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hodyreva [135]
3 years ago
13

Can someone please help me with this? ^^"

Mathematics
1 answer:
meriva3 years ago
7 0

Answer:

Not issue discussed plz full page

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3. A balancing balloon toy is in the shape of a hemisphere (half-sphere) attached to the base of a cone. If the toy is 4ft tall
Katen [24]

Answer:

The volume of the toy is V=5.23\ ft^3

Step-by-step explanation:

step 1

Find the volume of the hemisphere

The volume of the hemisphere is given by the formula

V=\frac{2}{3}\pi r^{3}

In this problem, the wide of the toy is equal to the diameter of the hemisphere

so

D=2\ ft

r=2/2=1\ ft ----> the radius is half the diameter

substitute

V=\frac{2}{3} \pi (1)^{3}=\frac{2}{3} \pi\ ft^3

step 2

Find the volume of the cone

The volume of the cone is given by

V=\frac{1}{3}\pi r^{2}h

we know that

The radius of the cone is the same that the radius of the hemisphere

so

r=1\ ft

The height of the cone is equal to subtract the radius of the hemisphere from the height of the toy

h=4-1=3\ ft

substitute the given values

V=\frac{1}{3}\pi (1)^{2}(3)=\pi\ ft^3

step 3

Find the volume of the toy

we know that

The volume of the toy, is equal to the volume of the cone plus the volume of the hemisphere.

so

V=(\frac{2}{3} \pi+\pi)\ ft^3

V=(\frac{5}{3}\pi)\ ft^3

assume

\pi=3.14

V=\frac{5}{3}(3.14)=5.23\ ft^3

5 0
3 years ago
Find the equation of the line passing through the points (3,-2) and (4,6) in slope intercept form.​
noname [10]

Hey!

------------------------------------------------

They already give us the slope, so we can solve the y-intercept.

We can use (3, -2).

y = 8x + b

-2 = 8(3) + b

-2 = 24 + b

-2 - 24 = 24 + b - 24

-26 = b

------------------------------------------------

Hence, the answer is \Large\boxed{\mathsf{y~=~8x~+~-26}}

------------------------------------------------

Hope This Helped! Good Luck!

4 0
3 years ago
F(x) = 4√(2x³-1)<br> F'(x) =.....?
goldfiish [28.3K]

Answer:

\large\boxed{f'(x)=\dfrac{12x^2}{\sqrt{2x^3-1}}}

Step-by-step explanation:

f(x)=4\sqrt{2x^3-1}=4\left(2x^3-1\right)^\frac{1}{2}\\\\f'(x)=4\cdot\dfrac{1}{2}(2x^3-1)^{-\frac{1}{2}}\cdot3\cdot2x^2=\dfrac{12x^2}{(2x^3-1)^\frac{1}{2}}=\dfrac{12x^2}{\sqrt{2x^3-1}}\\\\\text{used}\\\\\sqrt{a}=a^\frac{1}{2}\\\\\bigg[f\left(g(x)\right)\bigg]'=f'(g(x))\cdot g'(x)\\\\\bigg[nf(x)\bigg]'=nf'(x)\\\\(x^n)'=nx^{n-1}

6 0
3 years ago
Which rectangular equation is equivalent to the polar equation r csc theta=8?
Drupady [299]
Rectangular and polar forms are two forms of equations that translates to plot. In this case, the two forms are convertible to each other by the expressions:

r sin theta = y
r cos theta = x
x2 + y2 = r2

we are given the polar expression r csc theta = 8 and is asked to convert to rectangular form.

in this case, csc theta is equal to 1/ sin theta. thys
 r / sin theta = 8

in order to make use of the equations above, then

we multiply r to both numerator and denominator in the left side, that is 
r^2 / r sin theta = 8 
x2+y2 / y = 8
x  2 + y2 = 8y
3 0
3 years ago
3.5
Elena L [17]
It doesn’t s 19% I hope this helps bye
6 0
3 years ago
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