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diamong [38]
2 years ago
15

Amit solved the equation StartFraction 5 over 12 EndFraction = Negative StartFraction x over 420 EndFraction for x using the ste

ps shown below. What was Amit’s error?
StartFraction 5 over 12 EndFraction = Negative StartFraction x over 420 EndFraction. StartFraction 5 over 12 EndFraction (420) = Negative StartFraction x over 420 EndFraction (420). X = 175.

Mathematics
2 answers:
melomori [17]2 years ago
7 0

Answer:

d

-AKA-

The product of StartFraction 5 over 12 EndFraction and –420 should have been the value of x.

-AKA-

The product of 5/12 and –420 should have been the value of x.

Step-by-step explanation:

i did it on edge 2020

see..........

hope it helps        :)

mel-nik [20]2 years ago
3 0

Answer:

D

Step-by-step explanation:

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Given the graphed function below, which of the following ordered pairs are found on the inverse function?
Dmitrij [34]

Answer:

Its C (-9,-2)(-2,-1)(-1,0)(0,1)(7,2)

Step-by-step explanation:

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Santiago hopes to buy a 4 horse trailer for about $12000. Describe all the numbers that when rounded to the nearest hundred are
jok3333 [9.3K]
To round a number to the nearest hundred, we count two places to the left of the decimal point, or from the last digit if the number is a whole number.

If the second digit from the last digit is upto 5, we add 1 to the preceding digit and we complete the last two numbers with zeros.

Therefore, any number from 11,950 to 12,049, will result to 12,000 when rounded to the nearest hundred.
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2 years ago
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9,290 is the value of the first 9 ten times as great as the value of the second 9?
Semenov [28]

Answer:

No

Step-by-step explanation:

The first nine in on the place of the thousands, while the second nine is on the place of the tens. so the first nine is a hundred times as great as the second nine

8 0
2 years ago
Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u
soldier1979 [14.2K]

x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}

\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}

\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}

Therefore:

\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]

Recall that at t=25, p(25)=\dfrac { 5800 } {13} \approx 446.15

Therefore:

\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009

The quantity demanded per month of the tablet computers will be changing at a rate of 12 tablet computers/month correct to 1 decimal place.

8 0
2 years ago
Expresión for the calculation 5 times the number 25 and then subtract the quotient of 14 and 7
Rudik [331]

Answer:

(5*25) - (14/7)

<em>* means multiply and / means divide</em>


7 0
2 years ago
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