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yaroslaw [1]
3 years ago
8

Humans shed about three million particles of skin every five hours. Humans can shed about fifty thousand particles of skin every

five minutes. The approximate number of particles of skin a human can shed in five hours can be written in the form a × 10b, where a is 3 and b is____. The approximate number of particles of skin a human can shed in five minutes can be written in the form a × 10b, where a is 5 and b is ____. Humans shed approximately ____ times more particles of skin in five hours than in five minutes.
Mathematics
1 answer:
lukranit [14]3 years ago
7 0

Answer:

for 5 hours: b is 6

for 5 minutes: b is 4

60 times more skin in 5 hours than 5 minutes

Step-by-step explanation:

I think it should be written as a x 10^b, so I'm going to answer it that way.  

3 x 10^6 = 5 times 10^6

   10^6 = 1,000,000, so we have 3 x 1,000,000 = 3,000, 000

5 x 10^4 = 5 times 10^4

   10^4 = 10,000, so we have 5 x 10,000 = 50, 000

For part 3, divide 3 million by 50 thousand to find the scalar multiple

   3,000,000/50,000 = 300/5 = 60

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irinina [24]

Jogging is given as 20 hours.

Total hours is given as 50.

Jogging = 20/50 hours, which reduces to 2/5.

A full circle is 360 degrees.

Multiply 360 by 2/5:

360 x 2/5 = 720 / 5 = 144 degrees.

5 0
3 years ago
Help would be very much appreciated
harina [27]

\dfrac{f(2+h)-f(2)}{h}=\dfrac{(2+h)^2+2(2+h)-1-(2^2+2\cdot2-1)}{h}=\\\\=\dfrac{4+4h+h^2+4+2h-1-(4+4-1)}{h}=\dfrac{h^2+6h+7-7}{h}=\dfrac{h^2+6h}{h}=h+6

8 0
2 years ago
Remember to show work and explain. Use the math font.
MrMuchimi

Answer:

\large\boxed{1.\ f^{-1}(x)=4\log(x\sqrt[4]2)}\\\\\boxed{2.\ f^{-1}(x)=\log(x^5+5)}\\\\\boxed{3.\ f^{-1}(x)=\sqrt{4^{x-1}}}

Step-by-step explanation:

\log_ab=c\iff a^c=b\\\\n\log_ab=\log_ab^n\\\\a^{\log_ab}=b\\\\\log_aa^n=n\\\\\log_{10}a=\log a\\=============================

1.\\y=\left(\dfrac{5^x}{2}\right)^\frac{1}{4}\\\\\text{Exchange x and y. Solve for y:}\\\\\left(\dfrac{5^y}{2}\right)^\frac{1}{4}=x\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\dfrac{(5^y)^\frac{1}{4}}{2^\frac{1}{4}}=x\qquad\text{multiply both sides by }\ 2^\frac{1}{4}\\\\\left(5^y\right)^\frac{1}{4}=2^\frac{1}{4}x\qquad\text{use}\ (a^n)^m=a^{nm}\\\\5^{\frac{1}{4}y}=2^\frac{1}{4}x\qquad\log_5\ \text{of both sides}

\log_55^{\frac{1}{4}y}=\log_5\left(2^\frac{1}{4}x\right)\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\dfrac{1}{4}y=\log(x\sqrt[4]2)\qquad\text{multiply both sides by 4}\\\\y=4\log(x\sqrt[4]2)

--------------------------\\2.\\y=(10^x-5)^\frac{1}{5}\\\\\text{Exchange x and y. Solve for y:}\\\\(10^y-5)^\frac{1}{5}=x\qquad\text{5 power of both sides}\\\\\bigg[(10^y-5)^\frac{1}{5}\bigg]^5=x^5\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(10^y-5)^{\frac{1}{5}\cdot5}=x^5\\\\10^y-5=x^5\qquad\text{add 5 to both sides}\\\\10^y=x^5+5\qquad\log\ \text{of both sides}\\\\\log10^y=\log(x^5+5)\Rightarrow y=\log(x^5+5)

--------------------------\\3.\\y=\log_4(4x^2)\\\\\text{Exchange x and y. Solve for y:}\\\\\log_4(4y^2)=x\Rightarrow4^{\log_4(4y^2)}=4^x\\\\4y^2=4^x\qquad\text{divide both sides by 4}\\\\y^2=\dfrac{4^x}{4}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\y^2=4^{x-1}\Rightarrow y=\sqrt{4^{x-1}}

6 0
4 years ago
Shelley has a 1 1/2 pound supply of Cat Treats. Each day she gives her kitten 1/8 pound of the treats. How many days will the su
natta225 [31]

Answer:

The treats will last 12 days.

8 0
3 years ago
Figure out this 16r^3t^2/-4rt
ElenaW [278]
You mean to simplify as this is an expression not an equation, as it has no equal sign...

Factor our 4 first

(4r^3t^2)/(-rt)

Now factor out -rt

-4r^2t
8 0
3 years ago
Read 2 more answers
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