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Over [174]
1 year ago
7

• Write the equation of the line that passes through the points (-7,0) and(-3,-9). Put your answer in fully reduced point-slope

form.
Mathematics
1 answer:
coldgirl [10]1 year ago
3 0

We want to calculate the line that passes through the points (-7,0) and (-3,-9). Recall that the equation of a line is

y=mx+b

where m is the slope and b is the y intercept. We can calculate first the slope and then find the value of b. To do so, recall that given points (a,b) and (c,d) the slope of the line that passes through them is given by the formula

m=\frac{(d\text{ -b)}}{(c\text{ -a)}}=\frac{b\text{ -d}}{a\text{ -c}}

so by taking a=-7, b=0, c=-3 and d=-9 we get

m=\frac{0\text{ - (-9)}}{\text{ -7 -( -3)}}=\frac{9}{\text{ -4}}=\text{ -}\frac{9}{4}

so our equation becomes

y=\text{ -}\frac{9}{4}x+b

so know we want this line to pass through the point (-7,0), so whenever x= -7 then y=0 so we have the equation

0=\text{ -}\frac{9}{4}(\text{ -7)+b}

or equivalently

0=\frac{63}{4}+b

so if we subtract 63/4 on both sides, we get

b=\text{ -}\frac{63}{4}

so our equation would be

y=\text{ -}\frac{9}{4}x\text{ -}\frac{63}{4}

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Answer:

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s = 18

y = 9

f = 19

Step-by-step explanation:

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2 years ago
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Which of the following options best describes the function graphed below? A graph shows a curve that starts at the origin and go
Tamiku [17]
It's an increasing nonlinear. Exponential functions are graphed in basically any shape besides a line. This is an exponential function, therefore it is classified as nonlinear. By looking at the slope, from left to right, we can see that it is increasing, thus providing an answer that this is an increasing nonlinear function.
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3 years ago
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Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y
irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

y'(x)= x^2y(x)-\frac12y^2(x)

Putting the value of y'(x) in equation (1)

y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))

Substituting x =0 and h= 0.2

y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]

\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]    [∵ y(0) =3 ]

\Rightarrow y(0.2)\approx 2.7

Substituting x =0.2 and h= 0.2

y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]

\Rightarrow y(0.4)\approx  2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]

\Rightarrow y(0.4)\approx 1.9926

Substituting x =0.4 and h= 0.2

y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]

\Rightarrow y(0.6)\approx  1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]

\Rightarrow y(0.6)\approx 1.6593

Substituting x =0.6 and h= 0.2

y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]

\Rightarrow y(0.8)\approx  1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]

\Rightarrow y(0.6)\approx 0.8800

Substituting x =0.8 and h= 0.2

y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]

\Rightarrow y(1.0)\approx  0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]

\Rightarrow y(1.0)\approx 0.9152

Therefore the value of y(1)= 0.9152.

4 0
3 years ago
What is the slope of this line of best fit?
Bas_tet [7]

Answer:

The slope of the line of best fit is \frac{-6}{7}  ⇒ 2nd option

Step-by-step explanation:

The formula of the slope of a line is m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

<em>To find the slope of the best line fit choose two points their positions make the number of the points over the line equal to the number of the points below the line</em>

From the attached graph points (1 , 9) and (8 , 3) are the best choice

∵ The line passes through points (1 , 9) and (8 , 3)

∴ x_{1} = 1 and x_{2} = 8

∴ y_{1} = 9 and y_{2} = 3

- Substitute them in the formula of the slope

∴ m=\frac{3-9}{8-1}=\frac{-6}{7}

∴ The slope of the line of best fit is \frac{-6}{7}

6 0
3 years ago
What’s the area for 16 and 8 3/4
Alborosie

Answer:

Area = 140

16 x 8.75 = 140

Step-by-step explanation:

6 0
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