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dezoksy [38]
2 years ago
6

How to solve-2(p+-8)+-2=6

Mathematics
2 answers:
Fudgin [204]2 years ago
4 0

Answer:

p=4

Step-by-step explanation:

Simplify  p+-8  to  p−8:

−2(p−8)−2=6

Add 2 to both sides:

−2(p−8)=6+2

Simplify  6+2 to 8:

−2(p−8)=8

Divide both sides by -2:

p−8=−4

Add 8 to both sides:

p=−4+8

Simplify

p=4

natka813 [3]2 years ago
3 0

Answer:

p=4

Step-by-step explanation:

first change all the addition signs in to minus if the numbers are negative.    ex. p+-8 = p-8.

So the equation equals to -2(p-8)-2=6.

Next +2 to both sides to cancel out the -2 and you get -2(p-8)=8

Divide both sides by -2 and you get (p-8)=-4. At this point you can remove the parenthesis. So the equation is p-8=-4

Finally +8 to both sides and you get p=4

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Suppose that r varies directly with s and inversely with t, and r = 2 when s = 3 and t = 12. What is the value of r when s = 5 a
Dmitry_Shevchenko [17]
The statement suppose <span>that r varies directly with s and inversely with t is best represented as:

r </span>α s/t

To make it into equality, we insert a proportionality constant, k:

r = ks/t

Using the initial conditions, we solve for k.

r = ks/t
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3 years ago
Find values of  that satisfy the equation for 0º    360º and 0  θ  2 . Give answers in degrees and radians.
suter [353]

Answer:

\theta = 30^\circ, 330^\circ

\theta = \frac{\pi}{6}, \frac{11\pi}{6}

Step-by-step explanation:

Given [Missing from the question]

Equation:

cos\theta = \frac{\sqrt 3}{2}

Interval:

0 \le \theta \le 360

0 \le \theta \le 2\pi

Required

Determine the values of \theta

The given expression:

cos\theta = \frac{\sqrt 3}{2}

... shows that the value of \theta is positive

The cosine of an angle has positive values in the first and the fourth quadrants.

So, we have:

cos\theta = \frac{\sqrt 3}{2}

Take arccos of both sides

\theta = cos^{-1}(\frac{\sqrt 3}{2})

\theta = 30 --- In the first quadrant

In the fourth quadrant, the value is:

\theta = 360 -30

\theta = 330

So, the values of \theta in degrees are:

\theta = 30^\circ, 330^\circ

Convert to radians (Multiply both angles by \pi/180)

So, we have:

\theta = \frac{30 * \pi}{180}, \frac{330 * \pi}{180}

\theta = \frac{\pi}{6}, \frac{33 * \pi}{18}

\theta = \frac{\pi}{6}, \frac{11 * \pi}{6}

\theta = \frac{\pi}{6}, \frac{11\pi}{6}

8 0
2 years ago
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