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user100 [1]
3 years ago
6

Finding the Equation of a Line from a Graph and a Table plz help me

Mathematics
1 answer:
Leni [432]3 years ago
5 0

Answer:

slope (m)= 1/2

y-intercept: (0, -1)

equation: y=\frac{1}{2}x-1

Step-by-step explanation:

slope is \frac{rise}{run\\}=\frac{1}{2}

y-intercept is where the line crosses the vertical axis (y-axis) so its (0, -1)

using point slope form:

y-y1=m(x-x1)

y-(-1)=1/2(x-0)

y+1=1/2x

y=\frac{1}{2}x-1

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If the legs of a right triangle are given by x2 - y2 and 2xy, then which expression equals the hypotenuse? choose all that apply
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Hypothenus = sqrt ((x^2 - y^2)^2 + (2xy)^2) = sqrt(x^4 - 2x^2y^2 + y^4 + 4x^2y^2) = sqrt(x^4 + 2x^2y^2 + y^4) = sqrt(x^2 + y^2)^2 = x^2 + y^2
3 0
3 years ago
Which pattern is non-linear?
Zielflug [23.3K]

Answer:

<em>2</em><em>,</em><em> </em><em>4</em><em>,</em><em> </em><em>8</em><em>,</em><em> </em><em>1</em><em>6</em><em>,</em><em> </em><em>3</em><em>2</em>

<em> </em><em> </em><em> </em><em> </em><em>is</em><em> </em><em>non-linear</em><em> </em><em>pattern</em>

<em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

Step-by-step explanation:

--------------------------------------

1st

-4+6=2

2+6=8

8+6=14

14+6=20

3rd

3+3=6

6+3=9

9+3=12

12+3=15

4th

5+4=9

9+4 =13

13+4=17

17+4=21

<em>But</em><em> </em>

2nd

2×2=4

4×2=8

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5 0
3 years ago
Find sin θ if θ is in Quadrant III and tan θ = . 0.958
Leokris [45]
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sec^2 = 1  + tan^2 \\  \\ sec = \frac{1}{cos} \\  \\ sin^2 = 1 - cos^2
Also because the angle is in quadrant 3, sin must be negative.
Therefore
sin = - \sqrt{1 - \frac{1}{1 + tan^2}}
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3 years ago
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3 years ago
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erastova [34]

Answer:

4

Step-by-step explanation:

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we have now 2 more seats one adjacent to the driver and one rear (two combined).

so the total ways in which all five can be arranged is as follows.

driver, adjacent to him(1)  and three back.

driver adjacent to him (different person) and three back.

see the driver is always fixed so we can ignore him.

thus we when driver set fixed , on two remaining seats (adjacent to driver and the back )there can be 4 different combinations.

3 0
3 years ago
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