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2 years ago

Four students graphed one linear function each. Which student graphed a linear function with a y-intercept at -4?

Mathematics

1 answer:

balandron [24]2 years ago

7 0

Answer:

who ever line crosses the y-axis at -4.

Step-by-step explanation:

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The length of a rectangle is 2 meters more than its width. The area of the rectangle is 80 square meters. What is the length and

insens350 [35]

Answer:

Step-by-step explanation:

Formula

A = L * W

Givens

W = W

L = W + 2

Solution

Area = L*W

Area = (W+2)*W = 80 Remove the brackets.

Area = W^2 + 2W = 80 Subtract 80 from both sides.

Area = w^2+2W-80=80-80 Combine

Area = w^2 +2W-80 = 0 Factor.

Area = (w+10)(w - 8) = 0

W + 10 = 0 won't work

W = - 10 which isn't possible

W- 8 = 0

W = 8

L = 8 + 2 = 10

The answer looks like A

8 0

2 years ago

PLZ HELP GEOMETRY BELOW

katen-ka-za [31]

This is known as Einstein's proof, not because he was the first to come up with it, but because he came up with it as a 15 year old boy.

Here the problem is justification step 2. The written equation

BC ÷ DC = BC ÷ AC

is incorrect, and wouldn't get us our statement 2, which is correct.

For similar triangles we have to carefully pair the corresponding parts to get our ratios right:

ABC ~ BDC means AB:BD = BC:DC = AC:BC so BC/DC=AC/BC.

Justification 2 has the final division upside down.

7 0

2 years ago

What is the vertex of the graph of the function below? y = x2 + 6x + 5

Aleksandr-060686 [28]

Answer:

(-3,-4)

If you have a chart with the coordinates in it, it will be in the middle of alike coordinates.

7 0

2 years ago

For what value of c is the function defined below continuous on (-\infty,\infty)?

kozerog [31]

$f(x)= \left \{ {{x^2-c^2,x \ \textless \ 4} \atop {cx+20},x \geq 4} \right
$

It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
A function, f, is continuous at x = 4 if
 $\lim_{x \rightarrow 4} \ f(x) = f(4)$

In notation we write respectively
 $\lim_{x \rightarrow 4-} f(x) \ \ \ \text{ and } \ \ \ \lim_{x \rightarrow 4+} f(x)$

Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just 4c + 20.

On the other hand, for x < 4, f(x) = x^2 - c^2. Hence
$\lim_{x \rightarrow 4-} f(x) = \lim_{x \rightarrow 4-} (x^2 - c^2) = 16 - c^2$

Thus these two limits, the one from above and below are equal if and only if
4c + 20 = 16 - c²
Or in other words, the limit as x --> 4 of f(x) exists if and only if
4c + 20 = 16 - c²

$c^2+4c+4=0
\\(c+2)^2=0
\\c=-2$

That is to say, if c = -2, f(x) is continuous at x = 4.

Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers $(-\infty, +\infty)$