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Phoenix [80]
4 years ago
13

What are the necessary criteria for a line to be perpendicular to the given line and have the same y-intercept? The slope is and

contains the point (0, 2). The slope is and contains the point (0, −2). The slope is and contains the point (0, 2). The slope is and contains the point (0, −2).
Mathematics
2 answers:
xxMikexx [17]4 years ago
6 0
(-3,-4) (3, 0)
slope = (0 +4) / (3 + 3) = 4/6 = 2/3
perpendicular so slope = -3/2

line on graph
y = mx + b
0 = 2/3 (3) + b
0 = 2+ b
b = -2

y intercept (0,-2)

answer 
<span>The slope is -3/2 and contains the point (0, −2). </span>
Serga [27]4 years ago
6 0

Answer:

i believe the answer is d

Step-by-step explanation:

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Plz help me in this question
professor190 [17]

Answer:

Sorry

Step-by-step explanation:

I wish i could help!!

6 0
4 years ago
Which equation can be solved by using this expression?
Marat540 [252]

<u>Answer:</u>

The correct answer option is B. 2 = 3x + 10x^2

<u>Step-by-step explanation:</u>

We are to determine whether which of the given equations in the answer options can be solved using the following expression:

x=\frac{-3 \pm\sqrt{(3)^2+4(10)(2)} }{2(10)}

Here, a = 10, b = 3 and c=-2.

These requirements are fulfilled by the equation 4 which is:

12=3x+10x^2

Rearranging it to get:

10x^2+3x-2=0

Substituting these values of a,b,c in the quadratic formula:

x= \frac{-b \pm \sqrt{b^2-4ac} }{2a}

x= \frac{-3 \pm\sqrt{(3)^2-4(10)(-2)} }{y}

3 0
3 years ago
PLEASE HELP ME ON THIS!!!!
exis [7]
The last two choices are appropriate.

Both give you f(1) = 7, f(2) = 9.50, f(3) = 12.00, ... as they should.
7 0
4 years ago
Assume that x and y are both differentiable functions of t and find the required values of dy/dt and dx/dt. xy = 4 (a) Find dy/d
kompoz [17]

Answer:

a) -0.9375

b) 1.25

Step-by-step explanation:

We are given the following in the question:

f(x,y) = 4

where x and y are both differentiable functions of t.

a)  x = 8 and dx/dt = 15

\\xy = 4\\\\y = \dfrac{4}{x}\\\\y = \dfrac{4}{8}=\dfrac{1}{2}\\\\\dfrac{d(f(x,y))}{dt} = 0\\\\x\dfrac{dy}{dt} + y\dfrac{dx}{dt} = 0\\\\8\dfrac{dy}{dt}  + \dfrac{1}{2}(15) = 0 \\\\\dfrac{dy}{dt} = \dfrac{1}{8}\times \dfrac{-15}{2}\\\\\dfrac{dy}{dt} = -\dfrac{15}{16}\\\\\dfrac{dy}{dt}=-0.9375

b) x = 1 and dy/dt = –5

xy = 4\\\\y = \dfrac{4}{x}\\\\ y= 4\\\\\dfrac{d(f(x,y))}{dt} = 0\\\\x\dfrac{dy}{dt} + y\dfrac{dx}{dt} = 0\\\\(1)(-5)  + 4\dfrac{dx}{dt} = 0 \\\\\dfrac{dx}{dt} = \dfrac{5}{4}\\\\\dfrac{dx}{dt} = 1.25

3 0
4 years ago
Factor the Following:<br><br>1.) 8<br>2.) -12<br>3.) -3y²<br>4.) 6x²+36x<br>5.) 7x -14x²​
Alex787 [66]

Answer:

1.) 8 = 8

2.) -12 = -12

3.) -3y² = -3y²

4.) 6x² + 36x = 6x(x + 6)

5.) 7x - 14x² = 7x(1 - 2x)

~

3 0
3 years ago
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