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3 years ago

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A line with a negative slope intersects a horizontal line 3 units above the x-axis. Select each point that can not be the inters

ection of the two lines.
(−2,3)

(−3,5)

(3,−2)

(0,−3)

2 answers:

kykrilka [37]3 years ago

8 0

(3,-2) this is what I think the answer is

Ainat [17]3 years ago

6 0

It can’t not be (-2,3) or (-3,5) or (0,-3)

You might be interested in

A marker is randomly selected from a drawer that contains 20 green, 44 orange, and 30 blue markers. Which statement is true?.

Pavlova-9 [17]

Answer:

See below.

Step-by-step explanation:

Total number of markers: 20 + 44 + 30 = 94

p(green) = 20/94 = 0.213

p(orange) = 44/94 = 0.468

p(blue) = 30/94 = 0.319

Choose the correct answer by comparing your choices with the 3 probabilities above.

8 0

2 years ago

the length of a rectangle is 5 inches more then the width. the area of the rectangle is equal to 2 inches more than 4 times the

Svet_ta [14]

P = 2(L + W)
L = W + 5
A = 4P + 2

P = 2(W + 5 + W)
P = 2(2W + 5)
P = 4W + 10

A = 4P + 2
A = 4(4W + 10) + 2
A = 16W + 42

A = L * W
A = W(W + 5)
A = W^2 + 5W

W^2 + 5W = 16W + 42
W^2 + 5W - 16W - 42 = 0
W^2 - 11W - 42 = 0
(W + 3)(W - 14) = 0

W - 14 = 0
W = 14 <==

L = W + 5
L = 14 + 5
L = 19 <==

P = 2(19 + 14)
P = 2(33)
P = 66

A = L * W
A = 19 * 14
A = 266

answer : length = 19, width = 14....perimeter = 66....area = 266

5 0

3 years ago

Shelley wants to purchase a cell phone package, which includes a free phone. The cost of the package is $0.10 per minute, for th

GarryVolchara [31]

Answer:

The function would be linear because the rate of change is constant.

Step-by-step explanation:

Given

$Base= \$89.00$

$Rate = \$0.10$ per minute

See attachment for complete question and options

First, we write the function that calculates the cost (C(t)) for t minutes.

This is calculated as:

$Cost = Base + Rate * t$

$C(t) = 89.00 + 0.10 * t$

$C(t) = 89.00 + 0.10t$

Rewrite as:

$C(t) = 0.10t + 89.00$

A function that has the above format is referred to as a linear function which has the general format

$f(x) = mx + b$

Where

m represents the slope/rate and it is constant

From the list of given options, (a) is correct.

6 0

3 years ago

Need help (brainliest will be given)

ruslelena [56]

Answer:

-16

Step-by-step explanation:

−2(32)−

8

2

+6

=(−2)(9)−

8

2

+6

=−18−

8

2

+6

=−18−4+6

=−22+6

=−16

5 0

3 years ago

Read 2 more answers

(10 points) Consider the initial value problem y′+3y=9t,y(0)=7. Take the Laplace transform of both sides of the given differenti

Rashid [163]

Answer:

The solution

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3 t}$

Step-by-step explanation:

<u><em>Explanation</em></u>:-

Consider the initial value problem y′+3 y=9 t,y(0)=7

<em>Step(i)</em>:-

Given differential problem

y′+3 y=9 t

<em>Take the Laplace transform of both sides of the differential equation</em>

L( y′+3 y) = L(9 t)

<em>Using Formula Transform of derivatives</em>

<em> L(y¹(t)) = s y⁻(s)-y(0)</em>

<em> By using Laplace transform formula</em>

<em> </em> $L(t) = \frac{1}{S^{2} }$ <em> </em>

<em>Step(ii):-</em>

Given

L( y′(t)) + 3 L (y(t)) = 9 L( t)

$s y^{-} (s) - y(0) + 3y^{-}(s) = \frac{9}{s^{2} }$

$s y^{-} (s) - 7 + 3y^{-}(s) = \frac{9}{s^{2} }$

Taking common y⁻(s) and simplification, we get

$( s + 3)y^{-}(s) = \frac{9}{s^{2} }+7$

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

<em>Step(iii</em>):-

<em>By using partial fractions , we get</em>

$\frac{9}{s^{2} (s+3} = \frac{A}{s} + \frac{B}{s^{2} } + \frac{C}{s+3}$

$\frac{9}{s^{2} (s+3} = \frac{As(s+3)+B(s+3)+Cs^{2} }{s^{2} (s+3)}$

On simplification we get

9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Put s =0 in equation(i)

9 = B(0+3)

<em> B = 9/3 = 3</em>

Put s = -3 in equation(i)

9 = C(-3)²

<em>C = 1</em>

Given Equation 9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Comparing 'S²' coefficient on both sides, we get

9 = A s²+3 A s +B(s)+3 B +C(s²)

<em> 0 = A + C</em>

<em>put C=1 , becomes A = -1</em>

$\frac{9}{s^{2} (s+3} = \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}$

<u><em>Step(iv):-</em></u>

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

$y^{-}(s) =9( \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}) + \frac{7}{s+3}$

Applying inverse Laplace transform on both sides

$L^{-1} (y^{-}(s) ) =L^{-1} (9( \frac{-1}{s}) + L^{-1} (\frac{3}{s^{2} }) + L^{-1} (\frac{1}{s+3}) )+ L^{-1} (\frac{7}{s+3})$

<em>By using inverse Laplace transform</em>

<em></em> $L^{-1} (\frac{1}{s} ) =1$ <em></em>

$L^{-1} (\frac{1}{s^{2} } ) = \frac{t}{1!}$

$L^{-1} (\frac{1}{s+a} ) =e^{-at}$

<u><em>Final answer</em></u>:-

<em>Now the solution , we get</em>

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3t}$

5 0

3 years ago