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3 years ago

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A car dealership increased the price of a certain car by 13%. The original price was $47400 what is the new price?

2 answers:

PilotLPTM [1.2K]3 years ago

7 0

Answer:

$53,562

Step-by-step explanation:

garik1379 [7]3 years ago

5 0

Answer:

Step-by-step explanation:

To get the new price of the car multiply the original price by the percent increase in decimal form +1. 11% in decimal form is 0.11 = 11/100. Add one to this and you get 1.11 = 1 + 0.11. So now multiply $49,800 * 1.11 = $55,278. If the problem had said they decreased the price by 11% you would use 1 - 0.11 = 0.89 because 89% is 11% less than 100%.

hope that helps my man

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Using the Addition Method, solve for y in the following system of linear equations: 4x – 2y = 6 2x + y = 7

Lady_Fox [76]

Answer:

y = 2

Step-by-step explanation:

Addition Method:

4x - 2y = 6

2x + y = 7

-2(2x + y) = -2(7)

4x - 2y = 6

-4x - 2y = -14

-4y = -8

y = 2

Substitution Method:

4x - 2y = 6

2x + y = 7

-2y = 6 - 4x

2y = 4x - 6

y = 2x - 3

2x + (2x - 3) = 7

4x - 3 = 7

4x = 10

x = 2.5

2(2.5) + y = 7

5 + y = 7

y = 2

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4 0

2 years ago

Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant

Viktor [21]

Answer:

Therefore $k= \frac{ln2 }{18}$ , A=184

Step-by-step explanation:

Given function is

$T(t)=230 -e^{-kt}$

where T(t) is the temperature in °C and t is time in minute and A and k are constants.

She noticed that after 18 minutes the temperature of the pie is 138°C

Putting T(t) =138°C and t= 18 minutes

$138=230 -Ae^{-k\times 18}$

$\Rightarrow -Ae^{-18k}=138-230$

$\Rightarrow Ae^{-18k}=92$ .....(1)

Again after 36 minutes it is 184°C

Putting T(t) =184°C and t= 36 minutes

$184=230-Ae^{-k\times 36}$

$\Rightarrow Ae^{-36k}=230-184$

$\Rightarrow Ae^{-36k}=46$ .......(2)

Dividing (2) by (1)

$\frac{Ae^{-36k}}{Ae^{-18k}}=\frac{46}{92}$

$\Rightarrow e^{-18k}=\frac{46}{92}$

Taking ln both sides

$ln e^{-18k}=ln\frac{46}{92}$

$\Rightarrow -18k =ln (\frac12)$

$\Rightarrow -18k= ln1-ln2$

$\Rightarrow k= \frac{ln2 }{18}$

Putting the value k in equation (1)

$Ae^{-18\frac{ln2}{18}}=92$

$\Rightarrow A e^{ln2^{-1}}=92$

$\Rightarrow A.2^{-1}=92$

$\Rightarrow \frac{A}{2}=92$

$\Rightarrow A= 92 \times 2$

⇒A= 184.

Therefore $k= \frac{ln2 }{18}$ , A=184

7 0

3 years ago

For the given decision algorithm, find how many outcomes are possible. HINT [See Example 1.] Step 1: Step 2: Alternative 1: 1 ou

Kruka [31]

Answer:

the total number of outcome in step 1 = 1+2+3= 6

the total number of outcome in step 2 = 2 +2+3 = 7

hence total number of outcome = 6 x7

= 42

Step-by-step explanation:

7 0

3 years ago

How to solve the problem, and complete answer

rusak2 [61]

Notice the picture below

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5 0

3 years ago

(a) Determine a cubic polynomial with integer coefficients which has $\sqrt[3]{2} + \sqrt[3]{4}$ as a root.

Keith_Richards [23]

Answer:

(a) $x\³ - 6x - 6$

(b) Proved

Step-by-step explanation:

Given

$r = $\sqrt[3]{2} + \sqrt[3]{4}$$ --- the root

Solving (a): The polynomial

A cubic function is represented as:

$f = (a + b)^3$

Expand

$f = a^3 + 3a^2b + 3ab^2 + b^3$

Rewrite as:

$f = a^3 + 3ab(a + b) + b^3$

The root is represented as:

$r=a+b$

By comparison:

$a = $\sqrt[3]{2}$

$b = \sqrt[3]{4}$$

So, we have:

$f = ($\sqrt[3]{2})^3 + 3*$\sqrt[3]{2}*\sqrt[3]{4}$*($\sqrt[3]{2} + \sqrt[3]{4}$) + (\sqrt[3]{4}$)^3$

Expand

$f = 2 + 3*$\sqrt[3]{2*4}*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4$

$f = 2 + 3*$\sqrt[3]{8}*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4$

$f = 2 + 3*2*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4$

$f = 2 + 6($\sqrt[3]{2} + \sqrt[3]{4}$) + 4$

Evaluate like terms

$f = 6 + 6($\sqrt[3]{2} + \sqrt[3]{4}$)$

Recall that: $r = $\sqrt[3]{2} + \sqrt[3]{4}$$

So, we have:

$f = 6 + 6r$

Equate to 0

$f - 6 - 6r = 0$

Rewrite as:

$f - 6r - 6 = 0$

Express as a cubic function

$x^3 - 6x - 6 = 0$

Hence, the cubic polynomial is:

$f(x) = x^3 - 6x - 6$

Solving (b): Prove that r is irrational

The constant term of $x^3 - 6x - 6 = 0$ is -6

The divisors of -6 are: -6,-3,-2,-1,1,2,3,6

Calculate f(x) for each of the above values to calculate the remainder when f(x) is divided by any of the above values

$f(-6) = (-6)^3 - 6*-6 - 6 = -186$

$f(-3) = (-3)^3 - 6*-3 - 6 = -15$

$f(-2) = (-2)^3 - 6*-2 - 6 = -2$

$f(-1) = (-1)^3 - 6*-1 - 6 = -1$

$f(1) = (1)^3 - 6*1 - 6 = -11$

$f(2) = (2)^3 - 6*2 - 6 = -10$

$f(3) = (3)^3 - 6*3 - 6 = 3$

$f(6) = (6)^3 - 6*6 - 6 = 174$

For r to be rational;

The divisors of -6 must divide f(x) without remainder

i.e. Any of the above values must equal 0

<em>Since none equals 0, then r is irrational</em>

3 0

2 years ago