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lyudmila [28]
3 years ago
6

13 is equal to 5 less than the quotient of a number and 10 written as a variable expression.

Mathematics
1 answer:
Bingel [31]3 years ago
6 0
The answer:
13=5-(x÷10)
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The ____ is the horizontal distance between two points.
timama [110]

The length of line segment is the horizontal distance between two points

Suppose

a line has endpoints

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  • (x2,Y2)

The length of the line is

  • √(x_1-x_2)²+(y_1-y_2)²
6 0
1 year ago
Find the area of rectangle ABCD
satela [25.4K]

9514 1404 393

Answer:

  D. 12

Step-by-step explanation:

There are a number of ways to find the area of this rectangle. Perhaps the most straightforward is to find the lengths of the sides and multiply those. The distance formula is useful.

  d = √((x2 -x1)^2 +(y2 -y1)^2)

Using the two upper-left points, we find the length of that side to be ...

  d = √((3 -0)^2 +(3 -0)^2) = √(9 +9) = √18 = 3√2

Similarly, the length of the lower-left side is ...

  d = √((-2 -0)^2 +(-2 -0)^2) = √(4+4) = √8 = 2√2

Then the area of the rectangle is ...

  A = LW

  A = (3√2)(2√2) = 3·2·(√2)^2 = 3·2·2 = 12

The area of rectangle ABCD is 12.

_____

Other methods include subtracting the area of the corner triangles from the area of the bounding square:

  5^2 -2(1/2)(3·3) -2(1/2)(2·2) = 25 -9 -4 = 12

4 0
2 years ago
I need help plzz someone help meee :)
Eddi Din [679]

Answer:

Table 1 = Proportional

y is 7 times x

Table 2 = Not proportional

Step-by-step explanation:

5 0
2 years ago
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Answer plzzz first right answer gets brainlyiest answe okay I know I spelled it wrong
yan [13]
9.   she spends 1 hour total  and 1/2 hour on spanish. So she has another 1/2  on another subject ( as she spends equal time on each subject)

So number of subjects = 1 / 1/2 = 1 * 2 = 2


8 0
3 years ago
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
3 years ago
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