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3 years ago

Write an equation of the line with y-intercept 3 that is parallel to y=-3/4x-4

Mathematics

1 answer:

Bond [772]3 years ago

4 0

Answer:

y = -(3/4)x + 3

Step-by-step explanation:

Any line parallel to y = -(3/4)x - 4 will have the same equation, except that the y-intercept will be different. Here we need only replace that -4 intercept with 3, obtaining:

y = -(3/4)x + 3.

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X+3Y=6 turned into slope-intercept form

GarryVolchara [31]

Y=-1/3x+2

Use photomath app

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3 years ago

Which of the following graphs represents the function f(x) = x2 + x − 6?

Darya [45]

The answer is: D

Step-by-step explanation:

Direction: Opens Up

Vertex: ( −1/2, −25/4)

Focus: (−1/2, −6)

Axis of Symmetry: x = -12

Directrix: y = -132

3 0

3 years ago

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In a parallelogram ABCD point K belongs to diagonal BD so that BK:DK=1:4. If the extension of AK meets BC at point E, what is th

olya-2409 [2.1K]

Answer:

$\frac{BE}{EC} =\frac{1}{3}$

Step-by-step explanation:

In the diagram below we have

ABCD is a parallelogram. K is the point on diagonal BD, such that

$\frac{BK}{CK} =\frac{1}{4}$

And AK meets BC at E

now in Δ AKD and Δ BKE

∠AKD =∠BKE ( vertically opposite angles are equal)

since BC ║ AD and BD is transversal

∠ADK = ∠KBE ( alternate interior angles are equal )

By angle angle (AA) similarity theorem

Δ ADK and Δ EBK are similar

so we have

$\frac{AD}{BE} =\frac{DK}{BK}$

$\frac{AD}{BE} =\frac{4}{1}$

$\frac{BC}{BE}=\frac{4}{1}$ ( ABCD is parallelogram so AD=BC)

$\frac{BE+EC}{BE}=\frac{4}{1}$ ( BC= BE+EC)

$\frac{BE}{BE} +\frac{EC}{BE}=\frac{4}{1}$

$1+\frac{EC}{BE}=4$

$\frac{EC}{BE}=3$ ( subtracting 1 from both side )

$\frac{EC}{BE}=\frac{3}{1}$

taking reciprocal both side

$\frac{BE}{EC} =\frac{1}{3}$

8 0

3 years ago

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

2 years ago

Do i give in to the munchies bro

Vesna [10]

Answer:

wut thut u better tell me -_-

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers