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Reika [66]
2 years ago
10

Someone please help me I’ll give out brainliest please dont answer if you don’t know

Mathematics
1 answer:
DaniilM [7]2 years ago
4 0

Answer:

y = 2x + 5

Step-by-step explanation:

2 is the slope so it goes with the x and 5 is the y-intercept so it is either added or subtracted.

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<img src="https://tex.z-dn.net/?f=3%201%2F3%20times%204%202%2F5" id="TexFormula1" title="3 1/3 times 4 2/5" alt="3 1/3 times 4 2
frutty [35]

Answer:

86.8 what form do u want this answer?

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
What is the slope of a line that passes through the points (12,7) and (-14,19)?
grandymaker [24]

Answer:

H) -6/13  

Step-by-step explanation:

To find the slope of the line between a pair of two points, use the slope formula, m = \frac{y_2-y_1}{x_2-x_1}. Substitute the x and y values of (12,7) and (-14,19) into the formula and simplify like so:  

m = \frac{(19)-(7)}{(-14)-(12)} \\m =\frac{19-7}{-14-12} \\m = \frac{12}{-26} \\m = -\frac{6}{13}

So, Option H is correct.

5 0
2 years ago
Jill simplified the expression shown.
inn [45]
The answer is 3 because 4x-x=3x

8 0
3 years ago
Read 2 more answers
What are the x- and y- coordinates of point P on the directed line segment from K to J such that P is Three-fifths the length of
Mrrafil [7]

Answer:

\left ( \frac{3x_2+5x_1}{8},\frac{3y_2+5y_1}{8} \right )

Step-by-step explanation:

Given: P is Three-fifths the length of the line segment from K to J

To find: x- and y-coordinates of point P on the directed line segment from K to J

Solution:

Section formula:

Let point K and J be (x_1,y_1)\,,\,(x_2,y_2) such that the point p(x,y) divides KJ in ratio m:n

Then coordinates of point  P are given by \left ( \frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n} \right )

Take m:n=3:5

So,

coordinates of point P = \left ( \frac{3x_2+5x_1}{3+5},\frac{3y_2+5y_1}{3+5} \right )=\left ( \frac{3x_2+5x_1}{8},\frac{3y_2+5y_1}{8} \right )

7 0
3 years ago
Divid 15 into two parts such that the sum of their reciprocal is 3/10
emmasim [6.3K]
Let x and y be the 2 parts of 15 ==> x + y=15 (given)

Reciprocal of x and y ==> 1/x +1/y ==> 1/x + 1/y = 3/10 (given)


Let's solve  1/x + 1/y = 3/10 . Common denominator = 10.x.y (reduce to same denominator)

 ==> (10y+10x)/10xy = 3xy/10xy ==> 10x+10y =3xy

But x+y = 15 , then 10x+10y =150 ==> 150=3xy and xy = 50

Now we have the sum S of the 2 parts that is S = 15 and 
their Product = xy =50
Let's use the quadratic equation for S and P==> X² -SX +P =0
Or X² - 15X + 50=0, Solve for X & you will find:
The 1st part of 15 is 10 & the 2nd part is 5
4 0
3 years ago
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