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3 years ago

1. write an equation in point slope form that passes through the point (2,3) and slope of -4

1 answer:

4 0

For number 1, The equation for point slope form is y-y1=m(x-x1). This is called point-slope form because it uses one point, and the lines slope. We plug in the slope for m, and the point for (x1, y1), and then we have our answer for number 1! This is y-3=-4(x-2), in point-slope form. Slope intercept form is different however, requiring the slope of the line, and it's intercept. It is in the form y = mx+b. In the scenario by inputting the variables we already have we can solve for b very quickly. Using our point and slope we get 3=-8+b, or b = 11. This means that the equation in point slope form is y=-4x+11.

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Is 43 odd or even?but the answer don't have to be great i just need a answer

cupoosta [38]

43 is an odd number.

4 0

4 years ago

Write the difference of the following polynomials in standard form: (11-8y+2y^2) - (y^2 -15). Classify the difference by its deg

Mandarinka [93]

11-8y+2y^2-y^2+15
y^2-8y+15

7 0

3 years ago

Find the other endpoint given endpoint (-10,7) midpoint (6,2)

Makovka662 [10]

Answer:

(22;-3)

Step-by-step explanation:

if to calculate the coordinates of the given midpoint it needed to use:

$x_{midpoint}=\frac{x_{endpoint1}+x_{endpoint2}}{2};$

$y_{midpoint}=\frac{y_{endpoint1}+y_{endpoint2}}{2};$

then to calculate the coordinates of endpoint2, it needed to perform the substitution:

$\frac{-10+x_{endpoint2}}{2}=6,$

$2=\frac{7+y_{endpoint2}}{2},$

finally:

$x_{endpoint2}=22;$

$y_{endpoint2}=-3.$

3 0

3 years ago

Which point on the graph of g(x)=(1/5)^x? HELPP

Cloud [144]

Answer:

(-1,5) and $(3, \frac{1}{125})$ are points on the graph

Step-by-step explanation:

Given

$g(x) = \frac{1}{5}^x$

Required

Determine which point in on the graph

To get which of point A to D is on the graph, we have to plug in their values in the given expression using the format; (x,g(x))

A. (-1,5)

x = -1

Substitute -1 for x in $g(x) = \frac{1}{5}^x$

$g(x) = \frac{1}{5}^{-1}$

Convert to index form

$g(x) = 1/(\frac{1}{5})$

Change / to *

$g(x) = 1*(\frac{5}{1})$

$g(x) = 5$

This satisfies (-1,5)

Hence, (-1,5) is on the graph

B. (1,0)

x = 1

Substitute 1 for x

$g(x) = \frac{1}{5}^x$

$g(x) = \frac{1}{5}^1$

$g(x) = \frac{1}{5}$

(1,0) is not on the graph because g(x) is not equal to 0

C. $(3, \frac{1}{125})$

x = 3

Substitute 3 for x

$g(x) = \frac{1}{5}^x$

$g(x) = \frac{1}{5}^3$

Apply law of indices

$g(x) = \frac{1}{5} * \frac{1}{5} * \frac{1}{5}$

$g(x) = \frac{1}{125}$

This satisfies $(3, \frac{1}{125})$

Hence, $(3, \frac{1}{125})$ is on the graph

D. $(-2, \frac{1}{25})$

x = -2

Substitute -2 for x

$g(x) = \frac{1}{5}^x$

$g(x) = \frac{1}{5}^{-2}$

Convert to index form

$g(x) = 1/(\frac{1}{5}^2)$

$g(x) = 1/(\frac{1}{5}*\frac{1}{5})$

$g(x) = 1/(\frac{1}{25})$

Change / to *

$g(x) = 1*(\frac{25}{1})$

$g(x) = 25$

This does not satisfy $(-2, \frac{1}{25})$

Hence, $(-2, \frac{1}{25})$ is not on the graph

8 0

4 years ago

In ACE, G is the centroid and BE = 9. find BG and GE

rjkz [21]

Answer:

b-g is 3

g-e is 6

Step-by-step explanation:

good luck, have a good day, can i get brainliest

4 0

3 years ago