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BlackZzzverrR [31]
3 years ago
14

−6x+2x−5+2: simplified

Mathematics
1 answer:
Anestetic [448]3 years ago
3 0

Answer:

-4x-3

Step-by-step explanation:

all you do is combine like terms: -6x+2x=-4x and -5+2=-3

so the answer would be -4x-3

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MRS.Bonnette's recipe for banana bread calls for 3 and one fourths cups of sugar. She wants to make 6.5 batches so that she has
densk [106]

Answer:

21/ \frac {1}{8}

Step-by-step explanation:

Since we're given that a batch requires 3 and one fourths cups of sugar, this is written as

3/ \frac {1}{4}

If she wants 6.5 batches then we just cross multiply

1 batch=3/ \frac {1}{4}

6.5 batches=x

X=6.5\times 3/ \frac {1}{4}

X=6/ \frac {1}{2}\times 3/ \frac {1}{4}

X=\frac {13}{2}\times \frac {13}{4}

X=\frac {169}{8}

X=21/ \frac {1}{8}

6 0
3 years ago
Find the height of the triangle in which A=192 cm^2
romanna [79]

Step-by-step explanation:

one second 12345678910

6 0
2 years ago
7s0t-5/2-1m2 simplify
Mars2501 [29]
<span>The answer should be 14t - 2m2 - 5/2 , if I am correct</span>
6 0
3 years ago
Can someone help with this math question please
katrin2010 [14]
The answer is b

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5 0
3 years ago
Read 2 more answers
Linda goes water-skiing one sunny afternoon. After skiing for 15 min, she signals to the driver of the boat to take her back to
trasher [3.6K]
60 = a * (-30)^2
a = 1/15
So y = (1/15)x^2


abc)
The derivative of this function is 2x/15. This is the slope of a tangent at that point.
If Linda lets go at some point along the parabola with coordinates (t, t^2 / 15), then she will travel along a line that was TANGENT to the parabola at that point.
Since that line has slope 2t/15, we can determine equation of line using point-slope formula:
y = m(x-x0) + y0
y = 2t/15 * (x - t) + (1/15)t^2
Plug in the x-coordinate "t" that was given for any point.


d)
We are looking for some x-coordinate "t" of a point on the parabola that holds the tangent line that passes through the dock at point (30, 30).
So, use our equation for a general tangent picked at point (t, t^2 / 15):
y = 2t/15 * (x - t) + (1/15)t^2
And plug in the condition that it must satisfy x=30, y=30.
30 = 2t/15 * (30 - t) + (1/15)t^2
t = 30 ± 2√15 = 8.79 or 51.21
The larger solution does in fact work for a tangent that passes through the dock, but it's not important for us because she would have to travel in reverse to get to the dock from that point.
So the only solution is she needs to let go x = 8.79 m east and y = 5.15 m north of the vertex.
4 0
3 years ago
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