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Andrej [43]
3 years ago
10

PLZZ help ZOOM IN THE SEE IT CLEAR

Mathematics
2 answers:
masya89 [10]3 years ago
8 0
The answer is 4^45.
(4^9)^5=4^45 and 4^0=1
Anika [276]3 years ago
4 0
The answer is probably B
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Mary and Julie both have a goal of raising $1000 for charity. Mary raises $12 every 4 days. Julie raises $19 every 6 days.
baherus [9]
Julie would reach her goal first. Julie would reach her goal in approximately 316 days while it would take Mary approximately 333 days to reach it.

I know this because the steps to solve the problem are:

1.) 1,000/12 = 83.333 (to find how many days it would take her everyday)

2.) 83.333x4 = 333 (multiplying the answer by four makes it every four days)

3.) determine that it takes Mary 333 days

4.) 1,000/19 = 52.631 (just like step 1 but with different numbers)

5.) 52.631x6 = 316 (just like in step 2 but different numbers)

6.) determine it takes Julie 316 days

7.) 316 < 333 (takes Julie less time)
6 0
3 years ago
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In a two collum proof the left column states your reasoning
dsp73
Yes the left states your reasoning 
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3 years ago
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Can someone please help me???
olganol [36]

Answer:

1. 60000.12

2. 54000.15

3. 9600.48

4. 3840.48

Step-by-step explanation:

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7 0
3 years ago
The sum of the measures of angle M and angle R is 90°.
Fiesta28 [93]

Answer:

The value of x would be 5

Step-by-step explanation:

(for future reference if you have questions like this it’s better to fill in numbers for x to be able to answer the question)

So you want to fill in x with a number. In this case in order for this equation to be right you fill x with 5.

So 5(5)+10=35

And to check your answer to make sure it’s right you would add the measure of angle R which is 55 to the measure of angle M which is 35.

55+35 is 90. Which makes your answer for x correct :) hopefully this makes sense.

4 0
3 years ago
I need help please I would really appreciate it
Alchen [17]

Answer:

Your answer to the fig. would be 14cm^2.

Adding area of both rectangles,

(8+6) cm^2

= 14cm^2

Hope it helps!!

6 0
3 years ago
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