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zepelin [54]
3 years ago
11

Solve the following system equation graphically on the set of axes below y= x+4 and y= -3/2 - 1 PLEASE HELP MEEE

Mathematics
1 answer:
gizmo_the_mogwai [7]3 years ago
8 0

Answer:

y=4

y= -1

Step-by-step explanation:

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What is an equation of the line that passes through the point (−2,−3) and is perpendicular to the line x+3y=24?
NemiM [27]

Answer:

y=3x+3

I hope this helped you

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3 years ago
Hannah completed 40 problems in 32 minutes.<br> I have to put it in a unit rate pls help
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Answer: 1.25 problems/min

32 minutes --> 40 problems

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PLEASE HELP ME!
lana [24]

Answer:

28.7 meters

Step-by-step explanation:

Draw a diagram.  Let's say d is the distance between the anchor and the bottom of the tower, h is the height of the tower, and x is the length of the wire.

Using trigonometry, we can write two equations for the height of the tower:

h = d tan 65°

h = (d + 25) tan 35°

Setting these equal, we can solve for d.

d tan 65° = (d + 25) tan 35°

d tan 65° = d tan 35° + 25 tan 35°

d (tan 65° − tan 35°) = 25 tan 35°

d = 25 tan 35° / (tan 65° − tan 35°)

d ≈ 12.12

Now we can find x:

cos 65° = d / x

x = d / cos 65°

x ≈ 28.7

The wire is approximately 28.7 meters long.

5 0
3 years ago
Can a triangle have sides with lengths 3 cm, 9 cm, and
sveticcg [70]
No because 3 to the 2nd power is 9 and 9 to the second power is 81 which both of them added up is 90 and can’t be square rooted to 10. Therefore no
3 0
3 years ago
Integrate dx/3sinx+4cosx
german

\displaystyle\int\frac{\mathrm dx}{3\sin x+4\cos x}

A standard approach would be the tangent half-angle substitution:

t=\tan\dfrac x2\implies\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx

Then

\sin x=2\sin\dfrac x2\cos\dfrac x2\implies\sin x=\dfrac{2t}{1+t^2}

\cos x=\cos^2\dfrac x2-\sin^2\dfrac x2\implies\cos x=\dfrac{1-t^2}{1+t^2}

from which we get

\mathrm dx=\dfrac2{1+t^2}\,\mathrm dt

So the integral becomes

\displaystyle\int\frac{\frac2{1+t^2}}{\frac{6t}{1+t^2}+\frac{4(1-t^2)}{1+t^2}}\,\mathrm dt=\int\frac{\mathrm dt}{3t+2(1-t^2)}=-\int\frac{\mathrm dt}{2t^2-3t-2}

Rewrite the denominator as

2t^2-3t-2=(2t+1)(t-2)

and expand the integrand into its partial fractions:

\dfrac1{2t^2-3t-2}=\dfrac15\left(\dfrac1{t-2}-\dfrac2{2t+1}\right)

We have

\displaystyle-\frac15\int\frac1{t-2}-\frac2{2t+1}\,\mathrm dt=-\frac15(\ln|t-2|-\ln|2t+1|)+C

=\dfrac15\ln\left|\dfrac{2t+1}{t-2}\right|+C

=\dfrac15\ln\left|\dfrac{2\tan\frac x2+1}{\tan\frac x2-2}\right|+C

6 0
3 years ago
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