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3 years ago

Write an equation for the line that is parallel to the given line and passes through the given point. y = x + 8; (2, 16)

2 answers:

6 0

Y = x + 8...slope here is 1. A parallel line will have the same slope.

y = mx + b
slope(m) = 1
(2,16)...x = 2 and y = 16
now we sub and find b, the y int
16 = 1(2) + b
16 = 2 + b
16 - 2 = b
14 = b

so ur parallel equation is : y = x + 14

xxMikexx [17]3 years ago

5 0

The answer is A. y = x + 14

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svet-max [94.6K]

Answer:52

X+150=180. x=30

Y+82=180. y=98

X+y+z=180. Z=128

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Step-by-step explanation:

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4 years ago

All three sides of a triangle are initially 4 m in length. One of the triangle's sides is oriented horizontally. The triangle is

Arlecino [84]

Answer:

the new height of the triangle = 2.449

Step-by-step explanation:

Given that:

the sides of the triangle are 4m in length i.e they are equal. It shows that the triangle is known to be an equilateral triangle.

Let say the triangle is a triangle IJK

Let the length of the side to be i = 4

Definitely

IJ = JK = IK = i = 4

If a midline is drawn and cuts the equilateral triangle in two equal halves of a right-angle triangle. Then, suppose the midline is L

Then ;

$JL = \dfrac{1}{2} JK$

$JL = \dfrac{1}{2} i$

Let consider triangle IJL

(IL)² = (IJ)² - (JL)²

$(IL)^2 = i^2 - \dfrac{i^2}{4}$

$(IL)^2 = \dfrac{3i^2}{4}$

$(IL) =\sqrt{ \dfrac{3i^2}{4}}$

$IL =\sqrt{3} \dfrac{i}{2}$

Area of triangle IJK can be expressed as:

$Area = \dfrac{1}{2}\times Base \times Height$

$Area = \dfrac{1}{2}\times i \times \sqrt{3}\dfrac{i}{2}$

$Area = \sqrt{3}\dfrac{i^2}{4}$

where, i = 4

Then:

$Area = \sqrt{3}\dfrac{4^2}{4}$

$Area = \sqrt{3} \dfrac{16}{4}$

$Area =4 \sqrt{3}$

when the area is exactly half of the original triangle's area, the new height is :

$A = \dfrac{1}{2} \times Area$

$\sqrt{3} \dfrac{i^2}{4} = \dfrac{1}{2} \times 4\sqrt{3}$

$\dfrac{i^2}{4} = \dfrac{1}{2} \times 4$

$\dfrac{i^2}{4} =2$

$i^2 = 8$

$i= \sqrt{8}$

$i= \sqrt{4 \times 2}$

$i= 2 \sqrt{2}$

Finally, the new height of the new triangle is:

$IL =\sqrt{3} \dfrac{i}{2}$

$IL =\sqrt{3} \dfrac{2 \sqrt{2}}{2}$

$IL =\sqrt{3 \times 2}$

$IL =\sqrt{6}$

IL = 2.449 m

6 0

4 years ago

Find b in triangle TRG. Please help, I'm not sure how to solve for b.

Marina CMI [18]

Answer:

b=4.9

C. is the correct option.

Step-by-step explanation:

In triangle TGN,

Let <RTG be x

cosx= b/h

or, cosx = 2/a

Again,

In triangle TRG,

cosx = b/h

(for bigger right angled triangle)

cosx = a/(4+2)

cosx = a/6

Now

cosx = 2/a = a/6

or, 2/a = a/6

or, 12 = a²

so, a² = 12

Now,

For TRG,

h² = p²+b²

or, 6² = b² + a²(a²=12)

or, 6² = b² + 12

or, 36 = b² + 12

or, 36 -12 = b²

or, b² = 24

or, b = square root 24

so, b = 4.8989

so, b = 4.9

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2 years ago

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BabaBlast [244]

Answer:

$9,220,000(0.888)^t

Step-by-step explanation:

Model this using the following formula:

Value = (Present Value)*(1 - rate of decay)^(number of years)

Here, Value after t years = $9,220,000(1 -0.112)^t

Value after t years = $9,220,000(0.888)^t

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4 years ago

What is 16x with a base of 4?

Tju [1.3M]

4^2* x the answer to the question

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3 years ago

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