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navik [9.2K]
3 years ago
14

Rupesh is mowing grass to save money for a vacation. He charges $12 per yard. Rupesh already has $40 and wants

Mathematics
1 answer:
erik [133]3 years ago
6 0

Answer:

9

Step-by-step explanation:

1. has 40 dollars do 148-40

2.He needs 108

3.108/12

SOLVED YOU GET 9

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A town's population is 14,280, and it is increasing by 160 people every year. A nearby town has a population of 24,000, and it i
fgiga [73]

Answer:

27 years

Step-by-step explanation:

Given that :

Town A:

Initial population = 14280

Rate = increase by 160 per year

Let t = time in years

Population is thus :

14280 + 160t - - - (1)

Town B:

Initial population = 24000

Rate = decrease by 200 per year

P = 24000 - 200t - - - (2)

Equate both equations :

14280 + 160t = 24000 - 200t

160t + 200t = 24000 - 14280

360t = 9720

t = 9720 / 360

t = 27

In 27 years

7 0
2 years ago
Evaluate the definite integral using the graph of f(x)<br> (Image included)
Tanya [424]

a) The first integral corresponds to the area under y = f(x) on the interval [0, 3], which is a right triangle with base 3 and height 5, hence the integral is

\displaystyle \int_0^3 f(x) \, dx = \frac12 \times 3 \times 5 = \boxed{\frac{15}2}

b) The integral is zero since the areas under the curve over [3, 4] and [4, 5] are equal but opposite in sign. In other words, on the interval [3, 5], f(x) is symmetric and odd about x = 4, so

\displaystyle \int_3^5 f(x) \, dx = \int_3^4 f(x) \, dx + \int_4^5 f(x) \, dx = \int_3^4 f(x) \, dx - \int_3^4 f(x) \, dx = \boxed{0}

c) The integral over [5, 9] is the negative of the area of a rectangle with length 9 - 5 = 4 and height 5, so

\displaystyle \int_5^9 f(x) \, dx = -4\times5 = -20

Then by linearity, we have

\displaystyle \int_0^9 f(x) \, dx = \left\{\int_0^3 + \int_3^5 + \int_5^9\right\} f(x) \, dx = \frac{15}2 + 0 - 20 = \boxed{-\frac{25}2}

8 0
2 years ago
Find the three geometric means between 15 and 1215.
irina [24]
Hello,

g_m(15,1215)=√(5*1215)=9√15≈77,942286340....
====================================
6 0
2 years ago
Suppose
Marrrta [24]
F(3) = t4(3) = 2

The value of the function at the point of expansion is the first (constant) term of the Taylor series.
6 0
3 years ago
Find the arc measure of AB.<br> The measure of arc AB = ______ degrees
mariarad [96]
The picture is difficult to read, but it looks like
  ∠ADB = 2x²
  ∠ACB = 10x
These two angles have equal measure, so you have
  2x² = 10x
  2x(x -5) = 0 . . . . . subtract 10x and factor
  x = 0, or x = 5

The measure of arc AB is twice the measure of either inscribed angle, so would be
  arc AB = 2×10x
  = 2×10×5
  = 100 . . . . . degrees

Arc AB is 100°.
7 0
2 years ago
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