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maria [59]
2 years ago
5

What is the equation in slope-intercept form of a line that passes through the points (5,3) (4,-3)

Mathematics
1 answer:
Darya [45]2 years ago
8 0

Answer:

y=6x-27

Step-by-step explanation:

To find the slope of a line given 2 points; use this equation.

\frac{y_2-y_1}{x_2-x_1} = \frac{rise}{run}

(5,3) (4,-3)

x_1   y_1    \\x_2  y_2\\\\

\frac{3 - (-3)}{4-5} =\frac{-6}{-1}  = 6

slope = 6

as for y-intercept use this method in the attachment

  below

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7.
patriot [66]

Answer:

C

Step-by-step explanation:

Square root of 92 is 9.591663047

so round that to 9.59

5 0
3 years ago
Given the function f(x) = -2x2 + 4x - 7, find f(-4).<br> -55<br> -7<br> 25
Neko [114]

Answer:

f(-4)= -55

Step-by-step explanation:

We want to find f(x), or y, when x is equal to -4.

Therefore, we can substitute -4 in for x in the function.

f(x) = -2x^2 + 4x - 7

Substitute -4 in for x

f(-4) = -2(-4^2) + 4(-4) - 7

Evaluate the exponent

f(-4)=-2(16)+4(-4)-7

Multiply -2 and 16

f(-4)= -32+4(-4)-7

Multiply 4 and -4

f(-4)=-32-16-7

Subtract

f(-4)=-55

7 0
2 years ago
Read 2 more answers
Explain how you could find the difference of 1 and 0.1978.​
weqwewe [10]
You would just subtract .1978 from 1
8 0
3 years ago
Can someone help me? it’s due in a few hours
ioda

Answer:

cars value decreases in value exponentially

Step-by-step explanation:

cant really read it but say that the cars value decreases in value exponentially and compare that to however the other cars decrease in value over time

4 0
2 years ago
The line integral of (2x+9z) ds where the curve is given by the parametric equations x=t, y=t^2, z=t^3 for t between 0 and 1. Pl
Naya [18.7K]
Let r = (t,t^2,t^3)

Then r' = (1, 2t, 3t^2)

General Line integral is:
\int_a^b f(r) |r'| dt

The limits are 0 to 1
f(r) = 2x + 9z = 2t +9t^3
|r'| is magnitude of derivative vector \sqrt{(x')^2 + (y')^2 + (z')^2}

\int_0^1 (2t+9t^3) \sqrt{1+4t^2 +9t^4} dt

Fortunately, this simplifies nicely with a 'u' substitution.

Let u = 1+4t^2 +9t^4

du = 8t + 36t^3  dt

\int_0^1 \frac{2t+9t^3}{8t+36t^3} \sqrt{u}  du \\  \\ \int_0^1 \frac{2t+9t^3}{4(2t+9t^3)} \sqrt{u}  du \\  \\  \frac{1}{4} \int_0^1 \sqrt{u}  du

After integrating using power rule, replace 'u' with function for 't' and evaluate limits:
=\frac{1}{4} |_0^1 (\frac{2}{3}) (1+4t^2 +9t^4)^{3/2} \\  \\ =\frac{1}{6} (14^{3/2} - 1)
7 0
2 years ago
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