Write an equation for the quadratic function whose vertex is (âˆ’2,10) and passes through (1,28)

2. Juanita is asked to write the standard equation of a

Crazy boy [7]

Answer:

It is not. Also use ^2 to represent squaring a number.

Step-by-step explanation:

First, USE ^2 FOR SQUARED

It is not correct.

The standard equation of a circle is (x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius.

In your question the center is (-2,3) and the radius is the length of the line segment from the center to any pointon the circle. nd hey, it tells us a point on the circle is (-2,6).

Let's take it one part at a time. we know h and k so let's plug them in.

(x-h)^2 + (y-k)^2

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

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

now r is the length of the line segment from (-2,3) and (-2,6). To get from (-2,3) to (-2,6) hopefully it is pretty obvious you just move up (so int he y direction) 3 spaces. this makes r = 3. int he formula we want r^2 and that is 9. The result is that the circle has a formula of:

(x-h)^2 + (y-k)^2 = r^2

(x+2)^2 + (y-3)^2 = 9

The formula Juanita got is actually a hyperbola.

8 0

1 year ago

Two solutions of different concentrations of acid are mixed creating 40 mL of a solution that is 32% acid. One-quarter of the so

Sever21 [200]

Taking the 3 solutions as 3 different terms, we can create an equation as follows:

Solution 1 : 10mL with 20% acid
Solution 2 : 30mL with x% acid
Solution 3 : 40mL with 32% acid

Since solution 1 + solution 2 = solution 3, let us substitute the given values we have:

10(0.2) + 30(x) = 40(0.32)
2 + 30x = 12.8

To solve for the unknown concentration x, we subtract 2 from both sides:
2 + 30x - 2 = 12.8 - 2
30x = 10.8

Dividing both sides by 30:
30x/30 = 10.8/30
x = 0.36

Therefore the unknown solution is 36% acid.

4 0

1 year ago

Read 2 more answers

The perimeter of triangle ABC is 82 inches. The triangle is rotated about a line that passes through points B and D.

Musya8 [376]

Answer:

A cone with base radius of 17 in

Step-by-step explanation:

Given

Triangle ABC

Where

AB = BC = 24in

D = Midpoint of AC

Required

Which best describes the resulting three-dimensional figure?

First, the length of the third side of the triangle has to be calculated

The sides of triangle ABC are AB, AC and BC

Given that the perimeter = 82 in

AB + AC + BC = 82

Recall that AB = BC = 24in; The equation becomes

$24 + AC + 24 = 82$

Collect like terms

$AC = 82 - 24 - 24\\AC = 34$

Given that, the triangle is rotated at D, the midpoint of AC

The length of the midpoint has to be calculated

$D = \frac{1}{2}AC$

$D = \frac{1}{2} * 34$

$D = \frac{34}{2}$

$D = 17$

At this point, we can dismiss options C and D.

This is because; a triangular pyramid is formed by more than 1 triangles (at least 3 triangle) and in this case, we're dealing with only 1 triangle.

So, we have option A and B to select from.

When the triangle is rotated at point D, it generates a cone such that the radius of the cone is the distance at point D.

This implies that, a cone with base radius D is generated.

Recall that D = 17.

Hence, <em>A cone with base radius of 17 in</em>

7 0

1 year ago

Read 2 more answers

Beth wants to make exactly 60 blueberry muffins. Each batch makes 6 muffins, She has already made 2.5 batches. How many more bat

Katarina [22]

Answer: 4.5

You need that many more batches

8 0

2 years ago

Read 2 more answers

(1 point) Let P(t) be the performance level of someone learning a skill as a function of the training time t. The derivative dPd

monitta

Answer:

$P(t)=M+Ce^{-kt}$

Step-by-step explanation:

Given the differential model

$\dfrac{dP}{dt}=k[M-P(t)]$

We are required to solve the equation for P(t).

$\dfrac{dP}{dt}=kM-kP(t)\\$Add kP(t) to both sides\\\dfrac{dP}{dt}+kP(t)=kM\\$Taking the integrating factor\\e^{\int k dt} =e^{kt}\\$Multiply all through by the integrating factor\\\dfrac{dP}{dt}e^{kt}+kP(t)e^{kt}=kMe^{kt}\\\dfrac{dP}{dt}e^{kt}=kMe^{kt}\\(Pe^{kt})'=kMe^{kt} dt\\$Take the integral of both sides with respect to t\\\int (Pe^{kt})'=\int kMe^{kt} dt\\Pe^{kt}=kM \int e^{kt} dt\\Pe^{kt}=\dfrac{kM}{k} e^{kt} + C_0, C_0$ a constant of integration$

$Pe^{kt}=Me^{kt} + C\\$Divide both side by e^{kt}\\P(t)=M+Ce^{-kt}\\P(t)=M+Ce^{-kt}\\$Therefore:\\P(t)=M+Ce^{-kt}$

4 0

1 year ago