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Ahat [919]
3 years ago
14

3.

Mathematics
1 answer:
Lerok [7]3 years ago
5 0

3. Here we use the formula

A = P ( 1 + \frac{r}{n} )^ {nt}

And we have the following values given

P = $23400, r = 3%=0.03,n =2, t=10 years

So we will get

A = 23400(1+ \frac{0.03}{2} )^{2*10}
\\
A = 23400( \frac{2.03}{2} )^{20} = $31516.52

Question 4.

In this question , we have

P = $2310, R = 3.5% = 0.035 ,
\\
Number \ of \ days \ from \ april \12 to July \ 5 = 30+31+23 = 84 days

A = 2310(1+ \frac{0.035}{365})^{84}  = $2328.68

Interest is the difference of amount and principal, that is

I = 2328.68-2310 = $18.68

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The distance by road from town A to town B is 257 km. What is 50% of that distance?
Marina86 [1]

Answer: 128.5km

Step-by-step explanation:

Since we are given the information that the distance by road from town A to town B is 257 km. To get 50% of the distance, we simply have to multiply the distance given by 50%. This will be:

= 50% × 257km

= 50/100 × 257km

= 0.5 × 257km

= 128.5km

Therefore, 50% of the distance is 128.5km.

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2 years ago
Solve the SYSTEM: <br> 2x - 3y = 12<br> 3x - 2y = 13
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Answer:

x =69 y= 420

Step-by-step explanation:

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What does the slope of graph show about the Hardware stores selling price for the rope?
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If the price is increasing or decreasing

Step-by-step explanation:

The slope is a visual representation of the cost over time.

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Thad attempted 25 free throws and made 17 of them while Amy attempted 30 and made 20. Who shot the higher percentage?
ahrayia [7]

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2 years ago
1. if csc β = 7/3 and cot β = - 2√10 / 3, Find sec β
slava [35]

Step-by-step explanation:

1.

\tan \beta  =  \frac{1}{ \cot \beta }  =  -  \frac{3}{2 \sqrt{10} }  =  -  \frac{3 \sqrt{10} }{20}

\csc \beta  \tan \beta  =  \frac{1}{ \cos \beta  }  =  \sec \beta

Therefore,

\sec \beta  = ( \frac{7}{3} )( -  \frac{3 \sqrt{10} }{20} ) =  -  \frac{7 \sqrt{10} }{20}

2.

\csc y =  \frac{1}{ \sin y}  =  -  \frac{ \sqrt{6} }{2}

=  >  \sin y =  -  \frac{ \sqrt{6} }{3}

Use the identity

\cos y =   \sqrt{1 -  \sin ^{2} y}    \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \\ =  \sqrt{1 -  {( -  \frac{ \sqrt{6} }{3}) }^{2} }  =  -  \frac{ \sqrt{3} }{3}

We chose the negative value of the cosine because of the condition where cot y > 0. Otherwise, choosing the positive root will yield a negative cotangent value. Now that we know the sine and cosine of y, we can now solve for the tangent:

\tan \beta  =  \frac{ \sin y}{ \cos y} =( -  \frac{ \sqrt{6} }{3} )( -  \frac{3}{ \sqrt{3} } ) =  \sqrt{2}

3. Recall that sec x = 1/cos x, therefore cos x = 5/6. Solving for sin x,

\sin x =   \sqrt{1 -  \cos ^{2} x} =  \sqrt{ \frac{11}{6} }

Solving for tan x:

\tan x =  \frac{ \sin x}{ \cos x}  =  (\frac{ \sqrt{11} }{ \sqrt{6} } )( \frac{6}{5} ) =  \frac{ \sqrt{66} }{5}

5 0
3 years ago
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