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timurjin [86]
2 years ago
13

Please take a look at the picture

Mathematics
1 answer:
Pavlova-9 [17]2 years ago
5 0

Answer: C

Step-by-step explanation:

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A length is measured at 21 cm correct to two significant figures, what is the lower bound of the length and upper bound?
Andreyy89

Answer:

20.5 and 21.5

Step-by-step explanation:

It says 2 significant figures so the least it can possibly be is 0.5 less and the most it can be is 0.5 more as if it was, lets say 20.4 it would round down to 20 not 21.

7 0
3 years ago
What is the output when the input is 0
Cloud [144]

Answer:

When the inputs are 1 and 0, the output is zero.

7 0
3 years ago
PLEASE HELP FAST
skad [1K]

Answer:

The unit vector u is (-5/√29) i - (2/√29) j

Step-by-step explanation:

* Lets revise the meaning of unit vector

- The unit vector is the vector ÷ the magnitude of the vector

- If the vector w = xi + yj

- Its magnitude IwI = √(x² + y²) ⇒ the length of the vector w

- The unit vector u in the direction of w is u = w/IwI

- The unit vector u = (xi + yj)/√(x² + y²)

- The unit vector u = [x/√(x² + y²)] i + [y/√(x² + y²)] j

* Now lets solve the problem

∵ v = -5i - 2j

∴ IvI = √[(-5)² +(-2)²] = √[25 + 4] = √29

- The unit vector u = v/IvI

∴ u = (-5i - 2j)/√29 ⇒ spilt the terms

∴ u = (-5/√29) i - (2/√29) j

* The unit vector u is (-5/√29) i - (2/√29) j

8 0
3 years ago
What simple interest rate would allow $6000 to grow to an amount of $14550 in 10 years?
Harman [31]

Answer:

\boxed{ \bold{ \huge{  \boxed{ \sf{ \: 14.25 \: \% \: }}}}}

Step-by-step explanation:

Given,

Principal ( P ) = $ 6000

Amount ( A ) = $ 14550

Time ( T ) = 10 years

Rate ( R ) = ?

<u>Finding </u><u>the </u><u>Interest</u>

The sum of principal and interest is called an amount.

From the definition,

\boxed{ \sf{Amount =  \: Principal + Interest}}

plug the values

⇒\sf{14550 = 6000 + Interest}

Swap the sides of the equation

⇒\sf{6000 + Interest = 14550}

Move 6000 to right hand side and change its sign

⇒\sf{Interest = 14550 - 6000}

Subtract 6000 from 14550

⇒\sf{Interest = \: 8550 \: }

Interest = $ 8550

<u>Finding </u><u>the </u><u>rate </u>

{ \boxed{ \sf{Rate =  \frac{Interest \times 100}{Principal \times Time}}}}

plug the values

⇒\sf{ Rate = \frac{8550  \times 100}{6000 \times 10} }

Calculate

⇒\sf{Rate =  \frac{855000}{60000} }

⇒\sf{Rate = 14.25 \: \% \: }

Hope I helped!

Best regards!!

8 0
3 years ago
One month Reuben rented movies and video games for a total of . The next month he rented movies and video games for a total of .
Vesnalui [34]

Can you say the total cost? There's not enough to go by to answer this.

4 0
3 years ago
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