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victus00 [196]
3 years ago
15

Look at the following equation: –4x + y = 10

Mathematics
2 answers:
Zielflug [23.3K]3 years ago
3 0
-4x + y = 10
y = 4x + 10

Slope = 4
x=3, y = 4*3 + 10 = 22
Artyom0805 [142]3 years ago
3 0
The slope is 4. 
The value of y is 22.

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Step-by-step explanation:

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3 years ago
SOMEOne PLS EXPLAIN VERTICAL ASYMPTOTES TO ME IM DYING- i need to know how u get the asymptote(explanation) and answer T^TTTTTTT
Kitty [74]
A vertical asymptote is what you get when you try to divide by 0. To find where you get these, you need to look at the denominator and what values of x will make the denominator equal to 0.

In your denominator, you have (x+7)(x-5)(x-3).
What values of x makes (x+7)(x-5)(x-3)=0?
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Now the only place this can get trickier is if one of those factors — one of (x+7), (x-5), or (x-3) — also appears in the numerator. If that happens, then it’s more involved whether you have an asymptote or not. But that doesn’t happen in this example.

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4 0
2 years ago
Triangle ABC with vertices A (-1, 4), B( -4, 1), and C(-1, 1) is reflected across the y-axis and then translated 4 units to the
Anna71 [15]

Given:

Triangle ABC with vertices A (-1, 4), B( -4, 1), and C(-1, 1) is reflected across the y-axis and then translated 4 units to the left to form triangle A’B’C’.

To find:

The coordinates of the vertices of triangle A’B’C’.

Solution:

If triangle ABC reflected across y-axis, then

(x,y)\to (-x,y)

A(-1,4)\to A_1(1,4)

B(-4,1)\to B_1(4,1)

C(-1,1)\to C_1(1,1)

Coordinates of image after reflection across y-axis are A_1(1,4),B_1(4,1),C_1(1,1).

Then triangle translated 4 units to the left to form triangle A’B’C’.

(x,y)\to (x-4,y)

A_1(1,4)\to A'(-3,4)

B_1(4,1)\to B'(0,1)

C_1(1,1)\to C'(-3,1)

Therefore, the vertices of triangle A’B’C’ are A’ (-3, 4), B’ (0, 1), C’ (-3, 1).

Hence, the correct option is B.

6 0
3 years ago
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shepuryov [24]
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Romashka [77]

From the Figure :

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Point B is (6 , 6)

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\mathsf{\implies [(\frac{x_1 + x_2}{2})\;,\;(\frac{y_1 + y_2}{2})]}

\mathsf{\implies Midpoint\;of\;Line\;AB\;is\;[(\frac{-3 + 6}{2})\;,\;(\frac{-3 + 6}{2})]}

\mathsf{\implies Midpoint\;of\;Line\;AB\;is\;[(\frac{3}{2})\;,\;(\frac{3}{2})]}

\mathsf{\implies Midpoint\;of\;Line\;AB\;is\;(1.5,1.5)}

4 0
3 years ago
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