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bija089 [108]
2 years ago
9

Find the slope of a line which goes through the points (2,5) and (−4,3)

Mathematics
2 answers:
meriva2 years ago
8 0

Answer:

1/3

Step-by-step explanation:

natka813 [3]2 years ago
3 0

Answer:

slope (m) = 0.75

Step-by-step explanation:

Slope (m) =

ΔY

ΔX

=

1

3

= 0.33333333333333

θ =

arctan( ΔY ) + 180°

ΔX

= 198.43494882292°

ΔX = -4 – 2 = -6

ΔY = 3 – 5 = -2

Distance (d) = √ΔX2 + ΔY2 = √40 = 6.3245553203368

Equation of the line:

y = 0.33333333333333x + 4.3333333333333

or

y =  

1 x

3

+  13

3

When x=0, y = 4.3333333333333

When y=0, x = -13

please mark brainliest

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X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?
cluponka [151]

Answer:

x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}} and y=-1,\frac{1}{2}

The ordered pair solutions are (-\sqrt{\frac{7}{4}},0.5), (\sqrt{\frac{7}{4}},0.5), (-1,-1), and (1,-1).

Step-by-step explanation:

I'm assuming the system is \left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.:

x^2+y^2=2

x^2+(2x^2-3)^2=2

x^2+(4x^4-12x^2+9)=2

x^2+4x^4-12x^2+9=2

4x^4-11x^2+9=2

4x^4-11x^2+7=0

x^4-11x^2+28=0

(x^2-7)(x^2-4)=0

(4x^2-7)(x^2-1)=0

4x^2-7=0

4x^2=7

x^2=\frac{7}{4}

x=\pm\sqrt{\frac{7}{4}}

x^2-1=0

x^2=1

x=\pm1

y=2x^2-3

y=2(\pm\sqrt{\frac{7}{4}})^2-3

y=2({\frac{7}{4}})-3

y=\frac{7}{2}-3

y=\frac{1}{2}

y=2x^2-3

y=2(\pm1)^2-3

y=2(1)-3

y=2-3

y=-1

Therefore, x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}} and y=-1,\frac{1}{2}

The ordered pair solutions are (-\sqrt{\frac{7}{4}},0.5), (\sqrt{\frac{7}{4}},0.5), (-1,-1), and (1,-1).

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Answer:

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6 0
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Which of the following statements is true in a one-way ANOVA? a. The critical value of the test will be a value obtained from th
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Answer:

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Step-by-step explanation:

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The degrees of freedom associated with the sum of squares for treatments is equal to one less than the number of populations.

Let's say we are comparing the means of k population. The degree of freedom would be = k - 1

The correct option here is (d).

All of the above

3 0
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