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Vinvika [58]
3 years ago
15

Use the partial quotient to solve 673÷5

Mathematics
1 answer:
AleksAgata [21]3 years ago
8 0
Our answer would be 134.6
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Shelly compared the number of oak trees to the number of maple trees as a part of a study about hardwood trees in a woodlot. She
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There were 72 more maple trees before the bug problem then after because there was 108 maple trees before the bug problem and 36 maple trees after the bug problem
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Use the graph for problems 14 and 15<br><br>​
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9514 1404 393

Answer:

  14.  C

  15.  C

Step-by-step explanation:

14. The function is entirely in quadrants I and II, so the leading coefficient is positive. This eliminates choices A and B.

The horizontal asymptote is 0, not -1, eliminating choice D.

The curve is best described by the equation of choice C.

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15. The domain and range of an unadulterated exponential function are ...

  domain: all real numbers; range: y > 0 . . . . matches choice C

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3 years ago
Help me pleaseeee. due todayyy
zalisa [80]

Answer:

1.5

Step-by-step explanation:

The scale factor can be found by dividing the side length from triangle ABC by the corresponding side length from triangle ABC

so the scale factor would be equal to

12/8 and 9/6

9/6 = 1.5

12/8 =1.5

so we can conclude that the scale factor is 1.5

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3 years ago
Compare and contrast the effects that a proportional dimension change has on perimeter and area of figures
Kipish [7]
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Read 2 more answers
Given the following vector field and oriented curve​ C, evaluate ModifyingBelow Integral from nothing to nothing With Upper C Bo
sladkih [1.3K]

Answer:

The value of the line integral is \frac{157}{2}.

Step-by-step explanation:

Given a path C, with parametrization r(t) for t_0\leq t \leq t_1, we have that

\int_{C}F dr = \int_{t_0}^{t_1} F(r(t)) \cdot r'(t) dt where \cdot is the dot product between two vectors.

In our case, we have r(t) = (6t,11t^2), 0\leq t\leq 1. Then r'(t) = (6,22t). In this case we have that F(x,y) = (x,y). Then,F(r(t)) = (6t, 11t^2)

So

\int_{C}F dr = \int_{0}^{1} (6t,11t^2)\cdot(6,22t) dt = \int_{0}^{1} 36t+11\cdot 22 t^3= \left.(36\frac{t^2}{2}+11\cdot 22 \frac{t^4}{4})\right|_{0}^{1} = \frac{36}{2}+\frac{11\cdot 22}{4}= \frac{157}{2}

4 0
3 years ago
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