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LuckyWell [14K]
3 years ago
15

PLEASE HELP MEEE ASAP

Mathematics
1 answer:
lys-0071 [83]3 years ago
8 0

Step-by-step explanation:

first of all, g(x) has only negative functional result values (y) except for 0.

and a square is always positive.

so, the only possible right answers are the ones that include a minus ("-") sign.

the graph shows us that g(x) goes through the points (1, -3) and (-1, -3).

so, which equation turns an x = 1 into an y = -3 ?

therefore, the right answer must be D.

g(x) = -3x²

it works for both points :

-3 = -3×1² = -3 correct

-3 = -3×(-1)² = -3×1 = -3 correct

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-2/3 x 3/5 + 5/2 - 3/5 x 1/6
chubhunter [2.5K]

Answer:

sorry im on a time lapse, so the answer is 2  1/2. hope it helps

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Need help on this question please!
Alja [10]

Answer: she made 11 sparkelers and gavin made 21

Step-by-step explanation:

4 0
3 years ago
Cosa/2-tana +sina/1-cota=cosa+sina​
Inessa05 [86]

Answer:

Cosa/2-tana+sina

Step-by-step explanation:

Cosatana

7 0
1 year ago
What is the slope of the line that passes through:<br> (6,5) and (-1,-4)<br><br><br> Thank You!
My name is Ann [436]

Answer:

m=\frac{9}{7}

Step-by-step explanation:

Slope Formula: m=\frac{y_2-y_1}{x_2-x_1}

Simply plug in the 2 coordinates into the slope formula to find slope <em>m</em>:

m=\frac{-4-5}{-1-6}

m=\frac{-9}{-7}

m=\frac{9}{7}

6 0
3 years ago
a circular oil slick is expanding at a rate of 2m^2/h. Find the rate at which it's diameter is expanding when it's radius is 1.5
aleksandrvk [35]

Answer:  \frac{4\pi}{3} \text{ meter per hour}

Step-by-step explanation:

The circular oil slick is expanding at a rate of  2 m^2/h

Let A be the area of the circular oil slick,

So, the changes in  A with respect to time (t),

\frac{dA}{dt} = 2

\frac{d(\pi r^2)}{dt} = 2

2\pi r\frac{dr}{dt} = 2

\frac{dr}{dt} = \frac{1}{\pi r}  

Also, the change in diameter with respect to time(t),

\frac{d}{dt} (2 r) = 2 \frac{dr}{dt}

\frac{d}{dt} (2 r) = 2 \times \frac{1}{\pi r}

\frac{d}{dt} (2 r) = \frac{2}{\pi r}

For r = 1.5 m,

\frac{d}{dt} ( 2 r)]_{r=1.5} = \frac{2}{\pi \times 1.5}=\frac{20}{\pi \times 15}=\frac{4\pi }{3}\text{ meter per hour}




7 0
2 years ago
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