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zloy xaker [14]
3 years ago
8

Solve using elimination. 9x + 4y = -6 -7x + y = -20

Mathematics
1 answer:
zaharov [31]3 years ago
3 0
9x+4y=-6
4(-7x+y=-20)

9x+4y=-6
-(-28x+4y=-80)

37x=74

x=74/37

x=2


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4 years ago
Me need help pweaseee can anyone help
Kryger [21]

Answer:

B

Step-by-step explanation:

A 90 degree rotation would be a rotation of 90° about the origin. Knowing this, we can eliminate C since it is a 180° rotation. We can also eliminate A, as it would be a 0° or 360° rotation. This leaves us with B and C.

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6 0
3 years ago
(2, 7) and (-4,1)<br> In point slope form
GREYUIT [131]

Answer:

First you need to find the slope with this equation.

<h3>m=\frac{y2-y1}{x2-x1}  m=slope</h3><h3>m=\frac{1-7}{-4-2}</h3><h3>m=\frac{-6}{-6}</h3><h3>m=1</h3>

So the point slope form formula is:

<h3>y-y1=m(x-x1)</h3><h3>y-7=1(x-2)</h3><h3>y-7=x-2</h3>

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3 0
3 years ago
Write 3/50 as a decimal and as a percent<br><br> i need help
Alex17521 [72]

Answer:

0.06 and to change that to a percentage all you do is move the decimal point 2 places to the right so the answer would be <u>6</u><u> </u>percent

7 0
4 years ago
The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes
saul85 [17]

Answer:

a)0.08  , b)0.4  , C) i)0.84  , ii)0.56

Step-by-step explanation:

Given data

P(A) =  professor arrives on time

P(A) = 0.8

P(B) =  Student aarive on time

P(B) = 0.6

According to the question A & B are Independent  

P(A∩B) = P(A) . P(B)

Therefore  

{A}' & {B}' is also independent

{A}' = 1-0.8 = 0.2

{B}' = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B')  (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

P(\frac{B'}{A}) = \frac{P(B'\cap A)}{P(A)} = \frac{0.4\times 0.8}{0.8} = 0.4

Part c)

Assume the events are not independent

Given Data

P(\frac{{A}'}{{B}'}) = 0.4

=\frac{P({A}'\cap {B}')}{P({B}')} = 0.4

P({A}'\cap {B}') = 0.4 x P({B}')

= 0.4 x 0.4 = 0.16

P({A}'\cap {B}') = 0.16

i)

The probability that at least one of them is on time

P(A\cup B) = 1- P({A}'\cap {B}')  

=  1 - 0.16 = 0.84

ii)The probability that they are both on time

P(A\cap  B) = 1 - P({A}'\cup {B}') = 1 - [P({A}')+P({B}') - P({A}'\cap {B}')]

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

6 0
3 years ago
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