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4 years ago

11

The graph of a quadratic function has a minimum at (0, 2). The graph contains the point (-3, 11). What is another point on the g

raph?

2 answers:

NemiM [27]4 years ago

8 0

Answer:

Step-by-step explanation:

Graphs of quadratics are symmetrical about the vertex (minimum or maximum point).

The x value of the point (-3, 11) is 3 units to the left of the vertex (0, 2), the x value that is 3 units to the right of the vertex will have the same y value, so the point

(3, 11) is also on the graph

Degger [83]4 years ago

8 0

Answer:

(3, 11)

Step-by-step explanation:

imagine math

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The investor owns __ of stock A<br><br><br> The investor owns __ of stock <br> B

Angelina_Jolie [31]

9.5 +27b =4000

a=80 shares of stock A
b=120 shares of stock B

Let a be the number of shares of stock B

B the number of shares of stock B

A+B=200

6 0

3 years ago

At Ralph's fruit stand X pound of grapes cost 90 cents. You want to buy 5 pounds of grapes. How

diamong [38]

Answer:At Ralph’s fruit stand ½ pound of grapes cost 90 cents. You want to buy 5 pounds of grapes.

How much will they cost?

1. What are the two possible rates for this problem?

Solution: ½ pound/90 cents 90 cents/ ½ pound

2. Show each rate as a unit rate.

Solution: . pounds of per 1 cent 180 cents per 1 pound of grapes

3. What does each unit rate tell you?

Solution: The pounds of grapes for 1 cent

The number of cents for one pound of grapes

4. Which unit rate will help you solve the problem?

Solution: 180 per 1 pound of grapes

5. If it costs 180 cents to buy 1 pound of grapes, how much will 2 pounds cost? 4 pounds?

Complete the table below.

POUNDS

OF

GRAPES

COST IN CENTS

1 180

2 360

3 540

What pattern do you see? The cost in cents increases by 180 cents each time.

Multiply the pounds of grapes by 180 cents to get the total cost

6. Since you know the unit price, write a number sentence for the cost of five pounds of

grapes. Write an equation for the cost of any pounds of grapes.

(180 cents per pound)(5 pounds) = $9.00

(Unit rate)(pounds of grapes) = total cost

180x = y where x is the cost per pound of grapes and y is the number of pounds.

Step-by-step explanation:

8 0

3 years ago

Which expression represents the products of a and b decresed by 9?

Dmitriy789 [7]

Answer:

(a × b) - 9

Step-by-step explanation:

The product of a and b basically just means a being multiplied by b and then decreased by 9 just means to subtract that by 9.

I hope this helps you :D

8 0

3 years ago

Part A: The average rate of change between the 2nd and 3rd point is Select a Value Part A: The average rate of change between th

Fofino [41]

Answer:

$Rate = 6$

$Rate = 10$

$Rate = 18$

$Rate = 162$

Step-by-step explanation:

Given

See attachment for table

Solving (a): Rate of change between 2nd and 3rd point on A

The rate of change is calculated as:

$Rate = \frac{y_2 - y_1}{x_2 - x_1}$

In table A, the 2nd and 3rd point is:

$(x_1,y_1) =(1,7)$

$(x_2,y_2) =(2,13)$

So, the average rate of change is:

$Rate = \frac{13 - 7}{2 - 1}$

$Rate = \frac{6}{1}$

$Rate = 6$

Solving (b): Rate of change between 3rd and 4th point on A

In table A, the 3rd and 4th point is:

$(x_1,y_1) =(2,13)$

$(x_2,y_2) =(3,23)$

So, the average rate of change is:

$Rate = \frac{23 - 13}{3 - 2}$

$Rate = \frac{10}{1}$

$Rate = 10$

Solving (c): Rate of change between 2nd and 3rd point on B

In table B, the 2nd and 3rd point is:

$(x_1,y_1) =(2,11)$

$(x_2,y_2) =(3,29)$

So, the average rate of change is:

$Rate = \frac{29 - 11}{3 - 2}$

$Rate = \frac{18}{1}$

$Rate = 18$

Solving (d): Rate of change between 4th and 5th point on B

In table B, the 4th and 5th point is:

$(x_1,y_1) =(4,83)$

$(x_2,y_2) =(5,245)$

So, the average rate of change is:

$Rate = \frac{245 - 83}{5 - 4}$

$Rate = \frac{162}{1}$

$Rate = 162$

7 0

3 years ago

Please help . thank u

Ierofanga [76]

Hi there!

The measure for the angle B is 13.

A complementary angle equals to 90 degrees. You add (2x + 10) and (3x + 15) and equal it to 90. Combine like terms and balance out the problem like a 2 step variable equations.

Hope this helps !

5 0

3 years ago