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2 years ago

6=-w/8. How do I solve this equation . What is w

1 answer:

6 0

To “cancel out the the 8 you would multiply by 8 on both sides and then you would be left with 48=-w. With that you would find the opposite of 48 and w and w=-48

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I need help with #15

gladu [14]

15. MP(5,11) S(3,5) M(x,y)

(3+x)/2 = 5 x=7

(5+y)/2 =-11 y=-27

M(7,-27) is the other point

8 0

3 years ago

3 questions

frutty [35]

1) add 7 to both sides and get -5....so -5 >x
2) add 7 to both sides.....so get b<-5
3) add 17 to both side...so get z>1

6 0

3 years ago

Read 2 more answers

Help me plz I will give brainliest

Zarrin [17]

Answer:

Step-by-step explanation:

(35 - 20)/20

15/20 = .75

.75 * 100 == 75%

75%

6 0

2 years ago

Evaluate the integral. (remember to use absolute values where appropriate. Use c for the constant of integration.) 5 cot5(θ) sin

ozzi

$I=5\int \frac{cos^{4}\theta }{sin\theta }\times cos\theta d\theta \\\\I=5\int \left ( 1-sin^{2}\theta \right )^{2}\times \frac{cos\theta }{sin\theta }d\theta \\put\ \sin\theta =t\\\\dt=cos\theta d\theta \\\\I=5\int\frac{t^{4}+1-2t^{2}}{t}dt\ \ \ \ \ \ \ \ \ \ \because (a-b)^2=a^2+b^2-2ab\\\\I=5\left ( \int t^{3}dt + \int \frac{1}{t} -2\int t \right )dt$

by using the integration formula

we get,

$\\I=5\left ( \frac{t^{4}}{4} +logt -t^{2}\right )\\\\I=\frac{5}{4}t^{4}+5\log t-5t^{2}+c$

now put the value of t=\sin\theta in the above equation

we get,

$\int 5\cot^5\theta \sin^4\theta d\theta=\frac{5}{4}sin^{4}\theta+5\log \sin\theta - 5sin^{2} \theta+c$

hence proved

7 0

2 years ago

The position of an object moving along a straight line for

NikAS [45]

Answer:

at 2/3 seconds

Step-by-step explanation:

S1(t) = t³ + 2

Average speed, dS1/dt = 3t²

S2(t) = t²

Average speed, dS2/dt = 2t

The distance between the objects is

dS1/dt - dS2/dt

= 3t² - 2t

The time the distance between the two object is at minimum is when the distance is 0

That is, when

3t² - 2t = 0

t(3t - 2) = 0

t = 0 or 3t - 2 = 0

t = 0 or t = 2/3

3 0

3 years ago