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Firdavs [7]
3 years ago
14

8/2(2+2) PLEASE HELP ME!!!!! Stop reporting this!!! I need help!!!

Mathematics
2 answers:
Eva8 [605]3 years ago
5 0

Answer:

16

Step-by-step explanation:

you do 2+2 which is 4 then 8/2 is 4 and you multiple 4*4

Sergio [31]3 years ago
5 0
16 is the answer I’m positive
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Please help I’ll give you brainiest !
Salsk061 [2.6K]

Answer:

{290,315,340, 365, 390, 415, 440, 465} That's the answer. Its hard for me to explain how i got it.

4 0
3 years ago
If H is the midpoint of TE and TH = 11.22, then TE =
Alexus [3.1K]

Answer:

22.44

Step-by-step explanation:

If it’s the midpoint, just multiply it by 2 giving you full length

5 0
4 years ago
Finding the Sum of a System of Equations
s344n2d4d5 [400]
The Answer is : -3y = 3
4 0
4 years ago
Find the slope if it exits 2x-5y=6
77julia77 [94]
In standard form, slope is always the negative of the x coefficient over the y coefficient.

If you dont want to have that memorized, you can use algebra to get the equation of the line into slope-intercept form (y=mx+b)

Set it equal to y

2x - 5y = 6
2x = 5y + 6
2x - 6 = 5y
(2/5)x - 6/5 = y
Now it is in slope intercept form. In slope intercept form, the coefficient multiplying with x is the slope of the line. Therefore, the slope equals 2/5.
6 0
3 years ago
Read 2 more answers
In this problem we consider an equation in differential form Mdx+Ndy=0. (4x+2y)dx+(2x+8y)dy=0 Find My= 2 Nx= 2 If the problem is
zheka24 [161]

Answer:

f(x,y)=2x^2+4y^2+2xy=C_1\\\\Where\\\\y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )

Step-by-step explanation:

Let:

M(x,y)=4x+2y\\\\and\\\\N(x,y)=2x+8y

This is and exact equation, because:

\frac{\partial M(x,y)}{\partial y} =2=\frac{\partial N}{\partial x}

So, define f(x,y) such that:

\frac{\partial f(x,y)}{\partial x} =M(x,y)\\\\and\\\\\frac{\partial f(x,y)}{\partial y} =N(x,y)

The solution will be given by:

f(x,y)=C_1

Where C1 is an arbitrary constant

Integrate \frac{\partial f(x,y)}{\partial x} with respect to x in order to find f(x,y):

f(x,y)=\int\ {4x+2y} \, dx =2x^2+2xy+g(y)

Where g(y) is an arbitrary function of y.

Differentiate f(x,y) with respect to y in order to find g(y):

\frac{\partial f(x,y)}{\partial y} =2x+\frac{d g(y)}{dy}

Substitute into \frac{\partial f(x,y)}{\partial y} =N(x,y)

2x+\frac{dg(y)}{dy} =2x+8y\\\\Solve\hspace{3}for\hspace{3}\frac{dg(y)}{dy}\\\\\frac{dg(y)}{dy}=8y

Integrate \frac{dg(y)}{dy} with respect to y:

g(y)=\int\ {8y} \, dy =4y^2

Substitute g(y) into f(x,y):

f(x,y)=2x^2+4y^2+2xy

The solution is f(x,y)=C1

f(x,y)=2x^2+4y^2+2xy=C_1

Solving y using quadratic formula:

y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )

4 0
4 years ago
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