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RoseWind [281]
3 years ago
11

write an equation in point-slope form of the line having the given slope that contains the given point. m=−0.8, (14.5,12.8) sele

ct one: a. y−14.5=−0.8(x−12.8) b. y−12.8=−0.8(x−14.5) c. y=−0.8x−1.2 d. y 12.8=−0.8(x−14.5)
Mathematics
2 answers:
NARA [144]3 years ago
7 0
Y - y₁ = m(x -x₁)

slope m = -0.8,  from point (14.5, 12.8) ,  x₁ = 14.5, y₁ = 12.8

 y - y₁ = m(x -x<span>₁)

</span>y - 12.8 = -0.8(x - 14.5<span>)
</span>
<span>Option B </span>
Ganezh [65]3 years ago
4 0

Answer: b. (y-12.8)=-0.8(x-14.5)

Step-by-step explanation:

We know that the equation of a line in point slope form with slope m and passing through point (a,b) is given by :-

(y-b)=m(x-a)

Given : The slope of the line : m= -0.8

The point from which line is passing : (a,b) =(14.5,12.8)

Then , the equation in point slope form of the line having the given slope that contains the given point. m=−0.8, (14.5,12.8)  :-

(y-12.8)=-0.8(x-14.5)

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If f(x) = –x + 2, find f(4).
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Answer:

-2

Step-by-step explanation:

If you substitute x for 4, you have -4+2 to get -2.

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​ g(x)=−20−3x h(x)=( 2 1 ​ ) x ​ Evaluate. (g\circ h) (-2)=(g∘h)(−2)=left parenthesis, g, circle, h, right parenthesis, left par
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3 years ago
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Which expressions are equivalent to 4(x+1) + 7(x+3)? Select two answers.
Natalija [7]

Answer:

9(x+3) + 2(x+3) or 10(x+2) +(x+5)

Step-by-step explanation:

First you want to distribute the equation.

4(x+1) + 7(x+3) multiplied out is

4x + 4 + 7x + 21

Now we add like terms so it comes out to

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7 0
3 years ago
When a breeding group of animals is introduced into a restricted area such as a wildlife reserve, the population can be expected
jasenka [17]

Answer:

A. Initially, there were 12 deer.

B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. After 15 years, there will be 410 deer.

D. The deer population incresed by 30 specimens.

Step-by-step explanation:

N=\frac{12.36}{0.03+0.55^t}

The amount of deer that were initally in the reserve corresponds to the value of N when t=0

N=\frac{12.36}{0.33+0.55^0}

N=\frac{12.36}{0.03+1} =\frac{12.36}{1.03} = 12

A. Initially, there were 12 deer.

B. N(10)=\frac{12.36}{0.03 + 0.55^t} =\frac{12.36}{0.03 + 0.0025}=\frac{12.36}{y}=380

B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. N(15)=\frac{12.36}{0.03+0.55^15}=\frac{12.36}{0.03 + 0.00013}=\frac{12.36}{0.03013}= 410

C. After 15 years, there will be 410 deer.

D. The variation on the amount of deer from the 10th year to the 15th year is given by the next expression:

ΔN=N(15)-N(10)

ΔN=410 deer - 380 deer

ΔN= 30 deer.

D. The deer population incresed by 30 specimens.

8 0
3 years ago
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