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Citrus2011 [14]
3 years ago
12

8.71 x 10^7 in standard form! Please hurry.

Mathematics
1 answer:
olchik [2.2K]3 years ago
8 0

Answer:

87, 100,000

Step-by-step explanation:

Brainliest please! Visit my church's website at fbcinterlaken.org to learn more about God! It would mean a lot! Thank you and have a nice day! :)

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Which expression is equivalent to<br> ^3 sqrt 1/1000 c^9 d^12
tamaranim1 [39]

Answer:

\sqrt[3]{ \frac{1}{1000} {c}^{9}  {d}^{12}  }  \\  =  \frac{1}{10}  {c}^{3}  {d}^{4}

5 0
3 years ago
Suppose a local park is rectangular in shape and has a length of 210 yards and a width of 105 yards. What is the length of the d
Otrada [13]

Answer:

234.784 ~ 235

Step-by-step explanation:

One side = 210

other side = 105

a2+b2 = c2

210 ^2 + 105 ^2 = c2

55125 = c ^2

c = \sqrt{55125} = 234.784 ~ 235

hope its right!!!

7 0
3 years ago
Create a word problem using 30+2d as the answer
Tanzania [10]

Answer:

Sonya pays a flat fee of $30 for her phone bill and pays $2 per gigabyte she uses per month. Put Sonya's bill as an equation with <em>d</em> being how many gigabytes she uses per month.

Step-by-step explanation:

We get the flat fee of $30 from "30 + 2d" and the $2 per gigabyte from "30 + 2d" and since we want to find out Sonya's total bill, that's where the + in "30 + 2d" comes from. We want the variable to be <em>d</em> therefore <em>d</em> is how many gigabytes she uses per month.

4 0
3 years ago
A jet travels 510 miles in 5 hours. At this rate, how far could the jet fly in 15 hours? What is the rate of speed of the jet?
n200080 [17]

Answer:

1530

Step-by-step explanation:

We already know that 5 hours is 510 miles and to know how much is 15 hours you will add 510 3 times 510 + 510 + 510 what 1530 would give

8 0
3 years ago
Read 2 more answers
Hey so im also looking for the first answer and the choices were “ is not “ and “ is “
dexar [7]

so the investigator found the skid marks were 75 feet long hmmm what speed will that be?

s=\sqrt{30fd}~~ \begin{cases} f=\stackrel{friction}{factor}\\ d=\stackrel{skid}{feet}\\[-0.5em] \hrulefill\\ f=\stackrel{dry~day}{0.7}\\ d=75 \end{cases}\implies s=\sqrt{30(0.7)(75)}\implies s\approx 39.69~\frac{m}{h}

nope, the analysis shows that Charlie was going faster than 35 m/h.

now, assuming Charlie was indeed going at 35 m/h, then his skid marks would have been

s=\sqrt{30fd}~~ \begin{cases} f=\stackrel{friction}{factor}\\ d=\stackrel{skid}{feet}\\[-0.5em] \hrulefill\\ f=\stackrel{dry~day}{0.7}\\ s=35 \end{cases}\implies 35=\sqrt{30(0.7)d} \\\\\\ 35^2=30(0.7)d\implies \cfrac{35^2}{30(0.7)}=d\implies 58~ft\approx d

4 0
3 years ago
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