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3 years ago

WILL GIVE BRAINLIEST, SUPERB RATING, AND THANKS!

Mathematics

1 answer:

Vaselesa [24]3 years ago

8 0

A line on this coordinate plane is horizontal when every input (x) has the same output y).

This function always gives the same output with no matter what input you put it in.

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Use the figure for exercises 1-4 Name the intersection of plane S and line j

torisob [31]

Right angle i think

90 degree angle is right angle so jus Measure it

6 0

3 years ago

find the equation of a cubic function whose graph passes through points (3,0) and (1,4) and is tangent to x-axis at the origin

Tomtit [17]

Answer:

y = -2x^2(x - 3)

Step-by-step explanation:

Preliminary Remark

If a cubic is tangent to the x axis at 0,0

Then the equation must be related to y = a*x^2(x - h)

(3,0)

If the cubic goes through the point (3,0), then the equation will become

0 = a*3^2(3 - h)

0 = 9a (3 - h)

0 = 27a - 9ah

from which h = 3

From the second point, we get

4 = ax^2(x - 3)

4 = a(1)^2(1 - 3)

4 = a(-2)

a = 4 / - 2

a = -2

Answer

y = -2x^2(x - 3)

3 0

2 years ago

Find A and B if the graph of Ax + By = 12 passes through (2,1) and is parallel to the graph of 2x - 7y = 3

Ivenika [448]

Answer:

24-21

Step-by-step explanation:

2×12=24 7×3=21 I started from 2×2 towards 2×12 when I got the answer and I started 7×1 towards 7×3

4 0

2 years ago

1. Write the standard form of the line that passes through the given points. (7, -3) and (4, -8)

Sveta_85 [38]

Answer:

1. $-5x+3y+44=0$

2. $2x+y-2=0$

3. $2x+y-4=0$

Step-by-step explanation:

Standard form of a line is $Ax+By+C=0$ .

If a line passing through two points then the equation of line is

$y-y_1=m(x-x_1)$

where, m is slope, i.e., $m=\dfrac{y_2-y_1}{x_2-x_1}$ .

The line passes through the points (7,-3) and (4,-8). So, the equation of line is

$y-(-3)=\dfrac{-8-(-3)}{4-7}(x-7)$

$y+3=\dfrac{-5}{-3}(x-7)$

$y+3=\dfrac{5}{3}(x-7)$

$3(y+3)=5(x-7)$

$3y+9=5x-35$

$-5x+3y+9+35=0$

$-5x+3y+44=0$

Therefore, the required equation is $-5x+3y+44=0$ .

We need to find the equation of the line, in standard form, that has a y-intercept of 2 and is parallel to $2 x + y =-5$ .

Slope of the line : $m=\dfrac{-\text{Coefficient of x}}{\text{Coefficient of y}}=\dfrac{-2}{1}=-2$

Slope of parallel lines are equal. So, the slope of required line is -2 and it passes through the point (0,2).

Equation of line is

$y-2=-2(x-0)$

$y-2=-2x$

$2x+y-2=0$

Therefore, the required equation is $2x+y-2=0$ .

We need to find the equation of the line, in standard form, that has an x-intercept of 2 and is parallel to $2x + y =-5$ .

From part 2, the slope of this line is -2. So, slope of required line is -2 and it passes through the point (2,0).

Equation of line is

$y-0=-2(x-2)$

$y=-2x+4$

$2x+y-4=0$

Therefore, the required equation is $2x+y-4=0$ .

5 0

2 years ago

Find the values of the variables in this kite.

zavuch27 [327]

Answer:

x=7, y=16

Step-by-step explanation:

2x+5=x+12

2x=x+7

x=7

y-4=x+5

y-4=12

y=16

4 0

3 years ago