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4 years ago

Solve for Y py+qy= -4y+8

Mathematics

1 answer:

BlackZzzverrR [31]4 years ago

4 0

Answer:

This is a step by step procedure to get the value of y.

First: Move all terms to the left side and set equal to zero.

Second: Then set each factor equal to zero.

The application is:

Given: py+7=6y+q

-6y -7 -6y -7 = 0

(p-6)y = q-7

divide both sides by p-6

y=(q-7)/(p-6)

Answer is y = (q – 7) / (p – 6)

Step-by-step explanation:

i hope i helped : )

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3 years ago

Which of these values of x is NOT a solution of the equation tanx= 1?

Kobotan [32]

The tangent ratio is an illustration of the trigonometry identity

The value of x that is not a solution of tan(x) = 1 is 3 $\pi$ /4

<h3>How to determine the value that is not a solution</h3>

The trigonometry identity is given as:

tan(x) = 1

Next, we test the x values using a calculator.

When $x = -7\pi/4$ , we have:

tan(-7 $\pi$ /4) = 1

When x = 3 $\pi$ /4, we have:

tan(3 $\pi$ /4) = -1

When x = 5 $\pi$ /4, we have:

tan(5 $\pi$ /4) = 1

When x = $\pi$ /4, we have:

tan( $\pi$ /4) = 1

Hence, the value of x that is not a solution of tan(x) = 1 is 3 $\pi$ /4

Read more about trigonometry identity at:

brainly.com/question/7331447

5 0

3 years ago

Please help if you can. I will give Brainiest to best answer. This question is a Calculus/Trigonometry question. I ask that you

Ierofanga [76]

7 is incorrect. The answer should be -4. Here's how I'd derive it:

$\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}$

With $\pi$ , we should expect $\cos x$ . If $\tan x=\dfrac8{15}$ , then

$\sec x=-\sqrt{1+\tan^2x}=-\dfrac{17}{15}\implies\cos x=-\dfrac{15}{17}$

Also,

$\pi$

so we should expect $\tan\dfrac x2$ and

$\tan\dfrac x2=-\sqrt{\dfrac{1-\cos x}{1+\cos x}}=-4$

You seem to be taking

$\tan\dfrac x2=\dfrac{1+\cos x}{-\sin x}$

but this is not an identity.

###

8 is incorrect. Just a silly mistake, you swapped the order of the terms in the numerator. It should be

$\dfrac{\sqrt6-\sqrt2}4$

###

9. Use the identity from (7). $\dfrac{7\pi}{12}$ lies in the second quadrant, so

$\tan\dfrac{7\pi}{12}=-\sqrt{\dfrac{1-\cos\frac{7\pi}6}{1+\cos\frac{7\pi}6}}=-\sqrt{7+4\sqrt3}=-2-\sqrt3$

###

12.

$\dfrac{\cot x-1}{1-\tan x}=\dfrac{\frac{\cos x}{\sin x}-1}{1-\frac{\sin x}{\cos x}}$

$=\dfrac{\cos x(\cos x-\sin x)}{\sin x(\cos x-\sin x)}$

$=\dfrac{\cos x}{\sin x}$

$=\dfrac{\csc x}{\sec x}$

###

13.

$\dfrac{1+\tan x}{\sin x+\cos x}=\dfrac{1+\frac{\sin x}{\cos x}}{\sin x+\cos x}$

$=\dfrac{\cos x+\sin x}{\cos x(\sin x+\cos x)}$

$=\dfrac1{\cos x}$

$=\sec x$

###

14.

$\sin2x(\cot x+\tan x)=2\sin x\cos x\left(\dfrac{\cos x}{\sin x}+\dfrac{\sin x}{\cos x}\right)$

$=2\sin x\cos x\dfrac{\cos^2x+\sin^2x}{\sin x\cos x}$

$=2$

###

15.

$\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\dfrac{1-\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}}$

$=\dfrac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}$

$=\cos2\theta$

3 0

3 years ago

Keiko has scored 86, 91, 86, 61, and 87 on her previous five tests. what score does she need on her next test so that her averag

Andru [333]

X= grade on next test

Add up all scores, divide by 6 and this equals the average of 84

(86 + 91 + 86 + 61 + 87 + x)/6= 84
add in parentheses

(411 + x)/6= 84
multiply both sides by 6

411 + x= 504
subtract 411 from both sides

x= 93

ANSWER: 93 is needed on the next test to have an average of 84

Hope this helps! :)

8 0

3 years ago