All categories

3 years ago

Choose the equation below that represents the line passing through the point (-3,-1) with a slope of 4.

1 answer:

8 0

Y = mx + b
slope(m) = 4
(-3,-1)...x = -3 and y = -1
now we sub...we r looking for b, the y int.
-1 = 4(-3) + b
-1 = -12 + b
-1 + 12 = b
11 = b

so ur equation is : y = 4x + 11

You might be interested in

one positive number is 5 times another number. the difference between the two numbers is 764, find the numbers

damaskus [11]

<h3><u>The first number, x, is equal to 955.</u></h3><h3><u>The second number, y, is equal to 191.</u></h3>

x = 5y

x - y = 764

Because we have a value for x, we can plug it into the second equation to solve for y.

5y - y = 764

Combine like terms.

4y = 764

Divide both sides by 4.

y = 191

Because we have a value for y, we can solve for x.

x = 5(191)

x = 955

5 0

3 years ago

What is the answer to this problem

alexandr402 [8]

9514 1404 393

Answer:

5.7·10^-3

Step-by-step explanation:

The expanded scale shows point P lies 7/10 of the way between 5 and 6 times 10^-3. Its value is 5.7·10^-3.

3 0

2 years ago

Can someone pls help me on this

Readme [11.4K]

Answer:

Step-by-step explanation:

100 + 100 + 6x + 6 + 12x - 8 = 360 Equation. Collect like terms.

198 + 18x = 360 Subtract 198 from both sides

18x = 360 - 198

18x = 162 Divide by 18

x = 162/18

x = 9

8 0

2 years ago

The curves r1(t) = 2t, t2, t4 and r2(t) = sin t, sin 5t, 2t intersect at the origin. Find their angle of intersection, θ, correc

masya89 [10]

Answer:

Therefore the angle of intersection is $\theta =79.48^\circ$

Step-by-step explanation:

Angle at the intersection point of two carve is the angle of the tangents at that point.

Given,

$r_1(t)=(2t,t^2,t^4)$

and $r_2(t)=(sin t , sin5t, 2t)$

To find the tangent of a carve , we have to differentiate the carve.

$r'_1(t)=(2,2t,4t^3)$

The tangent at (0,0,0) is [ since the intersection point is (0,0,0)]

$r'_1(0)=(2,0,0)$ [ putting t= 0]

$|r'_1(0)|=\sqrt{2^2+0^2+0^2} =2$

Again,

$r'_2(t)=(cos t ,5 cos5t, 2)$

The tangent at (0,0,0) is

$r'_2(0)=(1 ,5, 2)$ [ putting t= 0]

$|r'_1(0)|=\sqrt{1^2+5^2+2^2} =\sqrt{30}$

If θ is angle between tangent, then

$cos \theta =\frac{r'_1(0).r'_2(0)}{|r'_1(0)|.|r'_2(0)|}$

$\Rightarrow cos \theta =\frac{(2,0,0).(1,5,2)}{2.\sqrt{30} }$

$\Rightarrow cos \theta =\frac{2}{2\sqrt{30} }$

$\Rightarrow cos \theta =\frac{1}{\sqrt{30} }$

$\Rightarrow \theta =cos^{-1}\frac{1}{\sqrt{30} }$

$\Rightarrow \theta =79.48^\circ$

Therefore the angle of intersection is $\theta =79.48^\circ$ .

8 0

3 years ago

DOES ANYONE TAKE FLVS CHEMISTRY?<br> Please answer only if you do, thanks :)

Elena-2011 [213]

Answer:Yes i do

Step-by-step explanation:

Your welcome :)

4 0

3 years ago

Read 2 more answers