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2 years ago

Darryl earns $10 per hour tutoring students. if Darryl tutored for 25 hours this month, how much money did he earn this month

Mathematics

2 answers:

Ksju [112]2 years ago

8 0

Answer:

danny makes 250 per month

Step-by-step explanation:

malfutka [58]2 years ago

4 0

Answer:

Darryl earned $250.00 in a month.

Step-by-step explanation:

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Simplify (4.4 × 10^17) + (3.3 × 10^14)

UkoKoshka [18]

Answer:

4.40033 x 10^17

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

J.k. Rowling and R.L. Stone are both ready hunger games . J.k. Reads 35 pages every 2 hours and R.L. Reads 45 pages every 3 hour

san4es73 [151]

Answer:

The rate of change for each reader is :

For J K Rowling : 17 and a half pages per hour

For R L Stone : 15 pages per hour

Step-by-step explanation:

Here the reading rate of each reader is given as:

J.K. Reads 35 pages every 2 hours.

So, the number of pages read by J K in 2 hours = 35 pages

⇒ The number of pages read per hour = 35 / 2 = 17 . 5 pages

So, the pages read by J K Rowling in 1 hour is 17 and a half pages.

R L Stone reads 45 pages every 3 hours.

So, the number of pages read by R L in 3 hours = 45 pages

⇒ The number of pages read per hour = 45 / 3 = 15 pages

So, the pages read by R L Stone in 1 hour is 15 pages.

Hence, the rate of change for each reader is :

For J K Rowling : 17 and a half pages per hour

For R L Stone : 15 pages per hour

5 0

3 years ago

Can you help me please?

klemol [59]

Answer:

the answer is true

Step-by-step explanation:

the x values don't repeat themselves so therefore it is a function

6 0

2 years ago

Work out the size of one of the exterior angles

hodyreva [135]

Given Information :-

⠀

A polygon with 10 sides ( Decagon )

⠀

To Find :-

⠀

The value of one of the exterior angles

⠀

Formula Used :-

⠀

$\qquad \diamond \: \underline{ \boxed{ \pink{ \sf Exterior ~angle = \dfrac {360^\circ}{no. ~of~sides}}}} \: \star$

⠀

Solution :-

⠀

Putting the given values, we get,

⠀

$\sf \dashrightarrow Exterior ~angle = \dfrac{360 ^\circ}{10} \: \: \\ \\ \\ \sf \dashrightarrow Exterior ~angle = \frac{36 \cancel{0}^\circ}{ \cancel{10}} \: \: \\ \\ \\ \sf \dashrightarrow Exterior ~angle = \underline{ \boxed{ \frak{ \red{36^\circ}}}} \: \star \\ \\$