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2 years ago

Write an equation of the line in slope-intercept form(2,-2)(4,-8)(3,8)(-2,8)

Mathematics

2 answers:

valina [46]2 years ago

7 0

Slope-intervept form: y = -3x + 4

May I have brainliest please? :)

Masteriza [31]2 years ago

5 0

Answer:

(2,-2) (4,-8) - y=-3x+4

(3,8) (-2,8) - y=0x+8 undefined I think

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Pa<br> Evaluate<br> when p=5 and q =13.<br> p+a

Ivenika [448]

The answer is 18 just add them

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2 years ago

Wirte the expression the quotient of the quantity k minus 8 and m

umka21 [38]

Answer:

k - m - 8

Step-by-step explanation:

pretty sure you just move the -8 to the back.

8 0

2 years ago

Need help please and Thank you

Novosadov [1.4K]

Top left:
7x-44 = 4x+4
7x-4x = 4+44
3x = 48
x = 16
8y-43 = 39+(7(16)-44)
8y = 39+68+43
8y = 150
y = 18.75

7 0

3 years ago

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

3 years ago

A train takes 1.5 hours to travel 525 km. Calculate the speed of the train in kilometers per hour.

kow [346]

Answer:

1.5 hours to 525 km

Speed in km/h=

$525 \div 1.5 = 350 \: kmph$

Step-by-step explanation:

8 0

3 years ago

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