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Dima020 [189]
3 years ago
10

Which of the following equations represents a line perpendicular to y = -3x + 4.5

Mathematics
2 answers:
Gnom [1K]3 years ago
5 0
The answer is
y=-13.5
Andrej [43]3 years ago
3 0
The perpendicular gradient to that would be 1/3x

So y= 1/3x + 4.5
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Help me please!
telo118 [61]

Answer:

f^{-1}(6) = 50

Step-by-step explanation:

Given

f(x) = \sqrt{2x} - 4

Required

Find f^{-1}(6)

First, we calculate the inverse function

f(x) = \sqrt{2x} - 4

Express f(x) as y

y = \sqrt{2x} - 4

Swap the positions of x and y

x = \sqrt{2y} - 4

Solve for y: Add 4 to both sides

4 + x = \sqrt{2y} - 4+4

4 + x = \sqrt{2y}

Square both sides

(4 + x)^2 = 2y

Divide both sides by 2

y = \frac{(4 + x)^2}{2}

Express y as an inverse function

f^{-1}(x) = \frac{(4 + x)^2}{2}

Next, solve for: f^{-1}(6)

Substitute 6 for x

f^{-1}(6) = \frac{(4 + 6)^2}{2}

f^{-1}(6) = \frac{(10)^2}{2}

f^{-1}(6) = \frac{100}{2}

f^{-1}(6) = 50

7 0
3 years ago
Prove that the diagonals of a rhombus are perpendicular using coordinate geometry
lozanna [386]

If we draw a rhombus OABC on a coordinate grid with one vertex at the point O (0,0)

and let its height be 1 then (using the Pythagoras theorem) the coordinates of the 4 points will be as follows:-

O = (0,0) , A = (1, 1), B = (sqrt2+ 1 , 1) , C = (sqrt2, 0)

Slope of the diagonal OB = 1 / (sqrt2 + 1)

Slope of the other diagonal AC = - 1 / (sqrt2 - 1)

Product of the slopes = - 1 / ( sqrt2 + 1)(sqrt2 - 1) = -1 / 1 = -1

This proves that the diagonals are perpendicular.

5 0
3 years ago
What postulate would you use to solve this triangle<br> congruence?<br> SAS<br> AAS<br> ASA<br> SSS
zavuch27 [327]
Im doing something similar to this you wanna help each other out?
4 0
3 years ago
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Use distributive property 5(6+b) *<br> Your answer
EleoNora [17]

Answer:

5b +30

Step-by-step explanation:

5 · 6 = 30

5 · b = 5b

8 0
3 years ago
Can someone help me with this If 5y*2=80 y=?
Temka [501]
<span>If 5y*2=80 then y is equal to 8</span>
8 0
3 years ago
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