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lozanna [386]
3 years ago
12

The sparky valley barbershop quartet made appearances in Quakertown and Doleystown. The hotel charge before taxes in Doleystown

was $500 higher than Quakertown. The tax in Quakertown is 3% and Doleystown is 6%. The total hotel tax from the two places is $480. How much did each hotel charge the Quartet before taxes?
Mathematics
1 answer:
joja [24]3 years ago
5 0

Answer:

The hotel at Qakertown charged $5000 before tax while the hotel at Doleystown charged $5,500 before tax

Step-by-step explanation:

In this question, we have a musical band that visited two different towns which hotel charge different amounts at different tax rate. We are now given the total amount in taxes and now asked to calculate what is charged in each of the towns.

Let’s proceed!

Since we do not know the amount charged in each of the towns, let the amount charged at the hotel at Quakertown be $x and the amount charged at the hotel at Doleytown be $y.

Doley charged $500 more then what is charged at Quaker.

Mathematically;

y = 500 + x •••••••••••••••(I)

Let’s now work with the taxes;

Tax rate at Doley is 6% or say 0.06; the amount of tax paid at Doley is thus 0.06y

Tax rate at Quaker is 3% or say 0.03; the amount of tax paid at Quaker is thus 0.03x

Mathematically;

0.06y + 0.03x= 480 •••••••••(ii)

We can substitute i into ii

0.06(500 + x ) + 0.03x = 480

30+ 0.06x + 0.03x = 480

30 + 0.09x = 480

0.09x = 480 - 30

0.09x = 450

x = 450/0.09

x = $5,000

y = 500 + x

y = 500 + $5000 = $5,500

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Solve the inequality 2n – 1 &lt; 39​
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Answer:

n < 20​

Step-by-step explanation:

2n – 1 < 39​

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3 0
3 years ago
Use Newton’s Method to find the solution to x^3+1=2x+3 use x_1=2 and find x_4 accurate to six decimal places. Hint use x^3-2x-2=
luda_lava [24]

Let f(x) = x^3 - 2x - 2. Then differentiating, we get

f'(x) = 3x^2 - 2

We approximate f(x) at x_1=2 with the tangent line,

f(x) \approx f(x_1) + f'(x_1) (x - x_1) = 10x - 18

The x-intercept for this approximation will be our next approximation for the root,

10x - 18 = 0 \implies x_2 = \dfrac95

Repeat this process. Approximate f(x) at x_2 = \frac95.

f(x) \approx f(x_2) + f'(x_2) (x-x_2) = \dfrac{193}{25}x - \dfrac{1708}{125}

Then

\dfrac{193}{25}x - \dfrac{1708}{125} = 0 \implies x_3 = \dfrac{1708}{965}

Once more. Approximate f(x) at x_3.

f(x) \approx f(x_3) + f'(x_3) (x - x_3) = \dfrac{6,889,342}{931,225}x - \dfrac{11,762,638,074}{898,632,125}

Then

\dfrac{6,889,342}{931,225}x - \dfrac{11,762,638,074}{898,632,125} = 0 \\\\ \implies x_4 = \dfrac{5,881,319,037}{3,324,107,515} \approx 1.769292663 \approx \boxed{1.769293}

Compare this to the actual root of f(x), which is approximately <u>1.76929</u>2354, matching up to the first 5 digits after the decimal place.

4 0
2 years ago
Look at the figure. Name the postulate or theorem you can use to prove the train goes congruent?
Nikolay [14]

Let us recall parallelogram properties, which states that opposite angles of parallelogram are congruent.

We can see from graph that side US is parallel to TR and measure of angle U equals to measure of angle R, therefore, quadrilateral drawn in our given graph is a parallelogram.

Since we know that opposite sides of parallelogram are congruent. In our parallelogram UT=SR and US=TR.

In our triangle STU and triangle TSR side TS=TS by reflexive property of congruence.

Therefore, our triangles are congruent by SSS congruence.        

6 0
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