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denpristay [2]
2 years ago
14

Which of the following expressions are equivalent to 48a^3-75a? Select all that apply. Answer 1) 3(48a^3-75a). Answer 2) 3a(16a^

2-25) Answer 3)3a(4a+5)(4a+5) Answer 4) 3a(4a+5)(4a-5) Answer 5) -3a(25-16a^2) Answer 6) -3a(5-4a)(5+4a)
Mathematics
2 answers:
Ainat [17]2 years ago
8 0
The answer is 4) 3a(4a+5)(4a-5)
lapo4ka [179]2 years ago
7 0

Answer: The answer is (2), (4), (5) and (6).

Step-by-step explanation: The given expression is

E=48a^3-75a.

We are to select the correct expressions from the given options that are equivalent to the above expression.

We have the following forms of the give expressipn:

E\\\\=48a^3-75a\\\\=3a(16a^2-25)\\\\=3a\left((4a)^2-5^2\right)\\\\=3(4a+5)(4a-5)\\\\=-3a(5+4a)(5-4a)\\\\=-3a(25-16a^2).

Thus, (2), (4), (5) and (6) are the correct options.

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Explain the tangent line problem
ale4655 [162]

The Tangent Line Problem  1/3How do you find the slope of the tangent line to a function at a point Q when you only have that one point? This Demonstration shows that a secant line can be used to approximate the tangent line. The secant line PQ connects the point of tangency to another point P on the graph of the function. As the distance between the two points decreases, the secant line becomes closer to the tangent line.
8 0
1 year ago
Jenna offers regular haircuts for $25 and haircuts plus coloring for $42.This weekend had a total of 24 clients and Jenna earned
shepuryov [24]
<u><em>Answer:</em></u>
Jenna did 16 regular haircuts
Jenna did 8 haircuts with coloring

<u><em>Explanation:</em></u>
Assume that the number of regular haircuts is x and the number of haircuts plus coloring is y

<u>We are given that:</u>
<u>1- Jenna did a total of 24 clients, this means that:</u>
x + y = 24
This can be rewritten as:
x = 24 - y ...............> equation I

<u>2- regular haircuts cost $25, haircuts plus coloring cost $42 and she earned a total of $736. This means that:</u>
25x + 42y = 736 ..........> equation II

<u>Substitute with equation I in equation II and solve for y as follows:</u>
25x + 42y = 736
25(24-y) + 42y = 736
600 - 25y + 42y = 736
17y = 136
y = 8

<u>Substitute with y in equation I to get x as follows:</u>
x = 24 - y
x = 24 - 8
x = 16

<u>Based on the above:</u>
Jenna did 16 regular haircuts
Jenna did 8 haircuts with coloring

Hope this helps :)
5 0
2 years ago
The physician tells a woman who usually drinks 5 cups of coffee daily to cut down on her coffee consumption by 75%. If this woma
Maru [420]

Answer:

Step-by-step explanation:

5 cups per day.....1 cup = 8oz.....so 5 cups = (5 * 8) = 40 oz

cut down by 75%....means ur only drinking 25%

25% of 40 oz =

0.25(40) = 10 oz per day <===

6 0
2 years ago
How much money do winners go home with from the television quiz show Jeopardy? To determine an answer, a random sample of winner
Liula [17]

Answer:

The 95% confidence interval would be given by (14444.04;33657.30)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Data: $26,650 $6,060 $52,820 $8,490 $13,660$25,840 $49,840 $23,790 $51,480 $18,960$990 $11,450 $41,810 $21,060 $7,860

We can calculate the mean and the deviation from these data with the following formulas:

\bar X= \frac{\sum_{i=1}^n x_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}

\bar X=24050.67 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=17386.13 represent the sample standard deviation

n=15 represent the sample size  

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=15-1=14

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,14)".And we see that t_{\alpha/2}=2.14

Now we have everything in order to replace into formula (1):

24050.67-2.14\frac{17386.13}{\sqrt{15}}=14444.04    

24050.67+2.14\frac{17386.13}{\sqrt{15}}=33657.30

So on this case the 95% confidence interval would be given by (14444.04;33657.30)    

6 0
2 years ago
Consider the initial value problem: 2ty′=8y, y(−1)=1. Find the value of the constant C and the exponent r so that y=Ctr is the s
VikaD [51]

The correct question is:

Consider the initial value problem

2ty' = 8y, y(-1) = 1

(a) Find the value of the constant C and the exponent r such that y = Ct^r is the solution of this initial value problem.

b) Determine the largest interval of the form a < t < b on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution.

c) What is the actual interval of existence for the solution obtained in part (a) ?

Step-by-step explanation:

Given the differential equation

2ty' = 8y

a) We need to find the value of the constant C and r, such that y = Ct^r is a solution to the differential equation together with the initial condition y(-1) = 1.

Since Ct^r is a solution to the initial value problem, it means that y = Ct^r satisfies the said problem. That is

2tdy/dt - 8y = 0

Implies

2td(Ct^r)/dt - 8(Ct^r) = 0

2tCrt^(r - 1) - 8Ct^r = 0

2Crt^r - 8Ct^r = 0

(2r - 8)Ct^r = 0

But Ct^r ≠ 0

=> 2r - 8 = 0 or r = 8/2 = 4

Now, we have r = 4, which implies that

y = Ct^4

Applying the initial condition y(-1) = 1, we put y = 1 when t = -1

1 = C(-1)^4

C = 1

So, y = t^4

b) Let y = F(x,y)................(1)

Suppose F(x, y) is continuous on some region, R = {(x, y) : x_0 − δ < x < x_0 + δ, y_0 −ę < y < y_0 + ę} containing the point (x_0, y_0). Then there exists a number δ1 (possibly smaller than δ) so that a solution y = f(x) to (1) is defined for x_0 − δ1 < x < x_0 + δ1.

Now, suppose that both F(x, y)

and ∂F/∂y are continuous functions defined on a region R. Then there exists a number δ2

(possibly smaller than δ1) so that the solution y = f(x) to (1) is

the unique solution to (1) for x_0 − δ2 < x < x_0 + δ2.

c) Firstly, we write the differential equation 2ty' = 8y in standard form as

y' - (4/t)y = 0

0 is always continuous, but -4/t has discontinuity at t = 0

So, the solution to differential equation exists everywhere, apart from t = 0.

The interval is (-infinity, 0) n (0, infinity)

n - means intersection.

7 0
1 year ago
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