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4 years ago

A student finds the slope of the line between (20,2) and (1,17). She writes 1-17/1-20 What mistake did she make?

Mathematics

2 answers:

Ivahew [28]4 years ago

7 0

Answer:

i would say its c

Step-by-step explanation:

i mean cause the points say (20,1) and (1,17) she mixed up the x and y value with (20,1) she put 1-20 and im guessing not 20-1 as it stands for the points on a graph

murzikaleks [220]4 years ago

4 0

I think that the correct answer would be A

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A sequence can be generated by using an = an-1 + 4, where a1 = 6 and n is a whole number greater than 1. What are the first four

Firlakuza [10]

Answer:

The first four terms are 4,8,12,16

Step-by-Step Explanation:

The given relation between the nth term and it's previous term is given by:

$\boxed{an = an - 1 + 4}$

GiveN:

a1 = 4

Now finding the other three terms of the AP with the given relation.

$a2 = a1 + 4$

Putting a1 = 4,

$a2 = 4 + 4 = 8$

Now, Third term:

$a3 = a2 + 4$

Putting a2 = 8,

$a3 = 8 + 4 = 12$

Now, Fourth term:

$a4 = a3 + 4$

Putting a3 = 12,

$a4 = 12 + 4 = 16$

Hence, The first four terms of the AP is 4, 8, 12 & 16.

7 0

3 years ago

If I make 11.50 an hour working full time (40 hours a week) how much money do i make a year after deducting 60 dollars of taxes

Kazeer [188]

Given:
hourly rate = 11.50
hours worked in a week = 40 hours
taxes per moth = 60

Every month may be 4 weeks or 5 weeks but we know that 1 year has 52.1429 weeks.

Let us compute for the whole year:

Earnings:
11.50 /1 hour * 40 hours/1week * 52.1429 weeks/1year
= 11.50 * 40 * 52.1429/year
= 23,985.734 per year

Taxes:
60 / per month * 12 months / 1 year
= 60 * 12/year
= 720 per year

Net earning per year
23,985.734 - 720 = 23,264.734

Net earning per month
23,264.734 ÷ 12 months = 1,938.81

You will earn 23,264.73 annually from a monthly earning of 1,938.81

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4 years ago

Use the given graph. Determine the period of the function.

vichka [17]

The "period" of a repeating function is the change in x between consecutive identical values...in this case 3 units.

4 0

3 years ago

Read 2 more answers

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

aliya0001 [1]

The Lagrangian

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^4+y^4+z^4-13)$

has critical points where the first derivatives vanish:

$L_x=2x+4\lambda x^3=2x(1+2\lambda x^2)=0\implies x=0\text{ or }x^2=-\dfrac1{2\lambda}$

$L_y=2y+4\lambda y^3=2y(1+2\lambda y^2)=0\implies y=0\text{ or }y^2=-\dfrac1{2\lambda}$

$L_z=2z+4\lambda z^3=2z(1+2\lambda z^2)=0\implies z=0\text{ or }z^2=-\dfrac1{2\lambda}$

$L_\lambda=x^4+y^4+z^4-13=0$

We can't have $x=y=z=0$ , since that contradicts the last condition.

(0 critical points)

If two of them are zero, then the remaining variable has two possible values of $\pm\sqrt[4]{13}$ . For example, if $y=z=0$ , then $x^4=13\implies x=\pm\sqrt[4]{13}$ .

(6 critical points; 2 for each non-zero variable)

If only one of them is zero, then the squares of the remaining variables are equal and we would find $\lambda=-\frac1{\sqrt{26}}$ (taking the negative root because $x^2,y^2,z^2$ must be non-negative), and we can immediately find the critical points from there. For example, if $z=0$ , then $x^4+y^4=13$ . If both $x,y$ are non-zero, then $x^2=y^2=-\frac1{2\lambda}$ , and

$xL_x+yL_y=2(x^2+y^2)+52\lambda=-\dfrac2\lambda+52\lambda=0\implies\lambda=\pm\dfrac1{\sqrt{26}}$

$\implies x^2=\sqrt{\dfrac{13}2}\implies x=\pm\sqrt[4]{\dfrac{13}2}$

and for either choice of $x$ , we can independently choose from $y=\pm\sqrt[4]{\frac{13}2}$ .

(12 critical points; 3 ways of picking one variable to be zero, and 4 choices of sign for the remaining two variables)

If none of the variables are zero, then $x^2=y^2=z^2=-\frac1{2\lambda}$ . We have

$xL_x+yL_y+zL_z=2(x^2+y^2+z^2)+52\lambda=-\dfrac3\lambda+52\lambda=0\implies\lambda=\pm\dfrac{\sqrt{39}}{26}$

$\implies x^2=\sqrt{\dfrac{13}3}\implies x=\pm\sqrt[4]{\dfrac{13}3}$

and similary $y,z$ have the same solutions whose signs can be picked independently of one another.

(8 critical points)

Now evaluate $f$ at each critical point; you should end up with a maximum value of $\sqrt{39}$ and a minimum value of $\sqrt{13}$ (both occurring at various critical points).