

Null hypothesis is that the cost is $650

Alternative hypothesis is that the cost is not $650

B) Null hypothesis; μ = 650

Alternative hypothesis; μ ≠ 650

C) Type I Error will be to make a claim that the repair cost is not $650, when it is actually $650.

D) Type II Error will be to make a claim that the repair cost is not $650, but fail to reject the claim that it is $650.

6 0

3 years ago

Express each ratio as a fraction in lowest terms

VladimirAG [237]

1/9
2/1
i am working on the last one but i hope i helped

7 0

3 years ago

6-2x=y 14+3x=y plzzz helllllllllllllllp

pashok25 [27]

Answer:

(- $\frac{8}{5}$ , $\frac{46}{5}$ )

Step-by-step explanation:

Given the2 equations

6 - 2x = y → (1)

14 + 3x = y → (2)

Substitute y = 14 + 3x into (1)

6 - 2x = 14 + 3x ( subtract 3x from both sides )

6 - 5x = 14 ( subtract 6 from both sides )

- 5x = 8 ( divide both sides by - 5 )

x = - $\frac{8}{5}$

Substitute x = - $\frac{8}{5}$ into either of the 2 equations for the corresponding value of y

Substituting into (1)

y = 6 - 2(- $\frac{8}{5}$ ) = $\frac{30}{5}$ + $\frac{16}{5}$ = $\frac{46}{5}$

solution is ( - $\frac{8}{5}$ , $\frac{46}{5}$ )

8 0

3 years ago

A circle is drawn inside a square so its circumference touches each of the four sides of the square. If the area of the circle i

Snowcat [4.5K]

Answer:

The length of the sides of the square is approximately 11.239 centimeters.

Step-by-step explanation:

Since the circle is inscribed in the square, the length of each side of the square ( $l$ ), in centimeters, is equal to the length of the diameter of the circle ( $D$ ), in centimeters. The area of the circle ( $A_{c}$ ), in square centimeters: