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3 years ago

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Which represents the solution(s) of the graphed system of equations, y = x2 + x – 2 and y = 2x – 2? (–2, 0) and (0, 1) (0, –2) a

nd (1, 0) (–2, 0) and (1, 0) (0, –2) and (0, 1) Mark this and return

2 answers:

MA_775_DIABLO [31]3 years ago

5 0

Answer:

<h2>(0, -2) (1, 0)</h2>

Step-by-step explanation:

I got it right on the test... Have a great day :)

Luda [366]3 years ago

3 0

ANSWER

$(0,-2), (1,0)$

EXPLANATION

The first equation is

$y = {x}^{2} + x - 2$

The second equation is

$y = 2x - 2$

We equate both equations to get,

${x}^{2} + x - 2 = 2x - 2$

${x}^{2} + x - 2x - 2 + 2 = 0$

Simplify

${x}^{2} - x = 0$

Factor

$x(x - 1) = 0$

Either

$x = 0$

Or

$x - 1 = 0$

$x = 1$

Put x=0 or x=1 into the second equation to get,

$y = 2(0) - 2 = - 2$

Or

$y = 2(1) - 2 = 0$

Therefore the solutions are;

$(0,-2), (1,0)$

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A kitchen makes 2944 ounces of soup each week. There are 128 ounces in one gallon. How many gallons of soup does the kitchen mak

zhannawk [14.2K]

Step 1:
Find number of gallons in one week
Total oz ÷ oz per gallon= gallons
=2944/128
= 23 gallons in one week

Step 2:
Multiply gallons made in one week by two.
=23 gallons per week * 2
=46 gallons made in 2 weeks

Hope this helps! :)

3 0

3 years ago

A triangle has sides represented by 7, x, and x−1. If the perimeter is 28, find the length of the longest side.

aleksley [76]

Answer:

x (11 unit) is the longest side

Step-by-step explanation:

Given:

Sides;

7, x, and x−1

Perimeter = 28

Find:

Longest side

Computation:

Perimeter = sum of all sides

So,

7 + x + (x -1) = 28

2x + 6 = 28

x = 11

So,

x -1 = 11 - 1 = 10 unit

x (11 unit) is the longest side

3 0

3 years ago

Answer 9 a and b and c

inn [45]

If she can only work 15 hours a week and earns $7 per hour, just multiply 7 by 15 and you'll have your answer :)

3 0

3 years ago

Please help me figure this one out !!!

strojnjashka [21]

These are "composite" functions.
g(x+a) means insert (x+a) into g(x) to replace every x:
g(x+a) - g(x) = -5(x+a)^2 +4(x+a) +5x^2 - 4x
= -5(x^2 + 2ax + a^2) +4x +4a + 5x^2 - 4x
Now just Multiply and sum up the answer.

5 0

3 years ago

Prove by mathematical induction that

postnew [5]

For $n=1$ , on the left we have $\cos\theta$ , and on the right,

$\dfrac{\sin2\theta}{2\sin\theta}=\dfrac{2\sin\theta\cos\theta}{2\sin\theta}=\cos\theta$

(where we use the double angle identity: $\sin2\theta=2\sin\theta\cos\theta$ )

Suppose the relation holds for $n=k$ :

$\displaystyle\sum_{n=1}^k\cos(2n-1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}$

Then for $n=k+1$ , the left side is

$\displaystyle\sum_{n=1}^{k+1}\cos(2n-1)\theta=\sum_{n=1}^k\cos(2n-1)\theta+\cos(2k+1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta$

So we want to show that

$\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

On the left side, we can combine the fractions:

$\dfrac{\sin2k\theta+2\sin\theta\cos(2k+1)\theta}{2\sin\theta}$

Recall that

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

so that we can write

$\dfrac{\sin2k\theta+2\sin\theta(\cos2k\theta\cos\theta-\sin2k\theta\sin\theta)}{2\sin\theta}$

$=\dfrac{\sin2k\theta+\sin2\theta\cos2k\theta-2\sin2k\theta\sin^2\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta(1-2\sin^2\theta)+\sin2\theta\cos2k\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta\cos2\theta+\sin2\theta\cos2k\theta}{2\sin\theta}$

(another double angle identity: $\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta$ )

Then recall that

$\sin(x+y)=\sin x\cos y+\sin y\cos x$

which lets us consolidate the numerator to get what we wanted:

$=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

and the identity is established.

8 0

3 years ago