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Yuri [45]
3 years ago
10

Find the y intercept of the line on the graph​

Mathematics
1 answer:
melisa1 [442]3 years ago
3 0

Answer:

(0, -4)

Step-by-step explanation:

because the y is the vertical line. (the one that is going up and down)

and it intercepts the blue line at (0, -4)

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Evaluate in the following expression when x = 4 and y = 5​
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3 years ago
Pls help me with this answer choice question! brainliest, rattings, thanks etc.
galina1969 [7]

Answer:

graph B

Step-by-step explanation:

Given

3y - 5x < - 6

Since the inequality is < then the line must be broken

If ≤ then line would be solid

The only possible graphs are B and D

Choose a test point in the shaded region of each graph and check validity

graph B (2, 0) ← substitute into left side of inequality

0 - 5(2) = 0 - 10 = - 10 < - 6 ← this is valid

graph D (0, 0 ) ← substitute into left side of inequality

0 - 0 = 0 > - 6 ← this is not valid

Hence graph B is the correct graph

3 0
4 years ago
HELP PLEASE!!! IM DOING A QUIZ!
Marat540 [252]

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equivalent = 30w 30(w) 30*w

the rest are non equivalent

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3 years ago
Ruby ate 1/3 (fraction) of a pizza, and Angie ate 1/5 (fraction) of the pizza. How much pizza did they eat in all?
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Ruby and Angie ate 8/15 of the pizza all together
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3 years ago
<img src="https://tex.z-dn.net/?f=%5Csf%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Ccfrac%7B%5Csqrt%7Bx-1%7D-2x%20%7D%7Bx-7%7D" id=
BARSIC [14]
<h3>Answer:  -2</h3>

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

Work Shown:

\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}\left(\sqrt{x-1}-2x\right) }{ \frac{1}{x}\left(x-7\right) }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}*\sqrt{x-1}-\frac{1}{x}*2x }{ \frac{1}{x}*x-\frac{1}{x}*7 }\\\\\\

\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}}*\sqrt{x-1}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}*(x-1)}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x}-\frac{1}{x^2}}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \frac{ \sqrt{0-0}-2 }{ 1-0 }\\\\\\\displaystyle L = \frac{-2}{1}\\\\\\\displaystyle L = -2\\\\\\

-------------------

Explanation:

In the second step, I multiplied top and bottom by 1/x. This divides every term by x. Doing this leaves us with various inner fractions that have the variable in the denominator. Those inner fractions approach 0 as x approaches infinity.

I'm using the rule that

\displaystyle \lim_{x\to\infty} \frac{1}{x^k} = 0\\\\\\

where k is some positive real number constant.

Using that rule will simplify the expression greatly to leave us with -2/1 or simply -2 as the answer.

In a sense, the leading terms of the numerator and denominator are -2x and x respectively. They are the largest terms for each, so to speak. As x gets larger, the influence that -2x and x have will greatly diminish the influence of the other terms.

This effectively means,

\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 } = \lim_{x\to\infty} \frac{ -2x }{ x} = -2\\\\\\

I recommend making a table of values to see what's going on. Or you can graph the given function to see that it slowly approaches y = -2. Keep in mind that it won't actually reach y = -2 itself.

5 0
3 years ago
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