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3 years ago

12

A cold front changes the temperature by -3 F each day. If the temperature started at 0 F, what will the temperature be after 5 d

ays?

1 answer:

Viefleur [7K]3 years ago

4 0

-15 F bc -3 x 5 is -15

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Solve the following system of equations

azamat

Answer:

(-3,5),(-3,-5),(3,5),(3,-5)

Step-by-step explanation:

i changed my answer :)

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2 years ago

Let R be the region bounded by the following curves. Let S be the solid generated when R is revolved about the given axis. Ifpo

steposvetlana [31]

The two curves $y=x$ and $y=x^{1/15}$ intersect at $x=0$ and $x=1$ , with $x^{1/15}\ge x$ over $0\le x\le1$ .

Washer method

$\displaystyle\pi\int_0^1\left(\left(x^{1/15}\right)^2-x^2\right)\,\mathrm dx=\pi\int_0^1\left(x^{2/15}-x^2\right)\,\mathrm dx=\pi\left(\frac{15}{17}-\frac13\right)=\boxed{\frac{28\pi}{51}}$

Shell method

We have $y=x^{1/15}\implies x=y^{15}$ . The curves $x=y$ and $x=y^{15}$ intersect at $y=0$ and $y=1$ , with $y\ge y^{15}$ over $0\le y\le1$ .

$\displaystyle2\pi\int_0^1y\left(y-y^{15}\right)\,\mathrm dy=2\pi\int_0^1\left(y^2-y^{16}\right)\,\mathrm dy=2\pi\left(\frac13-\frac1{17}\right)=\boxed{\frac{28\pi}{51}}$

6 0

3 years ago

Log14/3 +log11/5-log22/15=log

Tamiku [17]

Answer:

$log7$

Step-by-step explanation:

when adding logs, apply the log rule: $\log _a\left(x\right)+\log _a\left(y\right)=\log _a\left(xy\right)$

∴ $\log\left(\frac{14}{3}\right)+\log\left(\frac{11}{5}\right)=\log\left(\frac{14}{3}\cdot \frac{11}{5}\right)$

when subtracting logs, apply the log rule: $\log _a\left(x\right)\:-\:\log _a\left(y\right)=\log _a\left(\frac{x}{y}\right)$

$\log\left(\frac{14}{3}\cdot \frac{11}{5}\right)-\log\left(\frac{22}{15}\right)=\log\left(\frac{\frac{14}{3}\cdot \frac{11}{5}}{\frac{22}{15}}\right)\\\\=\log\left(7\right)$

5 0

2 years ago

Read 2 more answers

11. What is 25?<br> a. 10<br> b. 15<br> c. 32<br> d. 16

ikadub [295]

Answer:

for what source

Step-by-step explanation:

8 0

3 years ago

PLEASE HELP FASTTT!!!

nirvana33 [79]

Answer:

5 seconds

Step-by-step explanation:

Looking at your function (h(t) = -16t^2 + 48t + 160), I see that the peak height will be 196 feet, and that is achieved in 1.5 seconds.

h(1.5) = -16(1.5)^2 + 48(1.5) + 160

h(1.5) = -16(2.25)+ 48(1.5) + 160

h(1.5) = -36 + 48(1.5) + 160

h(1.5) = -36 + 72 + 160

h(1.5) = 36 + 160

h(1.5) = 196

Going down from that height, it would take 3.5 more seconds, so it would take 5 seconds in total

h(5) = -16(5)^2 + 48(5) + 160

h(5) = -16(25) + 48(5) + 160

h(5) = -400 + 48(5) + 160

h(5) = -400 + 240 + 160

h(5) = -400 + 400

h(5) = 0

8 0

2 years ago