I need help please

Plz help with this one toooo

Yuki888 [10]

Answer:y.5

Step-by-step explanation:5 times 3 = 15/2 = 7.5

6 0

3 years ago

What is the equation of the line that passes through the point (4,−2) and has a slope of −2?

AVprozaik [17]

Answer:y=-2x+6

Step-by-step explanation:

You want to find the equation for a line that passes through the point (4,-2) and has a slope of -2.

First of all, remember what the equation of a line is:

y = mx+b

Where:

m is the slope, and

b is the y-intercept

To start, you know what m is; it's just the slope, which you said was -2. So you can right away fill in the equation for a line somewhat to read:

y=-2x+b.

Now, what about b, the y-intercept?

To find b, think about what your (x,y) point means:

(4,-2). When x of the line is 4, y of the line must be -2.

Because you said the line passes through this point, right?

Now, look at our line's equation so far: . b is what we want, the -2 is already set and x and y are just two "free variables" sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the the point (4,-2).

So, why not plug in for x the number 4 and for y the number -2? This will allow us to solve for b for the particular line that passes through the point you gave!.

(4,-2). y=mx+b or -2=-2 × 4+b, or solving for b: b=-2-(-2)(4). b=6.

6 0

2 years ago

What is 2 1/2 + 5 1/2?

ZanzabumX [31]

Answer:

8

Step-by-step explanation:

add whole numbers first, and then add the fractions

2+5=7

1/2+1/2=1

7+1=8

8

4 0

3 years ago

Read 2 more answers

Evaluate m + 7.5 when m = 8.6

Nat2105 [25]

Answer:

16.1

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

B) there is a line with positive slope that is tangent to both circles. find the equation of this line, and (c) determine the po

IgorC [24]

Given that there is a line with positive slope that is tangent to both circles $x^2 + y^2 = 1$ and $(x-3)^2 + y^2 = 4$ .

Let the point of tangency in the first circle be (a, b) and the point of tangency in the second circle be (c, d), then the slope of the tangent to the first circle is given by

$2x+\left.2y \frac{dy}{dx}\right|_{(a, b)} =0 \\ \\ \Rightarrow \left.\frac{dy}{dx}\right|_{(a, b)} =- \frac{x}{y} =- \frac{a}{b}$

and the slope of the second circle is given by

$2(x-3)+\left.2y \frac{dy}{dx}\right|_{(c, d)} =0 \\ \\ \Rightarrow \left.\frac{dy}{dx}\right|_{(a, b)} =- \frac{x-3}{y} =- \frac{c-3}{d}$

From the first circle equation, we have

$x^2 = 1 - y^2 \\ \\ \Rightarrow a^2 = 1 - b^2$

and from the second equation, we have

 $(x-3)^2 = 4 - y^2 \\ \\ (c-3)^2 = 4 - d^2$ 

Since the two slopes above are for the same line, then

 $\frac{a}{b}= \frac{c-3}{d} \\ \\ \Rightarrow ad=b(c-3) \\ \\ \Rightarrow a^2d^2=b^2(c-3)^2 \\ \\ \Rightarrow(1-b^2)d^2=b^2(4-d^2) \\ \\ \Rightarrow d^2-b^2d^2=4b^2-b^2d^2 \\ \\ \Rightarrow d^2=4b^2 \\ \\ \Rightarrow d=\pm2b$

Thus, we have

$\frac{a}{b}= \frac{c-3}{2b} \\ \\ \Rightarrow a= \frac{c-3}{2}$

7 0

2 years ago

I need help please ​

I need help please