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ZanzabumX [31]
3 years ago
10

I need help with these

Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
5 0

Answer:

iii v i ii iv

Step-by-step explanation:

from a-e. just plug x value and y value into the equations (guess and check)

yw in advance :)

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A dealership is having a sale in which customers pay 85% of the retail price for a new vehicle. Keith is buying a vehicle with a
bazaltina [42]
85% of 36000 is 30600
9% of 30600 is 2754
30600 - 2754 = 27846

I'm fairly sure that's right, but I wouldn't say no to a second opinion
4 0
3 years ago
Graph the line.<br><br><br> y = −4x + 2
MrRissso [65]

Answer:

it rise over run so put 2y on graph count down 4 and move right 1

Step-by-step explanation:

8 0
3 years ago
4 divided by 5/9. Please answer in fraction form.
irina1246 [14]

Answer: 20/9

Step-by-step explanation:

Turn 4 into a fraction so 4/1

When dividing, flip the second fraction over then multiply them both, so

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4 0
3 years ago
Read 2 more answers
What are 3 completed examples for functions?
Lyrx [107]
Example 1

Write y = x2 + 4x + 1 using function notation and evaluate the function at x = 3.

Solution

Given, y = x2 + 4x + 1

By applying function notation, we get

f(x) = x2 + 4x + 1

Evaluation:


Substitute x with 3

f (3) = 32 + 4 × 3 + 1 = 9 + 12 + 1 = 22

Example 2

Evaluate the function f(x) = 3(2x+1) when x = 4.

Solution

Plug x = 4 in the function f(x).

f (4) = 3[2(4) + 1]

f (4) = 3[8 + 1]

f (4) = 3 x 9

f (4) = 27

Example 3

Write the function y = 2x2 + 4x – 3 in function notation and find f (2a + 3).

Solution

y = 2x2 + 4x – 3 ⟹ f (x) = 2x2 + 4x – 3


Substitute x with (2a + 3).

f (2a + 3) = 2(2a + 3)2 + 4(2a + 3) – 3

= 2(4a2 + 12a + 9) + 8a + 12 – 3
= 8a2 + 24a + 18 + 8a + 12 – 3
= 8a2 + 32a + 27
3 0
3 years ago
Plz with steps plzzzzzz
Stella [2.4K]

Answer:  -\frac{\sqrt{2a}}{8a}

=======================================================

Explanation:

The (x-a) in the denominator causes a problem if we tried to simply directly substitute in x = a. This is because we get a division by zero error.

The trick often used for problems like this is to rationalize the numerator as shown in the steps below.

\displaystyle \lim_{x\to a} \frac{\sqrt{3a-x}-\sqrt{x+a}}{4(x-a)}\\\\\\\lim_{x\to a} \frac{(\sqrt{3a-x}-\sqrt{x+a})(\sqrt{3a-x}+\sqrt{x+a})}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{(\sqrt{3a-x})^2-(\sqrt{x+a})^2}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{3a-x-(x+a)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{3a-x-x-a}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\

\displaystyle \lim_{x\to a} \frac{2a-2x}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2(-a+x)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2(x-a)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2}{4(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\

At this point, the (x-a) in the denominator has been canceled out. We can now plug in x = a to see what happens

\displaystyle L = \lim_{x\to a} \frac{-2}{4(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\L = \frac{-2}{4(\sqrt{3a-a}+\sqrt{a+a})}\\\\\\L = \frac{-2}{4(\sqrt{2a}+\sqrt{2a})}\\\\\\L = \frac{-2}{4(2\sqrt{2a})}\\\\\\L = \frac{-2}{8\sqrt{2a}}\\\\\\L = \frac{-1}{4\sqrt{2a}}\\\\\\L = \frac{-1*\sqrt{2a}}{4\sqrt{2a}*\sqrt{2a}}\\\\\\L = \frac{-\sqrt{2a}}{4\sqrt{2a*2a}}\\\\\\L = \frac{-\sqrt{2a}}{4\sqrt{(2a)^2}}\\\\\\L = \frac{-\sqrt{2a}}{4*2a}\\\\\\L = -\frac{\sqrt{2a}}{8a}\\\\\\

There's not much else to say from here since we don't know the value of 'a'. So we can stop here.

Therefore,

\displaystyle \lim_{x\to a} \frac{\sqrt{3a-x}-\sqrt{x+a}}{4(x-a)} = -\frac{\sqrt{2a}}{8a}\\\\\\

3 0
3 years ago
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