All categories

3 years ago

Which table shows the same relationship as y = -x^2 + 3x?

Mathematics

1 answer:

kkurt [141]3 years ago

8 0

Answer:

Step-by-step explanation:

given y = -x² + 3x.

let f(x) be y.

f(-2) = -(-2)² + 3(-2) = -10

f(-1) = -(-1)² + 3(-1) = -4

f(0) = -(0)² + 3(0) = 0

f(1) = -(1)² + 3(1) = 2

f(2) = -(2)² + 3(2) = 2

You might be interested in

An officeSupply company is shipping a case of paper to school. There are 10 reams of paper in the case. If each ream of paper aw

Alenkasestr [34]

------------------------------------
Find 10 reams
------------------------------------
1 ream = 32 ounces
10 reams = 32 x 10
10 reams = 320 ounces

------------------------------------
Convert ounce to pound:
------------------------------------
1 ounce = 0.0625
10 ounces = 0.0625 x 10
10 ounces = 0.625 pounds

------------------------------------------------------------------------
Answer: Each case weighs 0.625 pound.
------------------------------------------------------------------------

3 0

3 years ago

Slove for x, please.

KATRIN_1 [288]

Answer:

I do not know

Step-by-step explanation:

6 0

3 years ago

2x + (-x) + 3 + (-2)

meriva

Answer:

-x+3

Step-by-step explanation:

hope this helps

4 0

3 years ago

Let O be an angle in quadrant III such that cos 0 = -2/5 Find the exact values of csco and tan 0.

vivado [14]

well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus

$cos(\theta)=\cfrac{\stackrel{adjacent}{-2}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2 - (-2)^2}=b\implies \pm\sqrt{25-4}\implies \pm\sqrt{21}=b\implies \stackrel{III~Quadrant}{-\sqrt{21}=b}$

$\dotfill\\\\ csc(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{-\sqrt{21}}}\implies \stackrel{\textit{rationalizing the denominator}}{-\cfrac{5}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies -\cfrac{5\sqrt{21}}{21}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-\sqrt{21}}}{\underset{adjacent}{-2}}\implies tan(\theta)=\cfrac{\sqrt{21}}{2}$

4 0

2 years ago

What is the standard form of the following equation?

scZoUnD [109]

y = 2(x - 1)² + 4

y = 2(x² - 2x + 1) + 4

y = 2x² - 4x + 2 + 4

y = 2x² - 4x + 6

7 0

3 years ago