MATHEMATICS HIGH SCHOOL

Can somebody please help me with this question. Giving lots of points.

Answers

Answer 1

Answer:

Answer:

x-multiple choice

y- short response

23x+10y=86

28x+5y=76 /×(-2)

23x+10y=86

-56x-10y=-152

-33x=-66

x=2

10y=86-46

10y=40

y=4

multiple choice=2 points

short responce =4points

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Find f(a-2) when f(x)=x^2-1I forgot how to do these so please show all steps

Solve:-

7^5

Thanks!!!!!!!!!!!!!

Answers

These are exponents. To solve an exponent you need to multiply the base it self how many times its says in the power.

a²

a = base
2 = power

7^5 = 7 × 7 × 7 × 7 × 7 = 16807
7^5 = 16807

Solutions

7^5

=7 x 7 x 7 x 7 x 7

=16807

Z^5=-7776iFind the solution of the following equation whose argument is strictly between 270 and 360 degrees

Answers

Answer:

Z=+6

Step-by-step explanation:

Z^5=-7776i

Let's note that i I mathematics means negative one i.e

i = -1

So the equation is equal to

Z^5=-*-(7776)

Z^5 = 7776

Z= 5√7776

Since it's a divisible by 6

It's giving us a clue that 6 it's the answer.

Ok let's check the 5th root of 7776 in our calculator.

Z=+6

+6 is the solution to the equation

Z^5=-7776i

the function f is given by f(x) = 2x+3/x+1. Which of the following statements are true? i. the graph of f has a horizontal asymptote at y = 2 because lim x->[infinity] f(x) = 2 ii. the graph of f has a horizontal asymptote at y = 2 because lim x->[infinity] f(x) = 2 iii. the graph of f has a vertical asymptote at x = -1 because lim x->-1 f(x) = [infinity] a. I only b. Ill only c. I and Il only d. I, II, and III

Answers

Answer:

1. The graph of f has a horizontal asymptote at y -> 2 because lim f (x) = 2. 2 because lim f (x) = 2.

Step-by-step explanation:

Answer:

Hi,

Answer : option D

Step-by-step explanation:

the function f is given by f(x) = 2x+3/x+1.

Which of the following statements are true?

i. the graph of f has a horizontal asymptote at y = 2 because lim x->[infinity] f(x) = 2 is TRUE

ii. the graph of f has a horizontal asymptote at y = 2 because lim x->[infinity] f(x) = 2 is TRUE since I=II

iii. the graph of f has a vertical asymptote at x = -1 because lim x->-1 f(x) = [infinity] is TRUE

a. I only

b. Ill only

c. I and Il only

d. I, II, and III

A school play earned over 250 from selling tickets which cost $11 each what's the inqyeality model

Answers

250/11 All you need to do divided the 250 by 11 to get the answer :-)

Skye's two new aquariums each hold exactly 200 gallons of water. One aquarium will hold small fish and the other will hold large fish. Now hew needs new fish for his aquariums. She will buy 5 small fish for every 10 gallons of water in the aquarium. She will buy 8 large fish for every 40 gallons of water in the aquarium. What is the total number of fish Skye will have?

Answers

Answer=140 fish

200 gallons of water in the first aquarium.

200 gallons of water in the second aquarium.

The first will be for the small fishes.

5 small fish x fish
_____________=__________
10 gallons 200 gallons

cross multiply
10x=1000
divide both sides by 10
x=100 small fish

The first aquarium has 100 small fish.

The second aquarium holds large fish.

8 large fish x fish
____________=___________
40 gallons 200 gallons

cross multiply
40x=1600
divide both sides by 40
x=40 large fish

The second aquarium has 40 large fish.

100small fish+40 large fish=140 total fish

In terms of the first aquarium, the number of small fish Skye will have is:
200 gallons of water* (5 small fish/ 10 gallons of water)= 100 small fish.

In terms of the second aquarium, the number of large fish Skye will have is:
200 gallons of water* (8 large fish/ 40 gallons of water)= 40 large fish.

The total number of fish Skye will have is:
100 fish+ 40 fish= 140 fish.

Final answer: 140 fish.

Hope this would help~

I need help solving x^2+3xy-y^2=12 and x^2-y^2=-12

Answers

$substitute:x^2-y^2=-12\ to\ x^2+3xy-y^2=12\n\n3xy-12=12\n3xy=12+12\n3xy=24\ \ \ \ \ /:3\nxy=8\to x=(8)/(y)\n\nsubstitute\ to\ x^2-y^2=-12\n\n\left((8)/(y)\right)^2-y^2=-12\n\n(64)/(y^2)=y^2-12$

$(64)/(y^2)=(y^2-12)/(1)\n\ny^2(y^2-12)=64\n\ny^4-12y^2-64=0\n\nsubstitute:y^2=t > 0\n\nt^2-12t-64=0$

$a=1;\ b=-12;\ c=-64\n\Delta=b^2-4ac\to\Delta=(-12)^2-4\cdot1\cdot(-64)=144+256=400\n\nt_1=(-b-\sqrt\Delta)/(2a);\ t_2=(-b+\sqrt\Delta)/(2a)\n\n\sqrt\Delta=√(400)=20\n\nt_1=(12-20)/(2\cdot1)=(-8)/(2)=-4 < 0;\ t_2=(12+20)/(2\cdot1)=(32)/(2)=16\n\ny^2=16\Rightarrow y=\pm√(16)\Rightarrow y=-4\ or\ y=4\n\nx=(8)/(-4)=-2\ or\ x=(8)/(4)=2\n\nSolution:x=-2\ and\ y=-4\ or\ x=2\ and\ y=4$

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