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3 years ago

Which of the following ordered pairs could represent the y-intercept of a function.

Mathematics

1 answer:

snow_lady [41]3 years ago

6 0

<h3>Answer: (0,5)</h3>

This is because the x coordinate is 0. For any y intercept, the x coordinate is always 0 as this indicates the point is on the y axis.

Contrast this with something like (9,0) and we see that y = 0 indicating this point is an x intercept (it is on the x axis).

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The barrels weighed 834 pounds apiece. The ship's cargo included 618 barrels. Therefore, the barrels added pounds in weight to t

Mekhanik [1.2K]

I think you have to add the weights of the cargo ship

6 0

3 years ago

Factor the expression completely.<br>562 – 32

Dimas [21]

Answer:

530

Step-by-step explanation:

562 - 32 = 530

8 0

4 years ago

Read 2 more answers

What is the area of a triangle whose vertices are D(3,3), E(3,-1), and F(-2,-5)?

Elis [28]

There is a formula which employs the use of determinants and which helps us calculate the area of a triangle if the vertices are given as $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ . The formula is as shown below:

Area= $\frac{1}{2}\begin{vmatrix}x_1&y_1&1 \\ x_2&y_2&1\\ x_3&y_3&1\end{vmatrix}$

Now, in our case, we have: $(x_1,y_1)=(3,3)$

$(x_2,y_2)=(3,-1)$ , and

$(x_3,y_3)=(-2,-5)$

Thus, the area in this case will become:

Area= $\frac{1}{2}\begin{vmatrix}3&3&1 \\ 3&-1&1\\ -2&-5&1\end{vmatrix}$

Therefore, Area= $\frac{1}{2}\times [[3(-1\times 1-(-5)\times 1]-3[3\times 1-(-2)\times 1]+1[3\times -5-2]]= \frac{1}{2}\times -20=-10$

We know that area cannot be negative, so the area of the given triangle is <u>10 squared units</u>.

3 0

3 years ago

Determine the area of a rectangle with length of (x+3) and a width of (x-7)

Musya8 [376]

X+3 * x-7
=x^2- 4x - 21

5 0

3 years ago

Raylin went to the park with her friends and found a soccer ball. She picked up the soccer ball and kicked it as high as she cou

Tems11 [23]

Answer:

The maximum height is: $h(t_{max})=14.39 \: u$

The ball reaches the ground in 1.79 s.

Step-by-step explanation:

We need to take the derivative and equal to zero to find the time at the maximum height.

$h(t)=-16t^{2} +27t+3$ (1)

$\frac{dh(t)}{dt}=-32t +27=0$

$t_{max} =\frac{27}{32}$

$t_{max} =0.84\: u$

Now, we just need to put t(max) into equation (1) to find h(max)

$h(t_{max})=-16(0.84)^{2} +27(0.84)+3$

$h(t_{max})=14.39 \u$

If we want to get the time when the ball reaches the ground we just need to equal h(t) to zero.

$0=-16t^{2} +27t+3$

Let's solve this quadratic equation.

We will get two solutions and we must choose the positive value.

t1 = 1.79 u

t2 = -0.10 u

Therefore, the ball reaches the ground in 1.79 u.

I hope it helps you!

7 0

3 years ago