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shepuryov [24]
3 years ago
13

Which is the slope of the line that passes through the points (4,2) and (6,7)?

Mathematics
2 answers:
nikitadnepr [17]3 years ago
4 0

Answer:

5/2

Step-by-step explanation:

y2-y1

--------

x2-x1

in this case it is

7-2

------

6-4

which gives you 5/2

brainliest?

Ivahew [28]3 years ago
4 0

Answer:

the slope of the line is 5/2

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20 points, asap please!
devlian [24]
H(x)=-3,1 I think hope this helps
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3 years ago
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What is the vertex of the graph of the function below? <br><br>y = x2 + 8x + 12
Vera_Pavlovna [14]
Hello, 

The\,\,formula\,\,for\,\,the\,\,vertex\,\,of\,\,a\,\,quadratic\,\,equation\,\,is:  \\  V=( -\frac{b}{2a},f(-\frac{b}{2a}) )

That means that we must first obtain x, then we replace x in the function to get y, so:

x=- \frac{b}{2a}  \\  \\ x=- \frac{8}{2*(1)}  \\  \\ \boxed{x=-4} \\  \\ Reeplacing \,\,x: \\ y=(-4)^{2} +8*(-4)+12 \\ y=16-32+12 \\ \boxed{y=-4}

Answer: V=(-4 , -4)
5 0
3 years ago
Find the measures of the angles of the triangle whose vertices are A = (-3,0) , B = (1,3) , and C = (1,-3).A.) The measure of ∠A
alekssr [168]

Answer:

\theta_{CAB}=128.316

\theta_{ABC}=25.842

\theta_{BCA}=25.842

Step-by-step explanation:

A = (-3,0) , B = (1,3) , and C = (1,-3)

We're going to use the distance formula to find the length of the sides:

r= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}

AB= \sqrt{(-3-1)^2+(0-3)^2}=5

BC= \sqrt{(1-1)^2+(3-(-3))^2}=9

CA= \sqrt{(1-(-3))^2+(-3-0)^2}=5

we can use the cosine law to find the angle:

it is to be noted that:

the angle CAB is opposite to the BC.

the angle ABC is opposite to the AC.

the angle BCA is opposite to the AB.

to find the CAB, we'll use:

BC^2 = AB^2+CA^2-(AB)(CA)\cos{\theta_{CAB}}

\dfrac{BC^2-(AB^2+CA^2)}{-2(AB)(CA)} =\cos{\theta_{CAB}}

\cos{\theta_{CAB}}=\dfrac{9^2-(5^2+5^2)}{-2(5)(5)}

\theta_{CAB}=\arccos{-\dfrac{0.62}}

\theta_{CAB}=128.316

Although we can use the same cosine law to find the other angles. but we can use sine law now too since we have one angle!

To find the angle ABC

\dfrac{\sin{\theta_{ABC}}}{AC}=\dfrac{\sin{CAB}}{BC}

\sin{\theta_{ABC}}=AC\left(\dfrac{\sin{CAB}}{BC}\right)

\sin{\theta_{ABC}}=5\left(\dfrac{\sin{128.316}}{9}\right)

\theta_{ABC}=\arcsin{0.4359}\right)

\theta_{ABC}=25.842

finally, we've seen that the triangle has two equal sides, AB = CA, this is an isosceles triangle. hence the angles ABC and BCA would also be the same.

\theta_{BCA}=25.842

this can also be checked using the fact the sum of all angles inside a triangle is 180

\theta_{ABC}+\theta_{BCA}+\theta_{CAB}=180

25.842+128.316+25.842

180

6 0
3 years ago
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Given that ∠A≅∠B, Evelia conjectured that ∠A and ∠B are acute angles.
Valentin [98]

Answer:

answer is choice 4 because in te given angle A and angle B are congrunt and if they are acute the degree measure is 0-90

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3 years ago
Write the standard form of the line that passes through the given points. Include your work in your final answer. Type your answ
I am Lyosha [343]

Answer:

|x   y   1|

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x*(-3*1 - (-8)*1) - y*(7*1 - 4*1) + 1*(7*-8 - 4*-3) = 0

5x - 3y - 44 = 0

Step-by-step explanation:

6 0
2 years ago
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