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anyanavicka [17]
3 years ago
11

After three years, Elsie’s car had lost 1/3 of its original value. Two years later, it had lost an additional 1/4 of its origina

l value. If she bought the car for 3,600 dollars, what was the price it was worth after 5 years?
Mathematics
1 answer:
horrorfan [7]3 years ago
8 0

Answer:

1,500 dollars

Step-by-step explanation:

3,600 - 1/3(3600)=2,400

2,400 -1/4(3600)=$1,500

Hope this helps!

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ArbitrLikvidat [17]
When reflecting about the y axis, you x coordinate changes sign, so your answer would be D) (-2;-4)

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3 years ago
Solve the following system of equations: y=6-x<br> y=x^2-6x+6
nevsk [136]

Answer:

(0, 6 ) and (5, 1 )

Step-by-step explanation:

Given the 2 equations

y = 6 - x → (1)

y = x² - 6x + 6 → (2)

Substitute y = x² - 6x + 6 into (1)

x² - 6x + 6 = 6 - x ( subtract 6 - x from both sides )

x² - 5x = 0 ← factor out x from each term on the left side

x(x - 5) = 0

Equate each factor to zero and solve for x

x = 0

x - 5 = 0 ⇒ x = 5

Substitute these values into (1) for corresponding value of y

x = 0 : y = 6 - 0 = 6 ⇒ (0, 6 )

x = 5 : y = 6 - 5 = 1 ⇒ (5, 1 )

4 0
3 years ago
A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 meters
In-s [12.5K]

Answer:

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

Step-by-step explanation:

Given that,

A person stand 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 m/s.

From Pythagorean Theorem,

(The distance between car and person)²= (The distance of the car from intersection)²+ (The distance of the person from intersection)²+

Assume that the distance of the car from the intersection and from the person be x and y at any time t respectively.

∴y²= x²+10²

\Rightarrow y=\sqrt{x^2+100}

Differentiating with respect to t

\frac{dy}{dt}=\frac{1}{2\sqrt{x^2+100}}. 2x\frac{dx}{dt}

\Rightarrow \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}. \frac{dx}{dt}

Since the car driving towards the intersection at 13 m/s.

so,\frac{dx}{dt}=-13

\therefore \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}.(-13)

Now

\therefore \frac{dy}{dt}|_{x=24}=\frac{24}{\sqrt{24^2+100}}.(-13)

               =\frac{24\times (-13)}{\sqrt{676}}

               =\frac{24\times (-13)}{26}

               = -12 m/s

Negative sign denotes the distance between the car and the person decrease.

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

8 0
3 years ago
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