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3 years ago

Use the graph of the function y = -x2 + 2x + 8 to

Mathematics

2 answers:

Advocard [28]3 years ago

8 0

Answer:

It's C my guy.

Step-by-step explanation:

Took the assignment.

Furkat [3]3 years ago

5 0

Answer:

its c lol

Step-by-step explanation:

lol

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What amount of money do you have if you have 5 pennies, 13 nickels, 14 dimes, 2 quarters, and one 5 dollar bill?

BartSMP [9]

Answer:

$7.60

Step-by-step explanation:

Penny = $0.01
Nickel = $0.05
Dime = $0.10
Quarter = $0.25

( 5 × 0.01 ) + ( 13 × 0.05 ) + ( 14 × 0.10 ) + ( 2 × 0.25 ) + 5

-The 5 at the end represents the one five dollar bill.

Simplified: (solve what is in the parenthesis)

0.05 + 0.65 + 1.4 + 0.50 + 5 = $7.60

4 0

3 years ago

Determine the interquartile range for the data. 16, 19, 25, 20, 22, 21, 17, 20

Roman55 [17]

Answer:

the answer is 15 subtract the highest number from the lowest

5 0

3 years ago

Classify the quadrilateral by selecting its most specific name.

Lemur [1.5K]

Rectangle definitely not a parallelogram

7 0

3 years ago

###Please Help### for f(x)=1/x-5 and g(x)=x^2+2 Find the expression for g(x)

Bond [772]

Answer:

I don't get it

Step-by-step explanation:

8 0

3 years ago

Solve the following equation by factoring:9x^2-3x-2=0

olya-2409 [2.1K]

Answer:

The two roots of the quadratic equation are

$x_1= - \frac{1}{3} \text{ and } x_2= \frac{2}{3}$

Step-by-step explanation:

Original quadratic equation is $9x^{2}-3x-2=0$

Divide both sides by 9:

$x^{2} - \frac{x}{3} - \frac{2}{9}=0$

Add $\frac{2}{9}$ to both sides to get rid of the constant on the LHS

$x^{2} - \frac{x}{3} - \frac{2}{9}+\frac{2}{9}=\frac{2}{9}$ ==> $x^{2} - \frac{x}{3}=\frac{2}{9}$

Add $\frac{1}{36}$ to both sides

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{2}{9} +\frac{1}{36}$

This simplifies to

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{1}{4}$

Noting that (a + b)² = a² + 2ab + b²

If we set a = x and b = $\frac{1}{6}\right)$ we can see that

$\left(x - \frac{1}{6}\right)^2$ = $x^2 - 2.x. (-\frac{1}{6}) + \frac{1}{36} = x^{2} - \frac{x}{3}+\frac{1}{36}$

$\left(x - \frac{1}{6}\right)^2=\frac{1}{4}$

Taking square roots on both sides

$\left(x - \frac{1}{6}\right)^2= \pm\frac{1}{4}$

So the two roots or solutions of the equation are

$x - \frac{1}{6}=-\sqrt{\frac{1}{4}}$ and $x - \frac{1}{6}=\sqrt{\frac{1}{4}}$

$\sqrt{\frac{1}{4}} = \frac{1}{2}$

So the two roots are

$x_1=\frac{1}{6} - \frac{1}{2} = -\frac{1}{3}$

and

$x_2=\frac{1}{6} + \frac{1}{2} = \frac{2}{3}$

7 0

2 years ago