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Georgia [21]
3 years ago
10

Which digit is in the hundredths place? 18.36

Mathematics
2 answers:
V125BC [204]3 years ago
8 0
6 is in the hundredths place

ten, one, tenths, hundredths  (place value names)

hope this helps
pav-90 [236]3 years ago
4 0
The 6 is.
Afte the decimal point it goes tenths, hundredths, then thousandths and so on
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Island A is 150 miles from Island B. A ship captain travels 310 miles from Island A and then finds that he is off course and 200
wolverine [178]

Answer:

156.37°

Step-by-step explanation:

We solve this above question, using Cosine rule which is given as:

Cos C = a² + b² - c²/2ab

Let the Angle be represented as: x

Cos x = 310² + 200² - 150²/2 × 310 × 200

x = arc cos [ 310² + 200² - 150²/2 × 310 × 200]

x = 23.63°

The angle, in degrees, that he must turn through to head straight for Island B is given as:

180° - 23.63°

= 156.37°

4 0
3 years ago
Dwayne wants to buy a jump rope that costs $4, a board game that costs $7, and a playground ball that costs $3. He has saved $9
Natali5045456 [20]

Answer:

he needs $1

Step-by-step explanation:

add 4+7+3=14 then your gonna add what he already has which is 9+4= 13 so 14-13=1

4 0
3 years ago
Read 2 more answers
There are approximately 3,000 bass in a lake. The population grows at a rate of 2% per year. Round answers to the nearest tenth.
Tamiku [17]

Answer:

years 1–4: 62.4 bass per year

years 5–8: 67.6 bass per year

Step-by-step explanation:

If the population in year n is ...

 p(n) = 3000·1.02^n

then the average rate of change from year 1 to year 4 is ...

 (p(4) -p(1))/(4 -1) = 3000(1.02^4 -1.02^1)/3 = 1020·(1.02^3 -1) ≈ 62.4

The average rate of change for years 1–4 is 62.4 bass per year.

For years 5–8, the rate of change is similarly computed:

 (p(8) -p(5))/(8 -5) = 3000(1.02^8 -1.02^5)/3 = 1000·1.02^5·(1.02^3 -1) ≈ 67.6

The average rate of change for years 5–8 is 67.6 bass per year.

5 0
3 years ago
Read 2 more answers
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
Name each polynomial by degree amd number of terms<br>6p^4 + 10p^2
Volgvan
Two term quartic, or degree 4 polynomial
5 0
3 years ago
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