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Alecsey [184]
3 years ago
6

#11 Find the slope of a line that passes through the points (-1, -3) and (-6,5)

Mathematics
1 answer:
Elden [556K]3 years ago
3 0
Y2-y1/x2-x1
5-(-3)/(-6-(-1))
8/-5
-8/5 is the slope
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Andre tried to solve the equation 1/4(x + 12) = 2. What was his mistake? *
Umnica [9.8K]

Answer:

x=-4

Step-by-step explanation:

x+12=8    

x=8-12

Andre did not multiple both by four to get the exact whole number. He also, didnt subtract 12 which is not possible due to it being multiplied by 1/4.

5 0
3 years ago
plzz help me no link plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
boyakko [2]

Answer:

Zara ate 50 almonds last week

Step-by-step explanation:

part to whole ratio is 50 : 100

150 divided by 2 is 75

100 divided by 50 is equal to 2

Zara ate 50 almonds last week

pls mark brainliest if this helped :)

8 0
2 years ago
Find the distance between the point (3,1) and the line with the equation y=2x-5
True [87]

Answer:

B:  0

Step-by-step explanation:

Note that (3, 1) satisfies the equation y = 2x - 5:

1 = 2(3) - 5

1  =  6 -5  = 1

Therefore, the distance between this point and this line is zero.

B is the correct answer.

3 0
4 years ago
PLEASE HELP ASAP!!!!!
attashe74 [19]
For this case what we must do is use the law of cosines.
 We then have the following equation:
 c ^ 2 = a ^ 2 + b ^ 2 - 2 * a * b * cos (x)

 Where,
 a, b: sides of the triangle
 x: angle between sides a and b.
 Substituting values we have:
 c^2 = 90^2 + 75^2 - 2*90*75*cos(85)

 Clearing the value of c we have:
 c =  \sqrt{ 90^2 + 75^2 - 2*90*75*cos(85)}
 Answer:
 
An expression that is equivalent to how many feet the oak trees are from each other is:
 
c = \sqrt{ 90^2 + 75^2 - 2*90*75*cos(85)}
7 0
3 years ago
Try to sketch by hand the curve of intersection of the parabolic cylinder y = x2 and the top half of the ellipsoid x2 + 7y2 + 7z
vovikov84 [41]

Plug y=x^2 into the equation of the ellipsoid:

x^2+7(x^2)^2+7z^2=49

Complete the square:

7x^4+x^2=7\left(x^4+\dfrac{x^2}7+\dfrac1{14^2}-\dfrac1{14^2}\right)=7\left(x^2+\dfrac1{14}\right)^2+\dfrac1{28}

Then the intersection is such that

7\left(x^2+\dfrac1{14}\right)^2+7z^2=\dfrac{1371}{28}

\left(x^2+\dfrac1{14}\right)^2+z^2=\dfrac{1371}{196}

which resembles the equation of a circle, and suggests a parameterization is polar-like coordinates. Let

x(t)^2+\dfrac1{14}=\sqrt{\dfrac{1371}{196}}\cos t\implies x(t)=\pm\sqrt{\sqrt{\dfrac{1371}{196}}\cos t-\dfrac1{14}}

y(t)=x(t)^2=\sqrt{\dfrac{1371}{196}}\cos t-\dfrac1{14}

z=\sqrt{\dfrac{1371}{196}}\sin t

(Attached is a plot of the two surfaces and the intersection; red for the positive root x(t), blue for the negative)

4 0
4 years ago
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