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2 years ago

Find the length of leg x x=side a is 15side b is 12side x=

Mathematics

1 answer:

Paul [167]2 years ago

6 0

Answer:

x=9

Step-by-step explanation:

a=hyp=15

b=base=12

x=perp=?

using pathagorus theorem:

${hyp}^{2} = {base}^{2} + {perp}^{ 2} \\ {15}^{2} = {12}^{2} + {x}^{2} \\ 225 = 144 + x \\ 225 - 144 = {x}^{2} \\ 81 = {x}^{2} \\ x = 9$

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hoa [83]

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Step-by-step explanation:

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2 years ago

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Crazy boy [7]

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In a 3-digit number, the hundreds digit is one more than the one's digit, and the tens digit is twice the hundreds digit. If the

arsen [322]

3 digits
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2 years ago

The linear function f i with values f(-1) = 10 and f(5) = -1 is f(x)

zhenek [66]

Answer:

Use the two points to compute the slope, m, then use one of the points in the form y=m(x)+b to find the value of b.

Explanation:

The equation for the slope, m, of a line is:

m=y1−y0x1−x0 [1]

The equation f(2)=−1 tells us that x0=2andy0=−1; substitute this into equation [1]:

m=y1−−1x1−2 [2]

The equation f(5)=4 tells us that x1=5andy1=4; substitute this into equation [2]:

m=4−−15−2 [3]

m=53

Substitute 53 for m into the equation y=m(x)+b

y=53x+b [4]

Substitute 2 for x and -1 for y and the solve for b:

−1=53(2)+b

b=−133

Substitute −133 for b in equation [4]:

y=53x−133 [5]

Check:

−1=53(2)−133
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Hope this helps

3 0

2 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:

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

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3 years ago

Find the length of leg x x=side a is 15side b is 12side x=​

Find the length of leg x x=side a is 15side b is 12side x=