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3 years ago

If the following graph corresponds to an equation in the form of y= x^2+bx+c, What are the values of b and c?

Mathematics

2 answers:

Alexxx [7]3 years ago

5 0

Answer:

C. b = 1, c = -6

Step-by-step explanation:

In the equation y = x² + bx + c,

b refers the graph moving right and left

c refers to the graph moving up and down

Since the vertex is at -6, c = -6

I hope this helped! :)

Solnce55 [7]3 years ago

4 0

Answer:

Step-by-step explanation:

epic

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3 years ago

Read 2 more answers

Last June only 0.17 inches of rain fell all month. What is the difference between the average rainfall and the actual rainfall f

spin [16.1K]

Answer:

Part 1) 0.50 inches

Part 2) 1.14 inches

Part 3)

a) 1.25 inches

b) -0.32 inches

Step-by-step explanation:

The complete question in the attached figure

Part 1) Last June only 0.17 inches of rain fell all month. What is the difference between the average rainfall and the actual rainfall for the last June

Let

x ----> the average rainfall for the last June

y ----> the actual rainfall for the last June

The difference between the average rainfall and the actual rainfall for the last June is given by the expression

$(x-y)$

we have

$x=0.67\ in\\y=0.17\ in$

substitute

$0.67-0.17=0.50\ in$

Part 2) The departure from the average rainfall last July was -0.36 inches. How much rain fell last July?

we know that

The departure from the average is the difference between the actual amount of rain and the average amount of rain for a given month

Let

D -----> the departure from the average

x ----> the average rainfall for the last June

y ----> the actual rainfall for the last June

$D=y-x$

we have

$D=-0.36\ in\\x=1.5\ in$

substitute the given values

$-0.36=y-1.5$

solve for y

$y=1.5-0.36\\y=1.14\ in$

Part 3)

Part a) How much rain would have to fall in August so that the total amount of rain equals the average rainfall for these three months?

Part b) What would the departure from the average be in August in that

situation?

Part a) How much rain would have to fall in August so that the total amount of rain equals the average rainfall for these three months?

Find the average rainfall for these three months

Adds the average rainfall and divide by 3

$(0.67+1.5+1.57)/3=1.25\ in$

Part b) What would the departure from the average be in August in that

situation?

we know that

The departure from the average is the difference between the actual amount of rain and the average amount of rain for a given month

Let

D -----> the departure from the average

x ----> the average rainfall for the last June

y ----> the actual rainfall for the last June

$D=y-x$

we have

$x=1.57\ in\\y=1.25\ in$

substitute the given values

$D=1.25-1.57=-0.32\ in$

7 0

4 years ago

For the function f(x) = 8x7, find f-¹(x).

Zanzabum

Answer:

f^-1(x)= $\sqrt[7]{x} /8$

Step-by-step explanation:

f(x)=8x^7

x^7=f(x)/8

x= $\sqrt[7]{x} /8$

f^-1(x)= $\sqrt[7]{x} /8$

6 0

2 years ago

Read 2 more answers

The table shows a proportional relationship between the number of pounds of bananas purchased and the total cost of bananas.

Liula [17]

Answer:

Choice A, C and E

Step-by-step explanation:

Your table is poorly presented

Given

x y

$\begin{array}{cc} {3} & {1.47} \ \\ {5} & {2.45} \ \\ {9} & {4.41} \ \ \end{array}$

First, we calculate the constant of proportionality (k)

$k = \frac{y}{x}$

For (3,1.47)

$k = \frac{1.47}{3}$

$k = 0.49$

For (5,2.45)

$k = \frac{2.45}{5}$

$k = 0.49$

For (9,4.41)

$k = \frac{4.41}{9}$

$k = 0.49$

From the above calculations, the constant of proportionality is 0.49.

So, the usable rows must also have 0.49 as their constant of proportionality

Choice A

$(x,y) = (2,0.98)$

$k = \frac{0.98}{2}$

$k = 0.49$ --- This is usable

Choice B

$(x,y) = (7,4.45)$

$k = \frac{4.45}{7}$

$k = 0.64$ --- This is not usable

Choice C

$(x,y) = (6,2.94)$

$k = \frac{2.94}{6}$

$k = 0.49$ --- This is usable

Choice D

$(x,y) = (1,0.54)$

$k = \frac{0.54}{1}$

$k = 0.54$ ---- This is not usable

Choice E

$(x,y) = (8,3.92)$

$k = \frac{3.92}{8}$

$k = 0.49$ --- This is usable

5 0

3 years ago