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4 years ago

Can someone please help me with this?

Mathematics

1 answer:

horrorfan [7]4 years ago

3 0

Um this seems really confusing!

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Tiffany wants to calculate the volume of her globe. The globe is in the shape of a sphere. She measured the circumference of the

Elis [28]

Answer:

Step-by-step explanation:

We know that the volume of a sphere/globe is given as

V=4 /3πr^3

but the circumference is expressed as

C=2πr

solving for r given that C=24

24=2*3.142r

24=6.284r

r=24/6.284

r=3.8in

put r=3.6 in the expression for volume we have

V=4 /3π(3.8)^3

V=4 /3π(55

V=(220.59*3.142)/3

V=693.12/3

V=231in^3

The volume of the globe is 231in^3

3 0

3 years ago

Solve and check please help ty

raketka [301]

so wt's the question of this pic?

5 0

3 years ago

Rewrite the fraction as a decimal -33/25

sergiy2304 [10]

To rewrite something as a decimal, simply divide the numerator and denominator.
-33 / 25 = -1.32
So -33/25 = -1.32

3 0

3 years ago

How can you use geometry figures to solve real world problems

Otrada [13]

Bob has to tile a rectangular floor that is 10 by 20 feet
he has tiles with tiles that are 1 foot square
how many tiles does he need
area=legnth times wdith=10 times 20=200
200 square feet=200 tiles

4 0

3 years ago

If alpha and beta are the angles in the first quadrant tan alpha = 1/7 and sin beta =1/ root 10 then usind the formula sin (A +B

zysi [14]

Answer:

$\arcsin\left(\frac{129\sqrt{2}}{250}\right)\approx 0.8179$

Step-by-step explanation:

$\alpha \text{ and } \beta \text{ in Quadrant I}$

$\tan(\alpha)=\frac{1}{7} \text{ and } \sin(\beta)=\frac{1}{\sqrt{10}}=\frac{\sqrt{10} }{10}$

<u>Using Pythagorean Identities</u>:

$\boxed{\sin^2(\theta)+\cos^2(\theta)=1} \text{ and } \boxed{1+\tan^2(\theta)=\sec^2(\theta)}$

$\left(\frac{\sqrt{10} }{10} \right)^2+\cos^2(\beta)=1 \Longrightarrow \cos(\beta)=\sqrt{1-\frac{10}{100}} =\sqrt{\frac{90}{100}}=\frac{3\sqrt{10}}{10}$

$\text{Note: } \cos(\beta) \text{ is positive because the angle is in the first qudrant}$

$1+\left(\frac{1 }{7} \right)^2=\frac{1}{\cos^2(\alpha)} \Longrightarrow 1+\frac{1}{49}=\frac{1}{\cos^2(\alpha)} \Longrightarrow \frac{50}{49} =\frac{1}{\cos^2(\alpha)}$

$\Longrightarrow \frac{49}{50}=\cos^2(\alpha) \Longrightarrow \cos(\alpha)=\sqrt{\frac{49}{50} } =\frac{7\sqrt{2}}{10}$

$\text{Now let's find }\sin(\alpha)$

$\sin^2(\alpha)+\left(\frac{7\sqrt{2} }{10}\right)^2=1 \Longrightarrow \sin^2(\alpha) +\frac{49}{50}=1 \Longrightarrow \sin(\alpha)=\sqrt{1-\frac{49}{50}} = \frac{\sqrt{2}}{10}$