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harina [27]
3 years ago
9

Question 15

Mathematics
1 answer:
ycow [4]3 years ago
8 0

The perimeter of the smaller triangle is 8+12+16=36.  The big triangle has perimeter 54.  So our scale factor is 54/36=3/2.  So we scale up the short side, 8, by 3/2, giving our answer:

Answer: 12

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Find f(4) please, I need help
mixer [17]

Answer:

f(4) = 1

Step-by-step explanation:

f(4) = what?

So if we know that x, aka 4, is technicially f is greater then 2, so you can plug in x as 4. So that means 5(4) - 19 = 20 - 19 = 1, so f(4) = 1

7 0
3 years ago
Sammy bought 15.3 Liters of milk. She poured the milk equally into 5 bottles. There was 0.35 liters of milk left over. Calculate
Zanzabum

Answer:

The capacity of milk in 1 bottle is 2.99 liters

Step-by-step explanation:

Let us solve the question

∵ Sammy bought 15.3 liters of milk

∵ She poured the milk equally into 5 bottles

∵ There were 0.35 liters of milk leftover

→ At first, subtract the leftover from the quantity she bought to

   find the capacity of the five bottles

∴ The capacity of the 5 bottles = 15.3 - 0.35

∴ The capacity of the 5 bottles = 14.95 liters

→ To find the capacity of each bottle divide the total capacity by 5

∴ The capacity in one bottle = 14.95 ÷ 5

∴ The capacity in one bottle = 2.99 liters

∴ The capacity of milk in 1 bottle is 2.99 liters

8 0
3 years ago
I need help with these as well
Mkey [24]
The answer to c. Is 8t+12g
4 0
3 years ago
Use polar coordinates to find the volume of the given solid. Inside both the cylinder x2 y2 = 1 and the ellipsoid 4x2 4y2 z2 = 6
Anton [14]

The Volume of the given solid using polar coordinate is:\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta

V= \frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta

<h3>What is Volume of Solid in polar coordinates?</h3>

To find the volume in polar coordinates bounded above by a surface z=f(r,θ) over a region on the xy-plane, use a double integral in polar coordinates.

Consider the cylinder,x^{2}+y^{2} =1 and the ellipsoid, 4x^{2}+ 4y^{2} + z^{2} =64

In polar coordinates, we know that

x^{2}+y^{2} =r^{2}

So, the ellipsoid gives

4{(x^{2}+ y^{2)} + z^{2} =64

4(r^{2}) + z^{2} = 64

z^{2} = 64- 4(r^{2})

z=± \sqrt{64-4r^{2} }

So, the volume of the solid is given by:

V= \int\limits^{2\pi}_ 0 \int\limits^1_0{} \, [\sqrt{64-4r^{2} }- (-\sqrt{64-4r^{2} })] r dr d\theta

= 2\int\limits^{2\pi}_ 0 \int\limits^1_0 \, r\sqrt{64-4r^{2} } r dr d\theta

To solve the integral take, 64-4r^{2} = t

dt= -8rdr

rdr = \frac{-1}{8} dt

So, the integral  \int\ r\sqrt{64-4r^{2} } rdr become

=\int\ \sqrt{t } \frac{-1}{8} dt

= \frac{-1}{12} t^{3/2}

=\frac{-1}{12} (64-4r^{2}) ^{3/2}

so on applying the limit, the volume becomes

V= 2\int\limits^{2\pi}_ {0} \int\limits^1_0{} \, \frac{-1}{12} (64-4r^{2}) ^{3/2} d\theta

=\frac{-1}{6} \int\limits^{2\pi}_ {0} [(64-4(1)^{2}) ^{3/2} \; -(64-4(2)^{0}) ^{3/2} ] d\theta

V = \frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta

Since, further the integral isn't having any term of \theta.

we will end here.

The Volume of the given solid using polar coordinate is:\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta

Learn more about Volume in polar coordinate here:

brainly.com/question/25172004

#SPJ4

3 0
2 years ago
PLEASE HELP ME! I'v tried but still haven't got it!
spayn [35]

Answer:

Step-by-step explanation:

part a: time, student surf the internet  

part b

5 0
3 years ago
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