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2 years ago

Find the volume of this figure: cm 7 3 cm Cin 3

Mathematics

2 answers:

kenny6666 [7]2 years ago

7 0

I think it’s 63 because to find the volume you have to multiply all numbers

Vladimir79 [104]2 years ago

6 0

Answer:

4 i guess i dont rlly know

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HELP PLEASE A. 51.4 B. 50.0 C. 31.0 D. 39.5 E. None

Liono4ka [1.6K]

First, in order to find the length of side w, you must first find the angle W.
A triangle must be 180 degrees in total. 93 + 37 makes the triangle only have 130 at the moment.
180 - 130 = 50
So angle W is 50 degrees
Now using this, we can use law of sines.
$\frac{w}{sin50} = \frac{31}{sin37}$
Now multiply sin 50 on both sides. It can be written like so:
$\frac{31sin50}{sin37}$
Now plug this in a calculator.
Answer is 39.5

8 0

3 years ago

10 points Please help no Explanation needed A B C D

uysha [10]

Answer:

B * The sign stays the same *

Step-by-step explanation:

Hope this helped, Have a Wonderful Day

6 0

2 years ago

Read 2 more answers

Draw a quadrilateral with only one set of parallel

swat32

A trapezoid would work perfectly.

4 0

3 years ago

Read 2 more answers

Miranda enlarged a picture twice as shown below, each time using a scale factor of 3.

Grace [21]

Answer:

a,c,d

Step-by-step explanation:

5 0

2 years ago

The velocity of an automobile starting from rest is given by the equation below, where v is measured in feet per second and t is

maria [59]

Answer:

a. At t = 5 s

$a(5)=\frac{1785}{\left(6(5)+17\right)^2}=\frac{1785}{2209}\approx0.808 \frac{ft}{s^2}$

b. At t = 10 s

$a(10)=\frac{1785}{\left(6(10)+17\right)^2}=\frac{255}{847}\approx0.301 \frac{ft}{s^2}$

c. At t = 20 s

$a(20)=\frac{1785}{\left(6(20)+17\right)^2}=\frac{1785}{18769}\approx0.095 \frac{ft}{s^2}$

Step-by-step explanation:

We know that the velocity function is given by

$v(t)=\frac{105t}{6t+17}$

Acceleration is the rate of change of velocity so we take the derivative of the velocity function with respect to time.

$a(t)=\frac{dv}{dt}=\frac{d}{dt} (\frac{105t}{6t+17})$

$\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'\\\\105\frac{d}{dt}\left(\frac{t}{6t+17}\right)\\\\\mathrm{Apply\:the\:Quotient\:Rule}:\quad \frac{d}{{dx}}\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \frac{{\frac{d}{{dx}}f\left( x \right)g\left( x \right) - f\left( x \right)\frac{d}{{dx}}g\left( x \right)}}{{g^2 \left( x \right)}}$

$105\cdot \frac{\frac{d}{dt}\left(t\right)\left(6t+17\right)-\frac{d}{dt}\left(6t+17\right)t}{\left(6t+17\right)^2}\\\\105\cdot \frac{1\cdot \left(6t+17\right)-6t}{\left(6t+17\right)^2}\\\\a(t)=\frac{1785}{\left(6t+17\right)^2}$

a. At t = 5 s

$a(5)=\frac{1785}{\left(6(5)+17\right)^2}=\frac{1785}{2209}\approx0.808 \frac{ft}{s^2}$

b. At t = 10 s

$a(10)=\frac{1785}{\left(6(10)+17\right)^2}=\frac{255}{847}\approx0.301 \frac{ft}{s^2}$

c. At t = 20 s

$a(20)=\frac{1785}{\left(6(20)+17\right)^2}=\frac{1785}{18769}\approx0.095 \frac{ft}{s^2}$

3 0

3 years ago