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IrinaVladis [17]
3 years ago
5

Line q passes through (−5, 5) and is parallel to the line 2x + y + 1 = 0.

Mathematics
2 answers:
Fed [463]3 years ago
6 0
Parallel lines has equal slopes.
Equation of the line 2x + y + 1 = 0 in slope intercept form is y = -2x - 1 with slope of -2. Therefore, the slope is -2.
Paul [167]3 years ago
3 0

Answer:

-2

Step-by-step explanation:

2020 ed

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What is fo)?<br> O O only<br> 0 -6 only<br> 0-2, 1, 1, and 3 only<br> 0 -6, -2, 1, 1, and 3 only
mash [69]

Answer:

  -6 only

Step-by-step explanation:

f(0) is the value of y when x=0. This is the graph of a 4th-degree polynomial, so is a function. There is only one y-value for x=0. That is where the graph crosses the y-axis, at y = -6.

  f(0) = -6 . . . . only

8 0
3 years ago
Explain why the binomial 9.r– 3 is not<br> factorable. Please explain in sentence<br> form.
Aneli [31]

Answer:

Uhhh, it can be factored into 3 * ( 3r - 1 )

Step-by-step explanation:

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2 years ago
What is the reciprocal of seven
Montano1993 [528]
The reciprocal of seven (7) would be 0.142857143
5 0
3 years ago
Read 2 more answers
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netineya [11]

Answer:

the square root of 10000 is 100

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Step-by-step explanation:

6 0
3 years ago
For what values of q are the two vectors A = i + j + kq each other and B-iq-23 + 2kg perpendicular to
Sloan [31]

Answer:

The value of q are 0.781,-1.281.

Step-by-step explanation:

Given : Two vectors A=i+j+kq and B=iq-2j+2kq are perpendicular to each other.

To find : The value of q ?

Solution :

When two vectors are perpendicular to each other then their dot product is zero.

i.e. \vec{A}\cdot \vec{B}=0

Two vectors A=i+j+kq and B=iq-2j+2kq

(i+j+kq)\cdot (iq-2j+2kq)=0

(1)(q)+(1)(-2)+(q)(2q)=0

q-2+2q^2=0

2q^2+q-2=0

2q^2+q-2=0

Using quadratic formula,

q=\frac{-1\pm\sqrt{1^2-4(2)(-2)}}{2(2)}

q=\frac{-1\pm\sqrt{17}}{4}

q=\frac{-1+\sqrt{17}}{4},\frac{-1-\sqrt{17}}{4}

q=0.781,-1.281

Therefore, The value of q are 0.781,-1.281.

6 0
3 years ago
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