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4 years ago

5. For each quantity, decide whether circumference or area would be needed to calculate it. Explain or show your reasoning.

Mathematics

1 answer:

kolbaska11 [484]4 years ago

8 0

Answer:

a. The distance around a circular track. CIRCUMFERENCE

b. The total number of equally-sized tiles on a circular floor. AREA

c. The amount of oil it takes to cover the bottom of a frying pan. AREA

d. The distance your car will go with one turn of the wheels. CIRCUMFERENCE

Step-by-step explanation:

CIRCUMFERENCE : The circumference of a circle is the total length of its boundary.

AREA : Area of a circle is the total region covered buy that circle.

a. The distance around a circular track.

Here, we are covering the distance ONLY around the track and NOT inside it,

Hence, The CIRCUMFERENCE is calculated.

b. The total number of equally-sized tiles on a circular floor.

Here, we are covering the region or the total space occupied by all the tiles on the floor.

Hence, The AREA is calculated.

c. The amount of oil it takes to cover the bottom of a frying pan.

Here, we are covering the region or the total space occupied by all the oil on the bottom of the frying pan.

Hence, The AREA is calculated.

d. The distance your car will go with one turn of the wheels.

Here, the distance covered by wheel in 1 round

= Total length of the boundary of wheel

= The circumference of the wheel

Hence, The CIRCUMFERENCE is calculated.

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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:

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

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

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

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

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

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4 years ago

Please help! Geometry Dilations, Similar triangles & proportionality

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Answer:

Step-by-step explanation:

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3 years ago

Peter sold an article for $3600 gaining one sixth of its selling price. Find the cost price and the gain percent.

Ivahew [28]

Answer:

See the answers below

Step-by-step explanation:

Given data

Cost of artilce= $3600

Gain= 1/6 of 3600

Hence the Gain is

=1/6*3600

=0.166*3600

=$600

The Gain Percent is

=600/3600*100

=0.1666*100

=16.66%

5 0

3 years ago

Researches notice a grey whale cruising at a depth of 233 feet below sea level. It rises 73 feet and then comes up another 40 fe

musickatia [10]

Answer:

The whale is at a depth 120 feet below the sea level.

Step-by-step explanation:

A grey whale is cruising at a depth of 233 feet below sea level. The depth below the sea level has the negative position, thus the initial position of the whale is -233.

It rises 73 feet, the position of the whale becomes 73 feet higher. This means we should add 73.

Then it comes up another 40 feet, so we need to add 40.

Mathematically,

$-233+73+40=-160+40=-120$

The whale is at a depth 120 feet below the sea level.

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3 years ago

Find the sum of the first 50 terms in the sequence where a1=2.1 and a50=1.2

Elan Coil [88]

Answer:

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Step-by-step explanation:

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3 years ago

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