Answer this question please

Suppose a parabola has an axis of symmetry at x = -8 , a maximum height of 2, and passes through the point (-7, -1). Write the e

lidiya [134]

As it has a maximum value the coefficient of x^2 will be negative

The vertex will be at (-8,2) so in vertex form it is

y = a(x + 8)^2 + 2
and as it passes through (-7,-1) we have:

-1 = a(-7+8)^2 + 2

-1 = a + 2 so a = -3

answer is y = -3(x + 8)^2 + 2

in standard form this is y = -3x^2 -48x - 190

3 0

3 years ago

Solve and graph -4 + 2t - 14 - 18t > -6 - 100t

irina [24]

Okay so first u just add the terms on each side
2t-18>-6-100t
2(t-9)>2(-3-50t)
t-9>-3-50t
51t>6
t>6/51
t>2/17
so just take a number line put an empty circle like this : O on 2/17 and draw a line showing it continues to the right infinitely. Sorry if my math is wrong lol i am functioning on one hour of sleep.

7 0

3 years ago

Select all the equations where b= 11 is a solution. Choose 2 answers:

Allushta [10]

Answer:

what do you mean there is no pictures

5 0

3 years ago

Two trains leave a town at the same time heading in opposite directions. One train is traveling 12 mph faster than the other. Af

Flauer [41]

Answer:

Two trains leave a town at the same time heading in opposite directions. One train is traveling 12 mph faster than the other. After two hours, they are 232 miles apart. What is the average speed of each train?

52 mph; 64 mph

55 mph; 70 mph

92 mph; 104 mph

98 mph; 110 mph

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

What is true of the graph of two lines 3y-8=-5x and 6y=-10x+16

Murljashka [212]

Answer:

Both lines are equal (they are the same)

Step-by-step explanation:

Given

$3y - 8 = -5x$

$6y = -10x + 16$

Required

What is true about graph of both lines

Questions like this are better solved when there's option(s) to select from. However, some of the properties of line equation that I'll consider are to check if both lines are either parallel or perpendicular

To do this,

The first thing to do is to calculate the slope of both lines

$3y - 8 = -5x$

Add 8 to both sides

$3y - 8 + 8 = -5x + 8$

$3y = -5x + 8$

Divide both sided by 3

$\frac{3y}{3} = -\frac{5x}{3} + \frac{8}{3}$

$y = -\frac{5x}{3} + \frac{8}{3}$

The slope of the line is the coefficient of x;

$Slope = -\frac{5}{3}$

Solve for the y intercept; Let x = 0

$y = -\frac{5 * 0}{3} + \frac{8}{3}$

$y = 0 + \frac{8}{3}$

$y = \frac{8}{3}$

Solve for the x intercept; Let y = 0

$0 = -\frac{5x}{3} + \frac{8}{3}$

Subtract $\frac{8}{3}$ from both sides

$0 - \frac{8}{3} = -\frac{5x}{3} + \frac{8}{3} - \frac{8}{3}$

$- \frac{8}{3} = -\frac{5x}{3}$

Subtract both sides by $-\frac{3}{5}$

$-\frac{3}{5}*- \frac{8}{3} = -\frac{5x}{3} * -\frac{3}{5}$

$-\frac{3}{5}*- \frac{8}{3} = x$

$\frac{3}{5} * \frac{8}{3} = x$

$\frac{8}{5} = x$

$x = \frac{8}{5}$

------------------------------------------------------------------------------------------------------

$6y = -10x + 16$

Divide both sides by 6

$\frac{6y}{6} = -\frac{10x}{6} + \frac{16}{6}$

$y = -\frac{10x}{6} + \frac{16}{6}$

Simplify fractions to lowest term

$y = -\frac{5x}{3} + \frac{8}{3}$

The slope of the line is the coefficient of x;

$Slope = -\frac{5}{3}$

Solve for the y intercept; Let x = 0

$y = -\frac{5 * 0}{3} + \frac{8}{3}$

$y = 0 + \frac{8}{3}$

$y = \frac{8}{3}$

Solve for the x intercept; Let y = 0

$0 = -\frac{5x}{3} + \frac{8}{3}$

Subtract $\frac{8}{3}$ from both sides

$0 - \frac{8}{3} = -\frac{5x}{3} + \frac{8}{3} - \frac{8}{3}$

$- \frac{8}{3} = -\frac{5x}{3}$

Subtract both sides by $-\frac{3}{5}$

$-\frac{3}{5}*- \frac{8}{3} = -\frac{5x}{3} * -\frac{3}{5}$

$-\frac{3}{5}*- \frac{8}{3} = x$

$\frac{3}{5} * \frac{8}{3} = x$

$\frac{8}{5} = x$

$x = \frac{8}{5}$

-------------------------------------------------------------------------------------------------------

By comparing the slope, x intercept and y intercept of both lines;

It'll be observed that they have the same slope, x intercept and y intercept

This implies that both lines are equal; in other words, they are the same.

6 0

3 years ago