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4 years ago

What is the value for X, Y, and Z?

Mathematics

2 answers:

kvv77 [185]4 years ago

8 0

Answer:

x+132=180 [ linear pair]

x=180-132=48

y=132[ opposite angles]

z=x=48 [ opposite angles]

if we add all these four angles we get 360 degree.

Mark it as brainliest!!!

Oduvanchick [21]4 years ago

7 0

Answer:

_________

z is 48

x is 48

y is 132

_________

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A school has two kindergarten classes. There are 21 children in Ms. Toodle's kindergarten class. Of these, 17 are "pre-readers"�

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3 years ago

2x - 3y = 13 x + 2y = -4

-Dominant- [34]

Answer:

(2, - 3 )

Step-by-step explanation:

Given the 2 equations

2x - 3y = 13 → (1)

x + 2y = - 4 → (2)

Rearrange (2) expressing x in terms of y by subtracting 2y from both sides

x = - 4 - 2y → (3)

Substitute x = - 4 - 2y into (1)

2(- 4 - 2y) - 3y = 13 ← distribute and simplify left side

- 8 - 4y - 3y = 13

- 8 - 7y = 13 ( add 8 to both sides )

- 7y = 21 ( divide both sides by - 7 )

y = - 3

Substitute y = - 3 into (3) for corresponding value of x

x = - 4 - 2(- 3) = - 4 + 6 = 2

Solution is (2, - 3 )

6 0

3 years ago

the sum of the ages of a man and his wife is at least 104 years if the man is x years old and the wife is x - 4 years old find t

larisa [96]

Answer:

x ≥ 54

Step-by-step explanation:

x+x-4 ≥ 104

2x ≥ 104 + 4

2x ≥ 108

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6 0

3 years ago

What is the measurement of angle C? Part C

Genrish500 [490]

No idea

Forgot picture

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3 years ago

Read 2 more answers

Problem 4: Solve the initial value problem

pishuonlain [190]

Separate the variables:

$y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx$

Separate the left side into partial fractions. We want coefficients a and b such that

$\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}$

$\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}$

$\implies 1 = a(y-2)+b(y+1)$

$\implies 1 = (a+b)y - 2a+b$

$\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13$

So we have

$\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx$

Integrating both sides yields

$\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx$

$\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C$

$\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C$

$\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C$

$\dfrac{y-2}{y+1} = e^{3x + C}$

$\dfrac{y-2}{y+1} = Ce^{3x}$

With the initial condition y(0) = 1, we find

$\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12$

so that the particular solution is

$\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}$

It's not too hard to solve explicitly for y; notice that

$\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}$

Then

$1 - \dfrac3{y+1} = -\dfrac12 e^{3x}$

$\dfrac3{y+1} = 1 + \dfrac12 e^{3x}$

$\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}$

$y+1 = \dfrac6{2+e^{3x}}$

$y = \dfrac6{2+e^{3x}} - 1$

$\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}$

7 0

2 years ago