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ella [17]
3 years ago
7

$12 meal; 4.5% tax find total cost to the nearest cent

Mathematics
2 answers:
Lemur [1.5K]3 years ago
8 0

Answer:

12.54

Step-by-step explanation:

just multiply 12×0.045 and add the 12.

I hope and this helps.

Have a good day.

Good luck

xz_007 [3.2K]3 years ago
8 0
The answer would be 12.54 i believe sorry if i am wrong
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The square root of a number is between 8 and 9. Which of the following could be the value of that number? Select all that apply.
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What is the measure of ∠H?
jek_recluse [69]
B. 73
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What is the length of HG?
Leviafan [203]

Answer:

HG = 5

Step-by-step explanation:

Given

HE = 8mm

EF = 12mm

Area = 68mm^2

Find the length of HG

First, it should be noted that the displayed figure is a trapezium

<em>The area of a trapezium is calculated by multiplying the sum of parallel sides by half its height;</em>

In this case;

Area = \frac{HG + EF}{2} * HE

Substitute the values of Area, HE and EF

68 = \frac{HG + 12}{2} * 8

68 = (HG + 12) * 4

Divide both sides by 4

\frac{68}{4} = \frac{(HG + 12) * 4}{4}

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6 0
3 years ago
Find the equation of the line tangent to the graph of
garik1379 [7]

Answer:

\displaystyle y=\frac{2\sqrt{3}}{15}x+\frac{\pi-2\sqrt{3}}{6}

Step-by-step explanation:

We want to find the equation of the line tangent to the graph of:

\displaystyle y=\sin^{-1}\big(\frac{x}{5}\big)\text{ at } x=\frac{5}{2}

So, we will find the derivative of our equation first. Applying the chain rule, we acquire that:

\displaystyle y^\prime=\frac{1}{\sqrt{1-(\frac{x}{5})^2}}\cdot\frac{1}{5}

Simplify:

\displaystyle y^\prime=\frac{1}{5\sqrt{1-\frac{x^2}{25}}}

We can factor out the denominator within the square root:

\displaystyle y^\prime =\frac{1}{5\sqrt{\frac{1}{25}\big(25-x^2)}}

Simplify:

\displaystyle y^\prime=\frac{1}{\sqrt{25-x^2}}

So, we can find the slope of the tangent line at <em>x</em> = 5/2. By substitution:

\displaystyle y^\prime=\frac{1}{\sqrt{25-(5/2)^2}}

Evaluate:

\displaystyle y^\prime=\frac{1}{\sqrt{75/4}}=\frac{1}{\frac{5\sqrt{3}}{2}}=\frac{2\sqrt{3}}{15}

We will also need the point at <em>x</em> = 5/2. Using our original equation, we acquire that:

\displaystyle y=\sin^{-1}(\frac{1}{2})=\frac{\pi}{6}

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\displaystyle y-\frac{\pi}{6}=\frac{2\sqrt{3}}{15}(x-\frac{5}{2})

Distribute:

\displaystyle y-\frac{\pi}{6}=\frac{2\sqrt{3}}{15}x+\frac{-\sqrt{3}}{3}

Isolate. Hence, our equation is:

\displaystyle y=\frac{2\sqrt{3}}{15}x+\frac{\pi-2\sqrt{3}}{6}

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