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Katena32 [7]
2 years ago
11

What is the answer? pls

Mathematics
1 answer:
Alex777 [14]2 years ago
4 0

Answer:

C + D

\frac{3}{4} (x + 8 )= -12

\frac{3}{4} x +6 = -12

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18x6 = 108
110-108 =2
By weeks 6 she will only have $2 left
So being realistic, at week 7 she will have spent all her money.
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Need help with this pls asap
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5 0
2 years ago
Which of the following are point slope equations of the line going through (-3, -5) and (2, 4). Help is appreciated!​
ki77a [65]

Answer:

A and F

Step-by-step explanation:

first figure out the slope.

4-(-5)/2-(-3)= 9/5

slope: 9/5

y-y1=m(x-x1)

you substitute either coordinate for the second variables. we automatically take out any slopes that have a negative because the slope is positive.

C, D, E all have negative slopes so it is wrong.

B does not work because a negative times a negative is a positive. the equation shows it to be negative so it is wrong.

A & F fit the solution.

8 0
3 years ago
Y=2x-1<br><br> y=-3x+14<br><br> Solve for y
LUCKY_DIMON [66]
Y=2x-1
y=-3x+14

First you would substitue one of the y's for the other so
2x-1=-3x+14

no you would solve for x...
add one on both sides
2x=-3x+15

now add 3x to both sides which becomes
5x=15

divide 5 on both sides which gives you x=3

now to solve for y plug into any one of the equations(doesn't matter which one) from before 3 for x and solve
y=2(3)-1
y=6-1
y=5

And that is you answer y=5
3 0
3 years ago
Find parametric equations for the path of a particle that moves along the circle x2 + (y − 1)2 = 16 in the manner described. (En
ArbitrLikvidat [17]

Answer:

a) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t, b) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t, c) x = 4\cdot \cos \left(t+\frac{\pi}{2}  \right), y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right).

Step-by-step explanation:

The equation of the circle is:

x^{2} + (y-1)^{2} = 16

After some algebraic and trigonometric handling:

\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = 1

\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = \cos^{2} t + \sin^{2} t

Where:

\frac{x}{4} = \cos t

\frac{y-1}{4} = \sin t

Finally,

x = 4\cdot \cos t

y = 1 + 4\cdot \sin t

a) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t.

b) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t.

c) x = 4\cdot \cos t'', y = 1 + 4\cdot \sin t''

Where:

4\cdot \cos t' = 0

1 + 4\cdot \sin t' = 5

The solution is t' = \frac{\pi}{2}

The parametric equations are:

x = 4\cdot \cos \left(t+\frac{\pi}{2}  \right)

y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)

7 0
3 years ago
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