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3 years ago

11

Which statement is true

1 answer:

Klio2033 [76]3 years ago

5 0

Answer:

B

Step-by-step explanation:

this is true because if you rotate 180 degrees on a graph both the x and y will always become their opposite.

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If tan^3(theta) -1/tan(theta)-1 - sec^2(theta) + 1 = 0 find cot(theta)

Nutka1998 [239]

$\tan^3\theta-\dfrac1{\tan\theta-1}-\sec^2\theta+1=0$

Recall that $\tan^2\theta+1=\sec^2\theta$ , so we can write everything in terms of $\tan\theta$ :

$\tan^3\theta-\dfrac1{\tan\theta-1}-\tan^2\theta=0$

Let $x=\tan\theta$ , so that

$x^3-\dfrac1{x-1}-x^2=0$

With some rewriting we get

$x^3-x^2-\dfrac1{x-1}=0$

$x^2(x-1)-\dfrac1{x-1}=0$

$\dfrac{x^2(x-1)^2-1}{x-1}=0$

Clearly we cannot have $x=1$ , or $\tan\theta=1$ .

The numerator determines when the expression on the left reduces to 0:

$x^2(x-1)^2-1=0$

$x^2(x-1)^2=1$

$\sqrt{x^2(x-1)^2}=\sqrt1$

$|x(x-1)|=1$

$x(x-1)=1\text{ or }x(x-1)=-1$

Completing the square gives

$x(x-1)=x^2-x=x^2-x+\dfrac14-\dfrac14=\left(x-\dfrac12\right)^2-\dfrac14$

so that

$\left(x-\dfrac12\right)^2=\dfrac54\text{ or }\left(x-\dfrac12\right)^2=-\dfrac34$

The second equation gives no real-valued solutions because squaring any real number gives a positive real number. (I'm assuming we don't care about complex solutions.) So we're left with only

$\left(x-\dfrac12\right)^2=\dfrac54$

$\sqrt{\left(x-\dfrac12\right)^2}=\sqrt{\dfrac54}$

$\left|x-\dfrac12\right|=\dfrac{\sqrt5}2$

which again gives two cases,

$x-\dfrac12=\dfrac{\sqrt5}2\text{ or }x-\dfrac12=-\dfrac{\sqrt5}2$

$x=\dfrac{1+\sqrt5}2\text{ or }x=\dfrac{1-\sqrt5}2$

Then when $x=\tan\theta$ , we can find $\cot\theta$ by taking the reciprocal, so we get

$\boxed{\cot\theta=\dfrac2{1+\sqrt5}\text{ or }\cot\theta=\dfrac2{1-\sqrt5}}$

8 0

3 years ago

Ax + c = R (solve for x)

sp2606 [1]

Ax + c = R (subtract c from each side)

Ax + c - c = R - c

Ax = R - c (divide A from each side)

Ax/A = (R-c)/A

x = $\frac{R-c}{A}$ <-answer

4 0

3 years ago

Read 2 more answers

If 50% is 30 what is 75%

Gemiola [76]

Answer:

$50\% = 30 \\ 70\% = x \\ \\ 50\%x = 30 \times 70\% \\ x = \frac{30 \times 70\%}{50\%} \\ x = 42 \\ hence \: 70\% \: is \: 42$

6 0

3 years ago

When estimating a job to bid, a contractor’s estimator first determines the actual cost of labor using the function L(h) = 28.75

dybincka [34]

Answer:

M(h) = 73.18025h

Step-by-step explanation:

The composite function is ...

M(B(L(h))) = M(B(28.75h)) = M(1.78(28.75h)) = M(51.175h)

= 1.43(51.175h) = 73.18025h

The composite function is ...

M(h) = M(B(L(h))) = 73.18025h

7 0

3 years ago

Which equation represents the graph? (y-1)^2/49-x^2/81=1 (y-1)^2/7-x^2/9=1

Anon25 [30]

Answer: A, (y-1)^2/49-x^2/81=1

7 0

3 years ago