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2 years ago

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The mass of the sun is about 1.98x10^30 kg. The mass of the Earth is about 5.97x10^24 kg. Estimate how many times bigger the mas

s of the sun than the mass of earth. Round to the NEAREST TENTH!

1 answer:

lubasha [3.4K]2 years ago

8 0

Answer:

About 333000 times

Step-by-step explanation:

You might be interested in

Drew’s score on a math text was 92.He could earn an extra 1/2 point for each incorrect problem he corrected. Drew corrected 3 pr

svp [43]

Answer:

Point S represent the score 93.5 which is Drew's New Score.

Step-by-step explanation:

The score of Drew on a math test = 92

Point for each correction in the math test = 1/2 = 0.5

Now, the number of problems corrected = 3 x 0.5 = 1.5

So, Now, the new scale of Drew = Old Score + The Correction Score

= 92 + 1.5

= 93.5

or, Drew, new score = 93.5

Hence, point S represent the score 93.5 which is Drew's New Score.

8 0

2 years ago

You are making a new type of deodorant from two different mixtures the first mixture is 30% smell proof and the second mixture i

Readme [11.4K]

Answer:

600 L of 30% proof and 400 L of 80% proof is needed

Step-by-step explanation:

Let the amount of 30% smell proof be x while that of 80% smell proof is y

Then;

x + y = 1000 •••••(i)

Also;

30% of x + 80% of y = 50% of 1000

0.3x + 0.8y = 500 •••••(ii)

From i, x = 1000 - y

Put this into ii

0.3(1000-y) + 0.8y = 500

300 - 0.3y + 0.8y = 500

0.5y = 500-300

0.5y = 200

y = 200/0.5

y = 400 L

But x = 1000 - y

x = 1000 - 400 = 600 L

3 0

2 years ago

a person invest in 9000 in a bank. the bank pays 5% interest compounded semi annually. to the nearest tenth of a year, how long

QveST [7]

Answer:

10.1 years.

Step-by-step explanation:

It is given that,

Principal = 9000

Rate of interest = 5%

No. of times interest compounded = 2 times in an year

Amount after certain time = 14800

The formula for amount:

$A=P(1+\frac{r}{n})^{nt}$

where, P is principal, r is rate of interest, n is no. of times interest compounded in an year and t is time in years.

Substitute the given values in the above formula.

$14800=9000(1+\frac{0.05}{2})^{2t}$

$\frac{14800}{9000}=(1+0.025)^{2t}$

$1.644=(1.025)^{2t}$

Taking log both sides.

$\log(1.644)=\log(1.025)^{2t}$

$\log(1.644)=2t\log(1.025)$ $[\because \log a^b=b\log a]$

$\frac{\log(1.644)}{2\log(1.025)}=t$

$t=10.066$

$t=10.1$

Therefore, the required time is 10.1 years.

3 0

3 years ago

Generate the nest three terms of each arithmetic sequence shown below.

o-na [289]

Answer:

A)2,6,10

B)2,-6,-18

C)-1,-3,-5

Step-by-step explanation:

<u>A)a1=-2 and d=4</u>

We know that the arithmetic sequence formula is

$a_{n}=a_{1}+(n-1)d$

Now

$a_{2}=a_{1}+(2-1)d$

$a_{2}=a_{1}+(1)d$

Substituting the given value we get

$a_{2}= -2+(1)4$

$a_{2}= -2+4$

$a_{2}= 2$

------------------------------------------

Similarly

$a_{3}=a_{1}+(3-1)d$

$a_{3}=a_{1}+(2)d$

Substituting the given value we get

$a_{3}= -2+(2)4$

$a_{3}= -2+8$

$a_{3}= 6$

-------------------------------------------------

$a_{4}=a_{1}+(4-1)d$

$a_{4}=a_{1}+(3)d$

Substituting the given value we get

$a_{4}= -2+(3)4$

$a_{4}= -2+16$

$a_{4}= 10$

-------------------------------------------------------------------------------------------

<u>B</u><u> $a_n=a_{(n-1)}-8$ with $a_1=10$ </u>

$a_2=a_{(2-1)}-8$

$a_2=a_{1}-8$

Substituting the given value

$a_2= 10-8$

$a_2=2$

---------------------------------------------------------------------

$a_3=a_{(3-1)}-8$

$a_3=a_{2}-8$

Substituting the value

$a_3=2-8$

$a_3= -6$

---------------------------------------------------------------------

$a_4=a_{(4-1)}-8$

$a_4=a_{3}-8$

Substituting the value

$a_4= -6-8$

$a_4= -14$

-------------------------------------------------------------------------------------------

<u>C) $a_1=3, a_2=1$ </u>

Here the difference is -2

the arithmetic sequence formula is

$a_{n}=a_{1}+(n-1)d$

Now

$a_{3}=a_{1}+(3-1)d$

$a_{3}=a_{1}+(2)d$

Substituting the value we get

$a_{3}= 3+(2)-2$

$a_{3}= 3-4$

$a_{3}= -1$

------------------------------------------------------------------------------

$a_{4}=a_{1}+(4-1)d$

$a_{4}=a_{1}+(3)d$

Substituting the value we get

$a_{4}= 3+(3)-2$

$a_{4}= 3-6$

$a_{4}= -3$

----------------------------------------------------------------------------------

$a_{5}=a_{1}+(5-1)d$

$a_{5}=a_{1}+(4)d$

Substituting the value we get

$a_{5}= 3+(4)-2$

$a_{5}= 3-8$

$a_{5}= -5$

4 0

3 years ago

The radius of a circle is 10 inches. What is the angle measure of an arc bounding a sector with area 15 square inches?

vova2212 [387]

Answer:

times it by pi

Step-by-step explanation:

4 0

3 years ago