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3 years ago

How do i find the y-intercept of an eqaution that passes throught two points

1 answer:

3 0

They way to find the y-intercept from two points is to first find the slope, which can be derived from the formula $m=\frac{y_2-y_1}{x_2-x_1}$ .

Then, you can plug in one of the two points $(x_1,y_1)$ or $(x_2,y_2)$ into the formula y=mx+b to solve for b, which is your y-intercept.

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-12x2 + 45x + 12

bazaltina [42]

Answer:

3(4x + 1)(x - 4) Answer choice B

Step-by-step explanation:

-12x^2 + 45 + 12

You would take out the -3 in the equation.

-3(4x^2 -15x - 4)

You can then multiply the first 4 and the last -4 and get a -16. You can then find two numbers that would multiply to -16 and add to -15.

You would do this by listing the factors of -16, which the numbers 1 and -16 would work.

You would then substitute each value in for the 17x and add them together, which this is just separating the value 17x.

3(4x^2 - 16x + x - 4)

3(4x^2 - 16x)(x - 4)

You can then take ou ta four for the first function in the parentheses.

3(4x(x - 4) 1(x - 4))

Then you take the function from the outside and the inside.

3(4x + 1)(x - 4)

The answer to the question would be B.

8 0

2 years ago

What is 166.44 dived by 4. PLEASE USE LONG DIVISION. The person who USES LONG DIVISION TO GET THE ANSWER GETS 10 points and brai

ArbitrLikvidat [17]

Answer:

41.61

Step-by-step explanation:

Just divide.

$\begin{matrix}4\overline{|\smallspace166.44}\:\:\:\:\:\:\:\end{matrix}$

5 0

1 year ago

Read 2 more answers

VXYZ is a rectangle with a length of 20cm and a width of 12 cm. The length of the rectangle is increased by 30%. The width of th

katrin [286]

Step-by-step explanation:

the area is

length × width

increasing the length by 30% means multiplying the original length by 1.3.

increasing the width by 10% means multiplying the original width by 1.1.

so, the new area is (by using the old length and width)

length × 1.3 × width × 1.1 = length × width × 1.3×1.1 =

= old area × 1.3 × 1.1 = old area × 1.43

so, the area increases by 43%.

8 0

2 years ago

Https://docs.goo gle.com/forms/d/e/1FAIpQLSe6vil7bUd47FA8fnqS5sh23ZjJ1OUe5X2ZlogwND9MoURn0w/viewform?usp=sf_link

Galina-37 [17]

Answer:

uh.... Why??

Step-by-step explanation:

3 0

2 years ago

Find the direction cosines and direction angles of the vector. (Give the direction angles correct to the nearest degree.) 5, 1,

Dahasolnce [82]

Answer:

The direction cosines are:

$\frac{5}{\sqrt{42} }$ , $\frac{1}{\sqrt{42} }$ and $\frac{4}{\sqrt{42} }$ with respect to the x, y and z axes respectively.

The direction angles are:

40°, 81° and 52° with respect to the x, y and z axes respectively.

Step-by-step explanation:

For a given vector a = ai + aj + ak, its direction cosines are the cosines of the angles which it makes with the x, y and z axes.

If a makes angles α, β, and γ (which are the direction angles) with the x, y and z axes respectively, then its direction cosines are: cos α, cos β and cos γ in the x, y and z axes respectively.

Where;

cos α = $\frac{a . i}{|a| . |i|}$ ---------------------(i)

cos β = $\frac{a.j}{|a||j|}$ ---------------------(ii)

cos γ = $\frac{a.k}{|a|.|k|}$ ----------------------(iii)

And from these we can get the direction angles as follows;

α = cos⁻¹ ( $\frac{a . i}{|a| . |i|}$ )

β = cos⁻¹ ( $\frac{a.j}{|a||j|}$ )

γ = cos⁻¹ ( $\frac{a.k}{|a|.|k|}$ )

Now to the question:

Let the given vector be

a = 5i + j + 4k

a . i = (5i + j + 4k) . (i)

a . i = 5 [a.i is just the x component of the vector]

a . j = 1 [the y component of the vector]

a . k = 4 [the z component of the vector]

Also

|a|. |i| = |a|. |j| = |a|. |k| = |a| [since |i| = |j| = |k| = 1]

|a| = $\sqrt{5^2 + 1^2 + 4^2}$

|a| = $\sqrt{25 + 1 + 16}$

|a| = $\sqrt{42}$

Now substitute these values into equations (i) - (iii) to get the direction cosines. i.e

cos α = $\frac{5}{\sqrt{42} }$

cos β = $\frac{1}{\sqrt{42} }$

cos γ = $\frac{4}{\sqrt{42} }$

From the value, now find the direction angles as follows;

α = cos⁻¹ ( $\frac{a . i}{|a| . |i|}$ )

α = cos⁻¹ ( $\frac{5}{\sqrt{42} }$ )

α = cos⁻¹ ( $\frac{5}{6.481}$ )

α = cos⁻¹ (0.7715)

α = 39.51

α = 40°

β = cos⁻¹ ( $\frac{a.j}{|a||j|}$ )

β = cos⁻¹ ( $\frac{1}{\sqrt{42} }$ )

β = cos⁻¹ ( $\frac{1}{6.481 }$ )

β = cos⁻¹ ( 0.1543 )

β = 81.12

β = 81°

γ = cos⁻¹ ( $\frac{a.k}{|a|.|k|}$ )

γ = cos⁻¹ ( $\frac{4}{\sqrt{42} }$ )

γ = cos⁻¹ ( $\frac{4}{6.481}$ )

γ = cos⁻¹ (0.6172)

γ = 51.89

γ = 52°

Conclusion:

The direction cosines are:

$\frac{5}{\sqrt{42} }$ , $\frac{1}{\sqrt{42} }$ and $\frac{4}{\sqrt{42} }$ with respect to the x, y and z axes respectively.

The direction angles are:

40°, 81° and 52° with respect to the x, y and z axes respectively.

3 0

2 years ago