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My name is Ann [436]

3 years ago

7

Translate this sentence into an equation. 59 is the sum of 11 and Mai’s score

2 answers:

777dan777 [17]3 years ago

7 0

11 + Mai’s score (insert whatever variable you want here)= 59

trapecia [35]3 years ago

3 0

Answer:

11 + Mai's Score = 59

Step-by-step explanation:

You need to add 11 and Mai's score together to get 59, so with the values given we can make the equation 11 + Mai's Score = 59.

*depending on the question, Mai's score may need to be said as a letter variable, so:

If m = mai's score,

11 + m = 59

I hope this helped! :)

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Use cavalieri's principle to a circular pillar candle is 2.8 inches wide & 6 inches tall. find the volume of the candle.

wolverine [178]

The candle has the shape of a cylinder with height H=6 in, and radius of the circular base R= (2.8)/2 =1.4 (in).

The volume of the cylinder is Area(base)*height =

$\pi R^{2} *H=3.14* (1.4)^{2}*6= 37$ (in cubed)

According to Cavalieri's principle, the volume of the candle is equal to the volume of the cylinder we described.

Answer: 37 in cubed

5 0

3 years ago

I need help ASAP!!Please explain how to do the problem

hoa [83]

Answer:

i think its 30

Step-by-step explanation:

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8 0

3 years ago

Read 2 more answers

Whats the answer to this?

mina [271]

instead of P+2x = y it should be P-2x=y

so then it is P-2x/2 =y

6 0

4 years ago

The length of one base of a trapezoid is 44, the median is 36, and the other base is 2x + 10.

Nataliya [291]

Answer:

The answer is "9"

Step-by-step explanation:

Given value:

length of trapezoid = 44

Another length = 2x+10

median = 36

formula:

median =(first length + second length) /2

$36 = \frac{(44+ 2x+10)}{2}\\\\ 36 = \frac{54+2x}{2}\\\\36 \times 2 = 54 +2x\\\\72 =54 +2x\\\\2x= 72-54\\\\2x= 18\\\\x= 9$

6 0

3 years ago

John looked at his bank account and noticed that he had $2,029.90 in his account. All that he could remember is that he created

Over [174]

We will have that for the previous year, he had in the account:

$m_{y6}=2029.90-\frac{2029.90\cdot6}{100}\Rightarrow m_{y6}=1908.106$

For the year previous to that one, he had:

$m_{y5}=1908.106-\frac{1908.106\cdot6}{100}\Rightarrow m_{y5}=1793.61964$

For the previous year to that, he had:

$m_{y4}=1793.61964-\frac{1793.61964\cdot6}{100}\Rightarrow m_{y4}=1686.002462$

For the year previous to that one, he had:

$m_{y3}=1686.002462-\frac{1686.002462\cdot6}{100}\Rightarrow m_{y3}=1584.842314$

For the second year that the money was in the account, there was:

$m_{y2}=1584.842314-\frac{1584.842314\cdot6}{100}\Rightarrow m_{y2}=1489.751775$

And the orinial ammount of money that was in the account was:

$m_{y1}=1489.751775-\frac{1489.751775\cdot6}{100}\Rightarrow m_{y1}=1400.366669$

From this, we know that Jhon had originally $1400.37 in the bank account.

6 0

1 year ago