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1 year ago

Determine the equation of the line described. Put answer in the slope-intercept form, if possible.

Mathematics

1 answer:

julsineya [31]1 year ago

6 0

Step-by-step explanation:

do you have a picture of your work?

and if you have 9.9 in your graph and you get a point it can be valid otherwise it would not help you at all

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Really struggling on this

navik [9.2K]

Angle 4 would be 77degrees because angle two is vertical to angle four.

Angle 3 and Angle 1 are equal because they are vertical to each other.

You would subtract 180 (degrees) minus 77 (degrees) and get 103 (degrees).

So Angle 3 and Angle 1 would both be 103 degrees.

You would get 180 degrees from the line.

5 0

1 year ago

if 2 binders of the same size take up 3 inches on the shelf how much space will be needed for 28 of these binders?

Anna [14]

Answer:

Step-by-step explanation:

28x3=84

I hope i did that right

please give me Brainliest you dont have to tho

7 0

2 years ago

Find the value of ‘k’ if one zero of p(x) = kx2 – 14 x + 8 is 6 times the other.

Finger [1]

Answer:

k = 3

Step-by-step explanation:

one zero of p(x) = kx²– 14 x + 8 is 6 times the other.

let the zeros are ⇒ a and 6a

the equation kx²– 14 x + 8

can be written as k ( x²/k– 14/k x + 8/k)

So, a * 6a = 8/k ⇒ (1)

And a + 6a = 14/k ⇒ (2)

from (2) k = 14/(a + 6a) = 14/(7a) = 2/a

∴ a = 2/k

Substitute at (1)

a * 6a = 8/k

2/k * 12/k = 8/k

solve for k

24/k = 8

k = 24/8 = 3

the value of k = 3

And the zeros will be 2/3 and 4

5 0

3 years ago

81\3 using the distributive property

dalvyx [7]

The correct answer is 27

3 0

3 years ago

Find the slope of a line perpendicular to each given line

andre [41]

$k:\ y=m_1x+b_1\\l:\ y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}$

$115)\ -x-y=3\ \ \ |+x\\-y=x+3\ \ \ |\cdot(_1)\\y=-x-3\to m_1=-1\to\boxed{m_2=-\dfrac{1}{-1}=1}\\\\116)\ 0=-4x-y+5\ \ \ |+y\\y=-4x+5\to m_1=-4\to\boxed{m_2=-\dfrac{1}{-4}=\dfrac{1}{4}}\\\\117)\ 6-2y=x\ \ \ \ |-6\\-2y=x-6\ \ \ \ |:(-2)\\y=-\dfrac{1}{2}x+3\to m_1=-\dfrac{1}{2}\to\boxed{m_2=-\dfrac{1}{-\frac{1}{2}}=2}\\\\118)\ 5y=25+4x\ \ \ \ |:5\\y=5+\dfrac{4}{5}x\to m_1=\dfrac{4}{5}\to\boxed{m_2=-\dfrac{1}{\frac{4}{5}}=-\dfrac{5}{4}}$

$119)\ -15x-9y=9\ \ \ \ |+15x\\-9y=15x+9\ \ \ \ |:(-9)\\y=-\dfrac{15}{9}x-1\\y=-\dfrac{5}{3}x-1\to m_1=-\dfrac{5}{3}\to\boxed{m_2=-\dfrac{1}{-\frac{5}{3}}=\dfrac{3}{5}}$

8 0

3 years ago