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3 years ago

Over what interval is the graph of f(x) = -(X + 8)2 – 1 decreasing?

Mathematics

2 answers:

zloy xaker [14]3 years ago

7 0

Can you add a picture

gtnhenbr [62]3 years ago

3 0

Answer:

can u add a pic?

Step-by-step explanation:

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Please help. I don’t understand what to do

11111nata11111 [884]

$\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=23.9\\ h=100 \end{cases}\implies V=\cfrac{\pi (23.9)^2(100)}{3} \\\\\\ V=\cfrac{57121\pi }{3}\implies V\approx 59816.97\implies \stackrel{\textit{rounded up}}{V=59817} \\\\[-0.35em] ~\dotfill$

now, for the second one, we know the diameter is 10, thus its radius is half that or 5.

$\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=5\\ V=225 \end{cases}\implies 225=\cfrac{\pi (5)^2 h}{3}\implies 225=\cfrac{25\pi h}{3} \\\\\\ \cfrac{225}{25\pi }=\cfrac{h}{3}\implies \cfrac{9}{\pi }=\cfrac{h}{3}\implies \cfrac{27}{\pi }=h\implies 8.59\approx h\implies \stackrel{\textit{rounded up}}{8.6=h}$

6 0

3 years ago

An Olympic swimming pool is 50 m long, 25 m wide, and 1.5 m deep. How many Olympic pools would be filled by the dollar bills spe

Gelneren [198K]

Answer:

Number of dollars bills that can be fitted in Olympic pool = 1.67 billion.

Step-by-step explanation:

The dimensions of a dollar bill are 15.5956 cm by 6.6294 cm by 0.010922 cm. The U.S. federal government spent $3.8 trillion in fiscal year 2011.

An Olympic swimming pool is 50 m long, 25 m wide, and 1.5 m deep. How many Olympic pools would be filled by the dollar bills spent by the U.S. federal government in 2011

From above,

The volume for the dimension of the dollar bill = 15.5956 cm × 6.6294 cm × 0.010922 cm

The volume for the dimension of the dollar bill = 1.12922 cm³

The volume for the dimension of the dollar bill = 1.12922 × 10⁻⁶ m³

The volume of the pool = 50 m × 25 m × 1.5 m

The volume of the pool = 1875 m³

Number of dollars bills that can be fitted in pool = volume of the pool/volume of the dimension of dollar bills

Number of dollars bills that can be fitted in Olympic pool =1875 m³ / 1.12922 × 10⁻⁶ m³

Number of dollars bills that can be fitted in Olympic pool = 1,660,438,180

Number of dollars bills that can be fitted in Olympic pool = 1.67 billion.

8 0

3 years ago

Please, I need help ASAP!

enot [183]

Answer:

Step-by-step explanation:

You can eliminate B and D because it’s using $\frac{g}{f}$ , but the question is asking $\frac{f}{g}$ . So it’s either A or C. They are both written correctly, but out of the two, A is correct because just that x ≠ $\sqrt{7}$ part of it is what makes it true.

The reason being is that the only thing that won’t make $\frac{2x+1}{x^{2}-7 }$ true is when the denominator equals 0, because anything divided by 0 is undefined. So, when you plug $\sqrt{7}$ for x in the denominator you get: $\sqrt{7^{2} } -7$ . The $\sqrt{7^{2} }$ part just turns into 7, and 7 - 7 is 0. So you don’t want x to equal $\sqrt{7}$ .

<u>So the answer is A:</u> $\frac{2x+1}{x^{2}-7}$ , $x\neq$ ± $\sqrt{7}$

There you go my friend!

I hope you understand and that this helps with your question! :)

7 0

3 years ago

-2(x - 2) = -3 - 21 I need help with that question, I've been trying to solve it for hours but i cant find the answer

iVinArrow [24]

Answer:

x = 14

Step-by-step explanation:

you would first distribute the -2 with x - 2, which should give you -2x + 4, and you want to do the -3 - 21, which would give you -24, then subtract 4 from -2x + 4, then you should have -2x = -28, then divide it by -2, which should give you x = 14

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3 years ago

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andriy [413]

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that what i think they are

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3 years ago

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