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Damm [24]
3 years ago
6

If y=2x-8, what is the value of x when y=0? A)-4 B)0 C)4 D)8

Mathematics
2 answers:
Savatey [412]3 years ago
6 0
If y=0 then,
2x-8=0
2x=8
X=8/2
X=4
Answer is c
Gennadij [26K]3 years ago
6 0

Answer:

x=4

Step-by-step explanation:

y = 2x - 8

Let y =0

0 = 2x-8

Add 8 to each side

8 = 2x-8+8

8 = 2x

Divide each side by 2

8/2 = 2x/2

4 =x

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Step-by-step explanation:

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Which equation represents the line that passes through (-8,11) and
Elena-2011 [213]

Answer:

A) y = -5/8x + 6

Step-by-step explanation:

y2 - y`1 / x2 - x1

7/2 - 11 / 4 - (-8)

-15/2  / 12

= -5/8

y = -5/8x + b

11 = -5/8(-8) + b

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Angles ∠ABC= x and ∠CBD=(2y+4) are complementary. ∠CBD and ∠DBE=(3y+x) are supplementary.
Basile [38]

The missing question is find x , y and the measure of each angle

The values of x and y are x = 28 and y = 30

The measures of ∠ABC is 26°, ∠CBD is 64° and ∠DBE is 116°

Step-by-step explanation:

Let us revise the meaning of complementary angles and supplementary angles

  • Two angles are complementary if their sum is 90°
  • Two angles are supplementary if their sum is 180°

∵ ∠ABC = x

∵ ∠CBD = 2y + 4

∵ ∠ABC and ∠CBD are complementary

- That means their sum is 90°, add their values and equate

   the sum by 90

∴ x + (2y + 4) = 90

∴ x + 2y + 4 = 90

- Subtract 4 from both sides

∴ x + 2y = 86 ⇒ (1)

∵ ∠DBE = (3y + x)

∵ ∠CBD and ∠DBE are supplementary

- That means their sum is 180°, add their values and equate

   the sum by 180

∵ ∠CBD = (2y + 4)

∴ (2y + 4) + (3y + x) = 180

- Add like terms

∴ x + 5y + 4 = 180

- Subtract 4 from both sides

∴ x + 5y = 176 ⇒ (2)

Now we have a system of equations to solve them

Subtract equation (1) from equation (2) to eliminate x

∵ 3y = 90

- Divide both sides by 3

∴ y = 30

- Substitute the value of y in equation (1) to find x

∵ x + 2(30) = 86

∴ x + 60 = 86

- Subtract 60 from both sides

∴ x = 26

∵ ∠ABC = x

∴ The measure of angle ABC is 26°

∵ ∠CBD = 2y + 4

∴ ∠CBD = 2(30) + 4 = 60 + 4 = 64°

∴ The measure of angle CBD is 64°

∵ ∠DBE = 3y + x

∴ ∠DBE = 3(30) + 26 = 90 + 26 = 116°

∴ The measure of angle DBE is 116°

The values of x and y are x = 28 and y = 30

The measures of ∠ABC is 26°, ∠CBD is 64° and ∠DBE is 116°

Learn more:

You can learn more about the complementary angles and supplementary angles in brainly.com/question/10483199

#LearnwithBrainly

6 0
3 years ago
Triangle ABC has vertices A(-5, -2), B(7, -5), and C(3, 1). Find the coordinates of the intersection of the three altitudes
Darina [25.2K]

Answer:

Orthocentre (intersection of altitudes) is at (37/10, 19/5)

Step-by-step explanation:

Given three vertices of a triangle

A(-5, -2)

B(7, -5)

C(3, 1)

Solution A by geometry

Slope AB = (yb-ya) / (xb-xa) = (-5-(-2)) / (7-(-5)) = -3/12 = -1/4

Slope of line normal to AB, nab = -1/(-1/4) = 4

Altitude of AB = line through C normal to AB

(y-yc) = nab(x-xc)

y-1 = (4)(x-3)

y = 4x-11           .........................(1)

Slope BC = (yc-yb) / (xc-yb) = (1-(-5) / (3-7)= 6 / (-4) = -3/2

Slope of line normal to BC, nbc = -1 / (-3/2) = 2/3

Altitude of BC

(y-ya) = nbc(x-xa)

y-(-2) = (2/3)(x-(-5)

y = 2x/3 + 10/3 - 2

y = (2/3)(x+2)    ........................(2)

Orthocentre is at the intersection of (1) & (2)

Equate right-hand sides

4x-11 = (2/3)(x+2)

Cross multiply and simplify

12x-33 = 2x+4

10x = 37

x = 37/10  ...................(3)

substitute (3) in (2)

y = (2/3)(37/10+2)

y=(2/3)(57/10)

y = 19/5  ......................(4)

Therefore the orthocentre is at (37/10, 19/5)

Alternative Solution B using vectors

Let the position vectors of the vertices represented by

a = <-5, -2>

b = <7, -5>

c = <3, 1>

and the position vector of the orthocentre, to be found

d = <x,y>

the line perpendicular to BC through A

(a-d).(b-c) = 0                          "." is the dot product

expanding

<-5-x,-2-y>.<4,-6> = 0

simplifying

6y-4x-8 = 0 ...................(5)

Similarly, line perpendicular to CA through B

<b-d>.<c-a> = 0

<7-x,-5-y>.<8,3> = 0

Expand and simplify

-3y-8x+41 = 0 ..............(6)

Solve for x, (5) + 2(6)

-20x + 74 = 0

x = 37/10  .............(7)

Substitute (7) in (6)

-3y - 8(37/10) + 41 =0

3y = 114/10

y = 19/5  .............(8)

So orthocentre is at (37/10, 19/5)  as in part A.

8 0
3 years ago
Triangle JKL is fromed by plotting three points. What is the leanth of sides JK?
Phoenix [80]
<h3><u>Solution</u><u>:</u><u>-</u></h3>
  • J (1,5)
  • K (1,9)

\LARGE\leadsto\sf JK=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

\LARGE\leadsto\sf \sqrt {(1-1)^2+(9-5)^2}

\LARGE\leadsto\sf \sqrt {(0)^2+(4)^2}

\LARGE\leadsto\sf \sqrt {0+16}

\LARGE\leadsto\sf \sqrt {16}

\LARGE\leadsto\sf JK=4units

4 0
3 years ago
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