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3 years ago

What is the volume of this container?

Mathematics

1 answer:

anzhelika [568]3 years ago

7 0

hahhsiwnsoej!sjsiisujshsuhssss

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A.4 B.3 C.2 D.1 Help please

vazorg [7]

The number of exports is GREATER than the number of imports if the bar for exports (pink) is TALLER than the bar for imports (blue). It appears that happens only on Friday.

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4 years ago

After 6 months you have the same account balance as your friend and you have saved 280$. Write a linear equation that represents

astra-53 [7]

Answer:

y = 20x + 280

Step-by-step explanation:

The linear equation that represent your account balance is as follows:

Your friend

y = 25(6) + 250

= 400

And,

You

400 = m(6) + 280

6m = 400 - 280

6m = 120

m = 20

Now the equation of the account balance is

y = 20x + 280

Hence, the linear equation that represent your account balance is y = 20x + 280

4 0

3 years ago

2. A square-based tent has the cross-sectional

ollegr [7]

(a) Length of the height is 2.732 m

(b) Length of the base is 5.466 m

Explanation:

An image is attached for reference.

(a)

In ΔAOB,

$sin 30^o = \frac{AO}{AB} \\\\0.5 = \frac{AO}{2} \\\\AO = 1 m$

In ΔBGD,

$sin 60^o = \frac{BG}{BD} \\\\0.866 = \frac{BG}{2} \\\\BG = 1.732 m$

According to the figure, BG = OE = 1.732 m

Height of the tent, AE = AO + OE

= 1 m + 1.732 m

= 2.732 m

(b)

DF = ?

In ΔAOB,

$tan 30^o = \frac{AO}{OB} \\\\0.577 = \frac{1}{OB} \\\\OB = 1.733 m\\\\\\$

According to the figure, OB = GE = 1.733 m

In ΔBGD,

$tan 60^o = \frac{BG}{DG} \\\\1.732 = \frac{1.732}{DG}\\ \\DG = 1m$

According to the figure, DE = DG + GE

DE = 1 m + 1.733 m

DE = 2.733 m

Length of the base, DF = 2 X DE

DF = 2 X 2.733 m

DF = 5.466 m

8 0

3 years ago

Question 8 Find the unit vector in the direction of (2,-3). Write your answer in component form. Do not approximate any numbers

slamgirl [31]

Answer:

The unit vector in component form is $\hat{u} = \left(\frac{2}{\sqrt{13} },-\frac{3}{\sqrt{13}} \right)$ or $\hat{u} = \frac{2}{\sqrt{13}}\,i-\frac{3}{13}\,j$ .

Step-by-step explanation:

Let be $\vec u = (2,-3)$ , its unit vector is determined by following expression:

$\hat {u} = \frac{\vec u}{\|\vec u \|}$

Where $\|\vec u \|$ is the norm of $\vec u$ , which is found by Pythagorean Theorem:

$\|\vec u\|=\sqrt{2^{2}+(-3)^{2}}$

$\|\vec u\| = \sqrt{13}$

Then, the unit vector is:

$\hat{u} = \frac{1}{\sqrt{13}} \cdot (2,-3)$

$\hat{u} = \left(\frac{2}{\sqrt{13} },-\frac{3}{\sqrt{13}} \right)$

The unit vector in component form is $\hat{u} = \left(\frac{2}{\sqrt{13} },-\frac{3}{\sqrt{13}} \right)$ or $\hat{u} = \frac{2}{\sqrt{13}}\,i-\frac{3}{13}\,j$ .

6 0

3 years ago

What is the length of an arc cut off by an angle of 3 radians on a circle of radius 6 inches?

sattari [20]

1 inch is the answer

6 0

3 years ago