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2 years ago

1. What is the cross section formed by cutting a cylinder perpendicular to its base?

Mathematics

1 answer:

tatyana61 [14]2 years ago

3 0

Answer:

Depends on the cross section axis

place a cylinder can on a table , circular and side down .Look down at it and you see the other end so the cross section is circle

Turn it on its side and the cross section is rectangle

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F(x) =3x^2 -8 find f(-4)

grin007 [14]

Answer:

136

Step-by-step explanation:

plug in -4 in the x

f(-4)=3(-4)^2 - 8

f(-4)=-12^2 - 8

f(-4)=144-8

f(-4)=136

Done!

3 0

2 years ago

Evaluate 1 whole number 2/5 + 3/4 and give your answer to one significant figure

ehidna [41]

Answer:

The answer is

<h3>2 to 1 significant figure</h3>

Step-by-step explanation:

$1 \frac{2}{5} + \frac{3}{4}$

To solve the question first convert the mixed fraction to an improper fraction

That's

$1 \frac{2}{5} = \frac{7}{5}$

So we have

$\frac{7}{5} + \frac{3}{4}$

Find the LCM of the fractions

The LCM of 5 and 4 is 20

That's

$\frac{7}{5} + \frac{3}{4} = \frac{7(4) + 3(5)}{20} = \frac{28 + 15}{20}$

$= \frac{43}{20}$

= 2.15

We have the final answer as

<h3>2 to 1 significant figure</h3>

Hope this helps you

3 0

2 years ago

Find the circumference of a sphere's great circle with a radius of 4 m.

Ulleksa [173]

This is the concept of areas and circumference of the solid figures;
The circumference of the cylinder whose radius is 4 m will be given by;
C=2πr
C=2*π*4
C=8π m=25.133 m

3 0

3 years ago

If a 12-student class averaged 90 on a test, and a 20-student class averaged 80 on the test, then all 32 students averaged

Alexxx [7]

Let the marks of the students in class 1 be

$m_1, m_2, ....., m_1_2$

the average of these 12 marks is 90, so

$\frac{m_1+ m_2+ .....+ m_1_2}{12}=90$

which means

$m_1+ m_2+ .....+ m_1_2=90*12=1080$

similarly, let $n_1, n_2, ....., n_2_0$ be the marks of the students in the second class,

so $\frac{n_1+ n_2+ ..... n_2_0}{20}=80$

which means $n_1+ n_2+ ..... n_2_0=80*20=1600$

The 32 students averaged:

$\frac{(m_1+ m_2+ .....+ m_1_2)+(n_1+ n_2+ ..... n_2_0)}{12+20}= \frac{1080+1600}{32}= \frac{2680}{32}= 83.75$

5 0

3 years ago

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taurus [48]

C is the answer radical 77/22

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2 years ago

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