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Help please. <br> Give the slope of y - 3 = 4(x + 8) and a point on the line

Sauron [17]

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Make y the subject :
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y - 3 = 4(x + 8)
y - 3 = 4x + 32 // Apply distributive property
y = 4x + 35 // Add 3 to both sides

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Identify Slope :
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Slope = 4

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Find a point on the line :
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When x = 0,
y = 4(0) + 35 // Sub x = 0 into the equation
y = 35 // Combine like terms
Point = (0, 35)

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Answers :
(a) Slope = 4
(b) One point on the line is (0, 35)
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4 0

4 years ago

A house costs $600 more than twelve times the lot on which it was built. If the lot cost x dollars, what did the house cost? Whi

Mnenie [13.5K]

House=600+12lot
lot=x
house=600+12x

answer is first expresion

7 0

3 years ago

Read 2 more answers

Can someone help me plz

andrew-mc [135]

Use hypotenuse leg theorem.

7 0

3 years ago

The point (5, −2) is on the terminal ray of angle θ, which is in standard position. Without evaluating, explain how you would fi

Anni [7]

1. Check the picture attached.

let P, M, N be the points :P(5, -2), M(5, 0), N(0, -2)

ABP is the angle with measure ∅, where BA is the initial ray and BP the terminal ray.

2. m(PNM)=∅ as can be easily checked

In triangle PNM, NP=5, PM=2 and NM= $\sqrt{ 5^{2} + 2^{2} }= \sqrt{25+4}= \sqrt{29}$ units

3. let opp=opposite side, adj=adjecent side, hyp=hypothenuse

sin ∅ = opp/hyp= $\frac{2}{\sqrt{29}}$

cos ∅ = adj/hyp= $\frac{5}{\sqrt{29}}$

tan ∅ = opp/adj=2/5

cot ∅ = adj/opp=5/2

sec ∅ = hyp/adj= $\frac{\sqrt{29}}{5}$

csc ∅ = hyp/opp= $\frac{\sqrt{29}}{2}$

4 0

4 years ago

Read 2 more answers

Write the matrix equation as a system of linear equations without matrices. [Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column

ASHA 777 [7]

Answer:

$x-2y-4z=8\\5x-4y+7z=5\\-7x+5y+2z=-3$

Step-by-step explanation:

Given the matrix equation below:

$\left(\begin{array}{ccc}1&-2&-4\\5&-4&7\\-7&5&2\end{array}\right)\left(\begin{array}{ccc}x\\y\\z\end{array}\right) =\left(\begin{array}{ccc}8\\5\\-3\end{array}\right)$

We are required to write the matrix equation as a system of linear equations without matrices.

$\left(\begin{array}{ccc}1&-2&-4\\&&\\&&\end{array}\right)\left(\begin{array}{ccc}x\\y\\z\end{array}\right) =\left(\begin{array}{ccc}8\\&\\&\end{array}\right)$is equivalent to: x-2y-4z=8$

$\left(\begin{array}{ccc}\\5&-4&7\\&&\end{array}\right)\left(\begin{array}{ccc}x\\y\\z\end{array}\right) =\left(\begin{array}{ccc}&\\5\\&\end{array}\right)$ is equivalent to: 5x-4y+7z=5$

$\left(\begin{array}{ccc}&\\&\\-7&5&2\end{array}\right)\left(\begin{array}{ccc}x\\y\\z\end{array}\right) =\left(\begin{array}{ccc}&\\&\\-3\end{array}\right)$ is equivalent to: -7x+5y+2z=-3$

Therefore, we have the system of equations:

$x-2y-4z=8\\5x-4y+7z=5\\-7x+5y+2z=-3$

8 0

3 years ago