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2 years ago

What is the domain of the function shown in the graph below?

Mathematics

2 answers:

tigry1 [53]2 years ago

6 0

Answer:

All except one interval [-2,8]

Step-by-step explanation:

Notice that the function, wich an hyperbola, has a vertical asymptote in both -2 and 8 and there is nothing between this two values.

So we will exclude what's between them out of the domain.

Natali [406]2 years ago

5 0

All values except [-2,7]

the graph isn't proper, but it's obvious what the answer would be.

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Can anybody explain to me what this question mean so it says "8 more than a number n is written ___________ in algebraic form"

Umnica [9.8K]

Answer:

They want you to put it into an expression so that it could be expressed. So, 8 more than n would be 8+n.

Hope This Helps!!! :D

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2 years ago

Read 2 more answers

Simplify<br><br>Square root of 50 over 2 times -4 + square root of 121

kobusy [5.1K]

Answer:

-9

Step-by-step explanation:

$\sqrt{\dfrac{50}{2}}\cdot(-4)+\sqrt{121}=-4\sqrt{25}+11=-4\cdot5+11\\\\=-20+11=\boxed{-9}$

Your calculator can help with this.

4 0

2 years ago

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

Need your help ASAP please help

eduard

His average is 86 so it is below 90. to get his average i just added all of his grades up to get 517 and then I divided 517 by 6 because thats how many grades he has and I got 86.

On the Grade row in the table it would go 83, 94, 79, 90, 96, 75 in order

The Above/Below 90 row would go -7, 4, 11, 7, -6, -3 in order

Lamont needs to get a 99 on his next test to have an average of exactly 90.

I HOPE THIS HELPED! PLEASE RATE ME AND MAKE MY ANSWER MOST BRAINLIEST!!

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3 years ago

A biologist is researching a newly-discovered species of bacteria. He puts 1,000 bacteria into what he has determined to be a fa

Masteriza [31]

The bacteria grow from 1,000 bacteria into 12,500 bacteria in 12 hours. The exponent function would be
:an= ao(1+r)^t12.500= 1,000(1+r)^121,000(12.5)= 1,000(1+r)^1212.5= (1+r)^121+r= 1.23427r= 0.23Number of bacteria after 20hr would be:an= ao(1+r)^tan= 1,000(1+0.23)^20an= 1,000(62.821 )an= 62,821

8 0

2 years ago