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3 years ago

Can anyone pls helppppp

Mathematics

1 answer:

dalvyx [7]3 years ago

4 0

Hourly rate:
39(1.3r) = 541.80
(hours times hourly rate equals profit)

then solve
1.3r = 13.89
r = 10.69

the hourly rate is $10.69

to answer the bottom question, 390 + 117 = 507, but that’s not close to 541.80 so I have no idea what they’re asking there

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Sin10°-cos10° product and evaluate

Montano1993 [528]

Question should be sin 20 cos 10 + cos 20 sin 10

remember sin (a+b) = sin a cos b + cos a sin b ?

sin 20 cos 10 + cos 20 sin 10

= sin (20+10)

= sin 30

= 1/2

6 0

2 years ago

Read 2 more answers

Gest Inc. has provided the following data for the month of November. The balance in the Finished Goods inventory account at the

Helen [10]

Answer: The answer is a $226,000

Step-by-step explanation:

The adjusted cost of good sold Account

Balance at beginning 49,000

Add: Cost of goods manufactured 226,000

Add: Applied overhead on work in process 70,000

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345,000

Less: Balance b /d. 45,000

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300,000

Less: Actual manufacturing overhead. 74,000

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Balance on income statements. 226,000

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4 0

3 years ago

In the illustration below, the three cube-shaped tanks are identical. The spheres in any given tank

fredd [130]

Answer:

1) Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) $Amount \ of \, water \ remaining \ in \, the \ tank \ is \ \frac{x^3(6-\pi) }{6}$

Step-by-step explanation:

1) Here we have;

First tank A

Volume of tank = x³

The volume of the sphere = $\frac{4}{3} \pi r^3$

However, the diameter of the sphere = x therefore;

r = x/2 and the volume of the sphere is thus;

volume of the sphere = $\frac{4}{3} \pi \frac{x^3}{8}$ = $\frac{1}{6} \pi x^3$

For tank B

Volume of tank = x³

The volume of the spheres = $8 \times \frac{4}{3} \pi r^3$

However, the diameter of the spheres 2·D = x therefore;

r = x/4 and the volume of the sphere is thus;

volume of the spheres = $8 \times \frac{4}{3} \pi (\frac{x}{4})^3= \frac{x^3 \times \pi }{6}$

For tank C

Volume of tank = x³

The volume of the spheres = $64 \times \frac{4}{3} \pi r^3$

However, the diameter of the spheres 4·D = x therefore;

r = x/8 and the volume of the sphere is thus;

volume of the spheres = $64 \times \frac{4}{3} \pi (\frac{x}{8})^3= \frac{x^3 \times \pi }{6}$

Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) For the 4th tank, we have;

number of spheres on side of the tank, n is given thus;

n³ = 512

∴ n = ∛512 = 8

Hence we have;

Volume of tank = x³

The volume of the spheres = $512 \times \frac{4}{3} \pi r^3$

However, the diameter of the spheres 8·D = x therefore;

r = x/16 and the volume of the sphere is thus;

volume of the spheres = $512\times \frac{4}{3} \pi (\frac{x}{16})^3= \frac{x^3 \times \pi }{6}$

Amount of water remaining in the tank is given by the following expression;

Amount of water remaining in the tank = Volume of tank - volume of spheres

Amount of water remaining in the tank = $x^3 - \frac{x^3 \times \pi }{6} = \frac{x^3(6-\pi) }{6}$

$Amount \ of \ water \, remaining \, in \, the \ tank = \frac{x^3(6-\pi) }{6}$ .

5 0

3 years ago

Fwam answered 149 questions correctly on his multiple choice science final that had

aniked [119]

Answer:

he answered 74.5% of questions correctly

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2 years ago

Simplify this number 300mm:9m

DIA [1.3K]

Answer:

1 : 30

Step-by-step explanation:

300mm:9m

We need to change the meters to mm

1 meter is 1000 mm

so 9 m is 9000 mm

300mm:9000mm

Divide both sides by 300

300mm/300 : 9000/300

1 : 30

5 0

3 years ago