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1 year ago

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Which function will produce the same graph as the function shown below? v= (x+1)(x+3) O A. v=x2 + 4x +3 O B.V=x2 + 4x-3 O c. v+3

= x(x-1) O D.V-3=(x - 1)?

1 answer:

Strike441 [17]1 year ago

7 0

Given function is

$y=(x+1)(x+3)$

If we expand the function by multiplying, we have,

$y=x^2+4x+3$

Thus, we cans ay that the above two functiosn are equivalent and hence, they will produce the same graph.

Therefore, the correct asnwer is (A)

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Which equation can be used to find A, the area of the parallelogram shown?

aniked [119]

<h3>Answer: A = 6 x 7 </h3>

Explanation:

The area of a parallelogram is equal to base times height.

The base and height are always perpendicular to one another. Usually the base is horizontal along the ground (if not then you'd need to rotate your paper until it is, but that step is optional).

The base is 6 meters and the height is 7 meters.

Therefore, area = base x height = 6 x 7

The side of 11 meters is not used at all. It's likely put in there as a distraction.

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2 years ago

Read 2 more answers

ANYONE, PLEASE ANSWER ALL THESE QUESTIONS AND I WILL MARK HIM/HER BRAINLIEST + 57 PTS.

Lapatulllka [165]

Answer:

1.B

2.C

3.D

4.

a)1/9

b)5/12

5.2/5

Step-by-step explanation:

1.

Since there are 15 total outcomes of which 6 are desirable, the probability of randomly getting a red ball is 5/16 or choice B.

2.

Since there are a total of 8+7+6=21 outcomes, out of which 7 are neither red nor green, there is a 7/21=1/3 probability of that outcome, or answer choice C.

3.

Out of the 52 possible outcomes, 12 are face cards. Therefore, the probability of drawing a face card is 12/52=3/13, or answer choice D.

4.

a)

For each die, there are a total of 6 possible outcomes. This means that in total there are 6*6=36 possible combinations. Now, let's do some casework.

If the first die is a 3, then the second die must be a 6. If the first die is a 4, the second die must be a 5. If the first die is a 5, then the second die must be a 4. And if the first die is a 6, then the second die must be a 3. You cannot have any numbers less than 3, because then you would not be able to make the number 9. Therefore, there is a 4/36=1/9 chance of getting a sum of 9.

b)

All of the possible prime numbers that can be formed with the dice are 2, 3, 5, 7, and 11. For 2, there is only one possible situation where this can happen. For 3, there are 2 possible. For 5, there is 4-1, 3-2, 2-3, and 1-4, or 4 possibilities. For 7, there is 1-6, 2-5, 3-4, 4-3, 5-2, and 6-1, or 6 total possibilities. Finally, for 11, there is only 5-6 and 6-5, or 2 possible. In total, the probability of getting a prime sum is 15/36=5/12.

5.

The only numbers which are a multiple of 3 or 5 from 1 to 20 are: 3, 5, 9, 10, 12, 15, 18, and 20. This means that out of the 20 total possible options, 8 of them are a multiple of 3 or 5, which is a total chance of 2/5.

Hope this helps!

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3 years ago

A car and a truck are traveling north on a highway. The truck has a speed of 42 mph and the car has a speed of 37 mph. If the tr

irga5000 [103]

Answer:

5 mph north

Step-by-step explanation:

The relative velocity of the truck with respect to the car is ...

relative velocity = truck velocity - car velocity

relative velocity = 42 mph north - 37 mph north

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4 years ago

86) If 3 glasses of lemonade can be made from 8 lemons, exactly how

kkurt [141]

Answer:

37 and a 1/3

Step-by-step explanation:

3 glasses of lemonade=8lemons

1 glass of lemonade=2.66666667

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3 years ago

How do you solve this limit of a function math problem?

hram777 [196]

If you know that

$e=\displaystyle\lim_{x\to\pm\infty}\left(1+\frac1x\right)^x$

then it's possible to rewrite the given limit so that it resembles the one above. Then the limit itself would be some expression involving $e$ .

For starters, we have

$\dfrac{3x-1}{3x+3}=\dfrac{3x+3-4}{3x+3}=1-\dfrac4{3x+3}=1-\dfrac1{\frac34(x+1)}$

Let $y=\dfrac34(x+1)$ . Then as $x\to\infty$ , we also have $y\to\infty$ , and

$2x-1=2\left(\dfrac43y-1\right)=\dfrac83y-2$

So in terms of $y$ , the limit is equivalent to

$\displaystyle\lim_{y\to\infty}\left(1-\frac1y\right)^{\frac83y-2}$

Now use some of the properties of limits: the above is the same as

$\displaystyle\left(\lim_{y\to\infty}\left(1-\frac1y\right)^{-2}\right)\left(\lim_{y\to\infty}\left(1-\frac1y\right)^y\right)^{8/3}$

The first limit is trivial; $\dfrac1y\to0$ , so its value is 1. The second limit comes out to

$\displaystyle\lim_{y\to\infty}\left(1-\frac1y\right)^y=e^{-1}$

To see why this is the case, replace $y=-z$ , so that $z\to-\infty$ as $y\to\infty$ , and

$\displaystyle\lim_{z\to-\infty}\left(1+\frac1z\right)^{-z}=\frac1{\lim\limits_{z\to-\infty}\left(1+\frac1z\right)^z}=\frac1e$

Then the limit we're talking about has a value of

$\left(e^{-1}\right)^{8/3}=\boxed{e^{-8/3}}$

# # #

Another way to do this without knowing the definition of $e$ as given above is to take apply exponentials and logarithms, but you need to know about L'Hopital's rule. In particular, write

$\left(\dfrac{3x-1}{3x+3}\right)^{2x-1}=\exp\left(\ln\left(\frac{3x-1}{3x+3}\right)^{2x-1}\right)=\exp\left((2x-1)\ln\frac{3x-1}{3x+3}\right)$

(where the notation means $\exp(x)=e^x$ , just to get everything on one line).

Recall that

$\displaystyle\lim_{x\to c}f(g(x))=f\left(\lim_{x\to c}g(x)\right)$

if $f$ is continuous at $x=c$ . $\exp(x)$ is continuous everywhere, so we have

$\displaystyle\lim_{x\to\infty}\left(\frac{3x-1}{3x+3}\right)^{2x-1}=\exp\left(\lim_{x\to\infty}(2x-1)\ln\frac{3x-1}{3x+3}\right)$

For the remaining limit, write

$\displaystyle\lim_{x\to\infty}(2x-1)\ln\frac{3x-1}{3x+3}=\lim_{x\to\infty}\frac{\ln\frac{3x-1}{3x+3}}{\frac1{2x-1}}$

Now as $x\to\infty$ , both the numerator and denominator approach 0, so we can try L'Hopital's rule. If the limit exists, it's equal to

$\displaystyle\lim_{x\to\infty}\frac{\frac{\mathrm d}{\mathrm dx}\left[\ln\frac{3x-1}{3x+3}\right]}{\frac{\mathrm d}{\mathrm dx}\left[\frac1{2x-1}\right]}=\lim_{x\to\infty}\frac{\frac4{(x+1)(3x-1)}}{-\frac2{(2x-1)^2}}=-2\lim_{x\to\infty}\frac{(2x-1)^2}{(x+1)(3x-1)}=-\frac83$

and our original limit comes out to the same value as before, $\exp\left(-\frac83\right)=\boxed{e^{-8/3}}$ .

3 0

3 years ago