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Dennis_Churaev [7]

2 years ago

6

A collectible action figure sold for 193% of its original price. Write this percent as a decimal and as a mixed number or fracti

on in simplest form.

1 answer:

Lyrx [107]2 years ago

8 0

193%...(to turn to decimal divide by 100) = 1.93
1.93 as a mixed number = 1 93/100

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Consider the following quadratic equation x^2 =4x -5. How many solutions does it have?

aleksandrvk [35]

The answer for your question is C.

5 0

3 years ago

Read 2 more answers

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

Leni [432]

Answer: $\dfrac{2x^2-1}{x(x^2-1)}$

Step-by-step explanation:

The given function : $y=\ln(x(x^2 - 1)^{\frac{1}{2}})$

$\Rightarrow\ y=\ln x+\ln (x^2-1)^{\frac{1}{2}}$ [ $\because \ln(ab)=\ln a +\ln b$ ]

$\Rightarrow y=\ln x+\dfrac{1}{2}\ln (x^2-1)}$ [ $\because \ln(a)^n=n\ln a$ ]

Now , Differentiate both sides with respect to x , we will get

$\dfrac{dy}{dx}=\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})\dfrac{d}{dx}(x^2-1)$ (By Chain rule)

[Note : $\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}$ ]

$\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x-0)$

[ $\because \dfrac{d}{dx}(x^n)=nx^{n-1}$ ]

$=\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x) = \dfrac{1}{x}+\dfrac{x}{x^2-1}\\\\\\=\dfrac{(x^2-1)+(x^2)}{x(x^2-1)}\\\\\\=\dfrac{2x^2-1}{x(x^2-1)}$

Hence, the derivative of the given function is $\dfrac{2x^2-1}{x(x^2-1)}$ .

8 0

3 years ago

The perimeter of a shape shows a car racing track, how far does a car travel in a race which consists of 23 laps 1.2km , 1.2 km

inna [77]

Answer:

110.4km

Step-by-step explanation:

1 lap is equal to 4.8

4.8×23=110. 4

(Need someone to confirm)

6 0

2 years ago

Which table shows a proportional relationship between x and y?

Semenov [28]

Answer:

B

Step-by-step explanation:

A proportional relationship is a relationship which crosses through the origin (0,0) and which has a proportional constant. We can determine this either by finding (0,0) where x=0 and y=0 in the table or by dividing y/x. None of the tables contain (0,0) so we will divide y by x. We are looking for a table which when each y is divided by its x we have the same constant appearing.

<u>Table A</u>

$\frac{12}{3} \neq \frac{15}{6} \neq \frac{18}{8} \neq \frac{20}{10}$

These fractions are not equal. This is not proportional.

<u>Table B</u>

$\frac{0.5}{1}=\frac{1}{2} =\frac{3.5}{7} =\frac{4}{8}$

These fractions are equal and each shows the numerator to be half of the denominator. This is proportional.

<u>Table C</u>

$\frac{1}{3}\neq \frac{2.5}{7.5} \neq \frac{4}{15} \neq \frac{6}{20}$

These fractions are not equal. This is not proportional.

<u>Table D</u>

$\frac{7}{2} \neq \frac{9}{3}\neq \frac{11}{4} \neq \frac{13}{5}$

These fractions are not equal. This is not proportional.

3 0

3 years ago

Two cars simultaneously left Points A and B and headed towards each other, and met after 2 hours and 45 minutes. The distance be

zheka24 [161]

<h2>Hello!</h2>

The answer is:

$FirstCarSpeed=41mph\\SecondCarSpeed=55mph$

<h2>Why?</h2>

To calculate the speed of the cars, we need to write two equations in order to create a relation between the two speeds and be able to isolate one in function of the other.

So, let be the first car speed "x" and the second car speed "y", writing the equations we have:

For the first car:

$x_{FirstCar}=x_o+v*t$

For the second car:

We know that the speed of the second car is the speed of the first car plus 14 mph, so:

$x_{SecondCar}=x_o+(v+14mph)*t$

Now, we already know that both cars met after 2 hours and 45 minutes, meaning that positions will be the same at that moment, and the distance between A and B is 264 miles, so, we can calculate the relative speed between them:

If the cars are moving towards each other the relative speed will be:

$RelativeSpeed=FirstCarSpeed-(-SecondCarspeed)\\\\RelativeSpeed=x-(-x-14mph)=2x+14mph$

Then, since we know that they covered a combined distance which is equal to 264 miles of distance in 2 hours + 45 minutes, we have:

$2hours+45minutes=120minutes+45minutes=165minutes\\\\\frac{165minutes*1hour}{60minutes}=2.75hours$

Writing the equation, we have:

$264miles=(2x+14mph)*t\\\\264miles=(2x+14mph)*2.75hours\\\\2x+14mph=\frac{264miles}{2.75hours}\\\\2x=96mph-14mph\\\\x=\frac{82mph}{2}=41mph$

We have that the speed of the first car is equal to 41 mph.

Now, for the second car we have that:

$SecondCarSpeed=FirstCarSpeed+14mph\\\\SecondCarSpeed=41mph+14mph=55mph$

Hence, we have that:

$FirstCarSpeed=41mph\\SecondCarSpeed=55mph$

Have a nice day!

4 0

3 years ago

Read 2 more answers