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3 years ago

Determine whether the lines are parallel, intersect, or coincide y=-7x-+1 y=-7x-3

Mathematics

1 answer:

Diano4ka-milaya [45]3 years ago

7 0

Answer:

Parallel

Step-by-step explanation:

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ANSWER IS C

MA_775_DIABLO [31]

Answer:

x <2

Step-by-step explanation:

2.5 – 1.2x < 6.5 – 3.2x

Add 3.2x to each side

2.5 – 1.2x+3.2x < 6.5 – 3.2x+3.2x

2.5 +2x < 6.5

Subtract 2.5 from each side

2.5+2x-2.5<6.5-2.5

2x<4

Divide by 2

2x/2 < 4/2

x <2

8 0

2 years ago

Is square root of 15 a interger

cricket20 [7]

Answer:

Step-by-step explanation:

8 0

3 years ago

Easy math please help

insens350 [35]

1) False

Adjacent angles must share a common side/ray.

2) B and D

Adjacent angles are those that are directly next to each other and share a common side.

3) C

Angles 1 and 2 are congruent. Angles 3 and 4 are congruent. Both of these pairs have angles that are opposite each other.

4) B

Angles 1 and 2 add up to 180. Angles 3 and 4 add up to 180. Both of these pairs of angles are supplementary.

5) A

These angles are directly next to each other and share a common side.

Hope this helps!! :)

4 0

2 years ago

Please help! If you do you will get 'Brainliest'

salantis [7]

Answer:

D is your answer.

Step-by-step explanation:

if you do 2 numbers to the right you get 3 and now the half would go in between 3 which would make your answer 3 and one half

7 0

3 years ago

Read 2 more answers

15 points!!! GEOMETRY help please!!<br> Questions attached!

dolphi86 [110]

3. Look at the picture.

We have the right angle triangle. We know the sum of measures of angles in triangle is equal 180°. Therefore:

$x^o+90^o+43^o=180^o\\\\x^o+133^o=180^o\ \ \ |-133^o\\\\x^o=47^o$

$Answer:\ \boxed{47^o}$

4.
Look at the picture.

Use Pythagorean theorem:

$x^2+14^2=(10+x)^2\\\\x^2+196=10^2+2\cdot10\cdot x+x^2\ \ \ \ |-x^2\\\\196=100+20x\ \ \ |-100\\\\20x=96\ \ \ |:20\\\\x=4.8$

$Used:\ (a+b)^2=a^2+2ab+b^2$

$Answer:\ \boxed{4.8\ units}$

5.
TRUE: 1; 2; 4

6.
We find a slope of the line OP:

$m=\dfrac{y_2-y_1}{x_2-x_1}\\\\O(2;\ 6)\to x_1=2;\ y_1=6\\\\P(4;\ 3)\to x_2=4;\ y_2=3\\\\m=\dfrac{3-6}{4-2}=\dfrac{-3}{2}$

We have: $OP:\ y=-\dfrac{3}{2}x+b$

Now, we must find the slope of the line perpendicular to the line OP.
We know:

$k:y=m_1x+b;\ l:y=m_2x+c\\\\k\ \perp\ l\iff m_1m_2=-1$

therefore

$-\dfrac{3}{2}m_2=-1\ \ \ |\cdot\left(-\dfrac{2}{3}\right)\\\\m_2=\dfrac{2}{3}$

So. We have the answer! :)

$Answer:\ \boxed{y=\dfrac{2}{3}x+\dfrac{1}{3}}$

6 0

3 years ago