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4 years ago

Can someone help with this question?

Mathematics

1 answer:

Nina [5.8K]4 years ago

8 0

Answer:

1. 5

2. -3

3. -4

4. 0

Step-by-step explanation:

Substitute the value of x in the equation with each of the x's given in the table.

Example:

When x = -1

y = -1^2 - 4 x -1 = 1 - 4 = -3

When x = 4

y = 4^2 - 4 x 4 = 16 - 16 = 0

Please add brainiest

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Are the two lines parallel, perpendicular, or neither? Explain your answer. Show work needed to answer the question.

tatuchka [14]

Answer:

The lines are perpendicular

Step-by-step explanation:

The correct question is

Are the two lines parallel, perpendicular, or neither? Explain your answer

3x + 7y = 15

7x − 3y = 6

we know that

If two lines are parallel, then their slopes are equal

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of the slopes is equal to -1)

Remember that

The equation of the line in slope intercept form is equal to

$y=mx+b$

where m is the slope

b is the y-intercept

we have

$3x + 7y = 15$ ----> equation A

isolate the variable y and convert the equation in slope intercept form

$y=\frac{15-3x}{7}$

$y=\frac{15}{7}-\frac{3}{7}x$

The slope of the line A is $m_A=-\frac{3}{7}$

$7x-3y = 6$ ---> equation B

isolate the variable y and convert the equation in slope intercept form

$y=\frac{7x-6}{3}$

$y=\frac{7}{3}x-2$

The slope of the line B is $m_B=\frac{7}{3}$

Compare the slopes

$m_A\neq m_B$ ---> the lines are not parallel

$m_A* m_B=(-\frac{3}{7})*(\frac{7}{3})=-1$

The slopes are opposite reciprocal

therefore

The lines are perpendicular

8 0

3 years ago

Solve all: 2/3x+15=7 3x+8-x=7 4(2x-6)=2

Sonja [21]

Answer:

-1.3333

-1/2

3.25

Step-by-step explanation:

Ion need an explanation

8 0

3 years ago

A toy cannon ball is launched from a cannon on top of a platform. The equation h(t) =- 5<img src="https://tex.z-dn.net/?f=t%5E%7

DanielleElmas [232]

Answer:

Part A)

Part B)

About 2.9362 seconds.

Step-by-step explanation:

The equation $\displaystyle h(t)=-5t^2+14t+2$ models the height h in meters of the ball t seconds after its launch.

Part A)

To determine whether or not the ball reaches a height of 14 meters, we can find the vertex of our function.

Remember that the vertex marks the maximum value of the quadratic (since our quadratic curves down).

If our vertex is greater than 14, then, at some time t, the ball will definitely reach a height of 14 meters.

However, if our vertex is less than 14, then the ball doesn’t reach a height of 14 meters since it can’t go higher than the vertex.

So, let’s find our vertex. The formula for vertex is given by:

$\displaystyle (-\frac{b}{2a},h(-\frac{b}{2a}))$

Our quadratic is:

$\displaystyle h(t)=-5t^2+14t+2$

Hence: a=-5, b=14, and c=2.

Therefore, the x-coordinate of our vertex is:

$\displaystyle x=-\frac{14}{2(-5)}=\frac{14}{10}=\frac{7}{5}$

To find the y-coordinate and the maximum height, we will substitute this value back in for x and evaluate. Hence:

$\displaystyle h(\frac{7}{5})=-5(\frac{7}{5})^2+14(\frac{7}{5})+2$

Evaluate:

$\displaystyle \begin{aligned} h(\frac{7}{2})&=-5(\frac{49}{25})+\frac{98}{5}+2 \\ &=\frac{-245}{25}+\frac{98}{5}+2\\ &=\frac{-245}{25}+\frac{490}{25}+\frac{50}{25}\\&=\frac{-245+490+50}{25}\\&=\frac{295}{25}=\frac{59}{5}=11.8\end{aligned}$

So, our maximum value is 11.8 meters.

Therefore, the ball doesn’t reach a height of 14 meters.

Part B)

To find out how long the ball is in the air, we can simply solve for our t when h=0.

When the ball stops being in the air, this will be the point at which it is at the ground. So, h=0. Therefore:

$0=-5t^2+14t+2$

A quick check of factors will reveal that is it not factorable. Hence, we can use the quadratic formula:

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Again, a=-5, b=14, and c=2. Substitute appropriately:

$\displaystyle x=\frac{-(14)\pm\sqrt{(14)^2-4(-5)(2)}}{2(-5)}$

Evaluate:

$\displaystyle x=\frac{-14\pm\sqrt{236}}{-10}$

We can factor the square root:

$\sqrt{236}=\sqrt{4}\cdot\sqrt{59}=2\sqrt{59}$

Hence:

$\displaystyle x=\frac{-14\pm2\sqrt{59}}{-10}$

Divide everything by -2:

$\displaystyle x=\frac{7\pm\sqrt{59}}{5}$

Hence, our two solutions are:

$\displaystyle x=\frac{7+\sqrt{59}}{5}\approx2.9362\text{ or } x=\frac{7-\sqrt{59}}{5}\approx-0.1362$

Since our variable indicates time, we can reject the negative solution since time cannot be negative.

Hence, our zero is approximately 2.9362.

Therefore, the ball is in the air for approximately 2.9362 seconds.

5 0

3 years ago

Read 2 more answers

PLEASE ANSWER THIS QUESTION RIGHT The Jimenez family is bringing 12 packages of hamburger and hotdog buns to the neighborhood

sveta [45]

The system of equations is:

2x+1.5y=20
x+y=12

When you solve it you get:
x=4 hamburger bun packages
y=8 hot dog bun packages

3 0

3 years ago

For each algebraic equation, select the property that could be used to solve it: 1. x-3=-5 2. 6x=72 3. x/5=3

kykrilka [37]

X - 3 = -5........addition property....because u add -3 to both sides

6x = 72.....either division property...because u divide both sides by 6.
or multiplication property...because u multiply both sides by 1/6

x/5 = 3.....multiplication property....because u multiply both sides by 5

5 0

4 years ago