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vodomira [7]
2 years ago
13

Write the equation for a line perpendicular to a line passing through points (-1,2) and (1,-8)

Mathematics
1 answer:
soldi70 [24.7K]2 years ago
3 0

Answer:

Step-by-step explanation:

Slope m = (-8-2)/(1+1)=-10/2 = -5

Perpendicular slope-1/m = -1/-5 = 1/5

Equation of a line with perpendicular slope is

y-y1=(-1/m)(x-x1)

y - 2 = (1/5)(x+1)

y =2 +(1/5)(x+1)

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The figure below shows the correct construction of a segment bisector.
Maslowich

Answer:

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3 years ago
Jessica walks 7 times as far to school as Taylor does . Jessica walks 5 miles. How many miles does Taylor walk?
Kaylis [27]

Answer:

Step-by-step explanation:

Jessica: 5 miles

Taylor: 5/7 = .7143 miles

3 0
3 years ago
Can someone help me please and thankyou
Phantasy [73]
X° = 90°-29°
= 61°

y°= 90°- 61°
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8 0
2 years ago
Please help, this will just take a minute of your time ❤︎ (literally)<br>(Discriminant Worksheet)
igomit [66]

Answer:

\large\boxed{x=-\dfrac{7-\sqrt{97}}{4}\ \vee\ x=-\dfrac{7+\sqrt{97}}{4}}

Step-by-step explanation:

ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\\text{if}\ \Delta0,\ \text{then an equation has two solutions:}\ x=\dfrac{-b\pm\sqrt\Delta}{2a}\\\\\text{We have:}\\\\-2x^2-7x+10=4\qquad\text{subtract 4 from both sides}\\\\-2x^2-7x+6=0

a=-2,\ b=-7,\ c=6\\\\\Delta=(-7)^2-4(-2)(6)=49+48=97>0\\\\\sqrt\Delta=\sqrt{97}\\\\x=\dfrac{-(-7)\pm\sqrt{97}}{2(-2)}=\dfrac{7\pm\sqrt{97}}{-4}=-\dfrac{7\pm\sqrt{97}}{4}

6 0
3 years ago
an airplaine descends 1.5 miles to an elevation of 5.25 miles. find the elevation of the plane before its descent
Maksim231197 [3]

We can then write an equation representing this problem as:

e−1.5mi=5.25mi

Now, add 1.5mi to each side of the equation to solve for e while keeping the equation balanced:

e−1.5mi+1.5mi=5.25mi+1.5mi

e−0=6.75mi

e=6.75mi

The plane's starting elevation was 6.75 miles

Hope this helps!

8 0
3 years ago
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