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2 years ago

Whats this answer i need it fast uwu

Mathematics

1 answer:

IrinaVladis [17]2 years ago

3 0

Find slope (y2-y1)/(x2-x1)
Pick points: (1,10) , (2,8)
(8-10)/(2-1) = -2/1 = -2
Y = -2x + b
10 = -2 + b, b = 12
It would be the first response
12 = initial while -2 is rate of change

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Green thumb landscaping want to build a $124000.00 greenhouse in 2 years. The company sets up a sinking fund with payments made

DENIUS [597]

Answer:

$<em>150,858.5</em>

Step-by-step explanation:

The formula for calculating compound interest is expressed as;

A = P(1+r/n)^nt

P is the Principal = $124000.00

r is the rate = 12% = 0.12

t is the total time = 2 years

n is the time of compounding = 1/4 = 0.25(quarterly)

Substitute into the formula;

A= 124000(1+0.12/(0.25))^(0.25)(2)

A = 124000(1+0.48)^0.5

A = 124000(1.48)^0.5

A = 124000(1.2166)

A = 150,858.5

<em>The amount after 2 years if compounded quarterly is 150,858.5</em>

8 0

3 years ago

What is the slope of a line parallel to the line containing (-6,1) and (3,-2)

timama [110]

Answer:

$m = -\frac{3}{9}$

Step-by-step explanation:

Use the following formula:

slope: $m = (y_{2} - y_{1})/(x_{2} - x_{1})$

Let:

$(x_{1} , y_{1}) = (-6 , 1)\\ (x_{2} , y_ {2}) = (3 , -2)\\$

Plug in the corresponding numbers to the corresponding variables:

$m = \frac{-2 - 1}{3 - (-6)}\\\\m = \frac{-3}{3 + 6}\\\\m = \frac{-3}{9}$

$m = -\frac{3}{9}$ is your answer.

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3 years ago

Help me Please....................

MakcuM [25]

9514 1404 393

Answer:

1.3363

Step-by-step explanation:

The basic idea here is to find an expression for the direction vector between a point on L1 and a point on L2. Then, solve for the points on L1 and L2 that make that vector perpendicular to both lines L1 and L2. (The dot product of direction vectors is zero.) The distance between the points found is the shortest distance between the lines.

Let P be a point on L1. Then the parametric equation for P is ...

P = (6t, 0, -t) . . . . . . origin + t × direction vector

Let Q be a point on L2. The direction vector for L2 is given by the difference between the given points. It is (4-1, 1-(-1), 6-1) = (3, 2, 5). Then the parametric equation for Q is ...

Q = (3s+1, 2s-1, 5s+1) . . . . (1, -1, 1) + s × direction vector

The direction vector for PQ is ...

Q -P = (3s+1-6t, 2s-1, 5s+1+t)

The dot product of this and the two lines' direction vectors will be zero:

(3s+1-6t, 2s-1, 5s+1+t)·(6, 0, -1) = 0 = 13s -37t +5 . . . perpendicular to L1

(3s+1-6t, 2s-1, 5s+1+t)·(3, 2, 5) = 0 = 38s -13t +6 . . . perpendicular to L2

The solution to these equations is ...

s = -157/1237

t = 112/1237

Then (Q-P) becomes (94, -1551, 564)/1237, and its length is ...

|PQ| = √(94² +1551² +564²)/1237 ≈ 1.3363

The distance between the two lines is about 1.3363 units.

8 0

2 years ago

(PLESE HELP) Factor the following expression. Simplify your answer.

mars1129 [50]

The factor of the expression $5p(p + 2)^{\frac{2}{3} } + 4(p + 2)^{\frac{1}{3} }$ is $(p + 2)^{\frac{2}{3} } (5p^{3} + 10p^{2} + 20p + 4)$

To answer the question, we need to know what factorization is

<h3>What is factorization?</h3>

Factorization is the process of breaking down an expressing into a simpler form containing its factors.

Since

$5p(p + 2)^{\frac{2}{3} } + 4(p + 2)^{\frac{1}{3} }$

Since $(p + 2)^{\frac{1}{3} }$ is common, we factor it out. So, we have

$(p + 2)^{\frac{2}{3} } (5p(p + 2)^{2} + 4)$

Expanding the bracket, we have

$(p + 2)^{\frac{2}{3} } (5p(p^{2} + 2p + 4) + 4) = (p + 2)^{\frac{2}{3} } (5p^{3} + 10p^{2} + 20p + 4)$

So, the factor of the expression $5p(p + 2)^{\frac{2}{3} } + 4(p + 2)^{\frac{1}{3} }$ is $(p + 2)^{\frac{2}{3} } (5p^{3} + 10p^{2} + 20p + 4)$

Learn more about factorization here:

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1 year ago

Construct three parallel lines. Then construct two arbitrary non parallel transversal of the parallel lines. Make sure that the

zzz [600]

The image lines constructed are called parallel lines. They are parallel because they are all equidistant to each other at every point and at no point between any of the lines will there be an intersection of the three lines.

<h3>What does it mean for two or more lines to be equidistant from each other?</h3>

Equidistance is simply another word for equal in distance. That is, parallel lines maintain equal distance from each other at all points along them.

See the attached construction as well as more insights about parallel lines from the link below.

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7 0

1 year ago