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3 years ago

14

Chad will drive 672 more miles . He continues to drives at the same rate . How many hours will it take Chad to drive the 672 mil

es?

1 answer:

tekilochka [14]3 years ago

8 0

12 hours, you can use proportions, miles/hours. 56/1=672/x. x=12.

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How many ways can a dance instructor send 4 of her 11 students to a summer dance program?

Sedaia [141]

Answer:

d) 330

Step-by-step explanation:

The number of ways can a dance instructor send 4 of her 11 students to summer dance program is C(11 , 4)

That is $11_{C_{4} }$ ways

by using formula $n_{C_{r} } = \frac{n!}{(n-r)!r!}$

now simplification

$11_{C_{4} } = \frac{11!}{(11-4)!4!}$

= $\frac{11X10X9X8X7!}{(7!)4X3X2X1}$

after cancelling 7! and on simplification

= $\frac{11X10X9X8}{4X3X2X1}$

after cancellation we get 330 ways

The number of ways can a dance instructor send 4 of her 11 students to summer dance program is 330

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3 years ago

Find the slope of the line through the pair of points: (14, 10) (14, -3)

vivado [14]

Answer:

Slope is zero.

Step-by-step explanation:

Use <u>y2-y1</u>

x2-x1

to find slope.

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3 years ago

Find the values for a and b that would make the equality true.

Crank

Answer:

a = - 4, b = 5

Step-by-step explanation:

Expand the left side and compare the coefficients of like terms on the right side.

- 3(2x² + ax + b)

= - 6x² - 3ax - 3b

Comparing like terms with - 6x² + 12x - 15

x - term → - 3a = 12 ( divide both sides by - 3 )

a = - 4

constant term → - 3b = - 15 ( divide both sides by - 3 )

b = 5

6 0

3 years ago

t takes Phillip 7 hours to proof a chapter of Hawkes Learning Systems' Intermediate Algebra book and it takes Kim 3 hours. How l

Leya [2.2K]

Answer:

2.1 hrs

Step-by-step explanation:

In 1 hr, Phillip would have proofed 1/7 of the chapter. In 1 hr, Kim would have proofed 1/3 of the chapter. So, together in 1 hr, they would have proofed 10/21 of the chapter. So, it would take 21/10 or 2.1 hrs to proof the whole chapter together.

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3 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