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3 years ago

Jim picks a five digit even number.

Mathematics

1 answer:

melamori03 [73]3 years ago

3 0

Answer:

Step-by-step explanation:

68240

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How do I find the midpoint

Hunter-Best [27]

To find the midpoint of a line use the midpoint formula which is M = (x1+x2/2 , y1 +y2/2) After plugging in the coordinates for your variables you should get (-1+5/2, -1+2/2) This simplifies to (2, 1/2) which is the coordinates for the middle point of the base XY.

5 0

2 years ago

Find the factorization of each of the following number:<br> A.18 B.24 C.38 D.81

const2013 [10]

Answer:

A: (3^2)*2

B: (2^3)*3

C: 2*19

D: (3^2)*9

Step-by-step explanation:

We have:

A) 18= 3*3*2= (3^2)*2

B) 24=2*2*2*3=(2^3)*3

C) 38 = 2*19

D) 81 =3*3*9=(3^2)*9

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2 years ago

4x^2 y+8xy'+y=x, y(1)= 9, y'(1)=25

jarptica [38.1K]

Answer with explanation:

$\rightarrow 4x^2y+8x y'+y=x\\\\\rightarrow 8xy'+y(1+4x^2)=x\\\\\rightarrow y'+y\times\frac{1+4x^2}{8x}=\frac{1}{8}$

--------------------------------------------------------Dividing both sides by 8 x

This Integration is of the form ⇒y'+p y=q,which is Linear differential equation.

Integrating Factor

$=e^{\int \frac{1+4x^2}{8x} dx}\\\\e^{\log x^{\frac{1}{8}+\frac{x^2}{2}}\\\\=x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}$

Multiplying both sides by Integrating Factor

$x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\times [y'+y\times\frac{1+4x^2}{8x}]=\frac{1}{8}\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\\\\ \text{Integrating both sides}\\\\y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\frac{1}{8}\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx \\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=-[x^{\frac{9}{8}}]\times\frac{ \Gamma(0.5625, -x^2)}{(-x^2)^{\frac{9}{16}}}\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=(-1)^{\frac{-1}{8}}[ \Gamma(0.5625, -x^2)]+C-----(1)$

When , x=1, gives , y=9.

Evaluate the value of C and substitute in the equation 1.

6 0

3 years ago

Find the inverse of ????−7 −2

Masja [62]

Answer:

$A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

Step-by-step explanation:

$A=\begin{bmatrix}-7 & -2\\ 4 & 1\end{bmatrix}$

$det\ A=-7\times 1-(-2)\times 4\\\Rightarrow det\ A=1$

$A^{-1}=\frac{1}{det\ A}\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A^{-1}=\frac{1}{1}\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

$\therefore A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

$A.A^{-1}=\begin{bmatrix}-7 & -2\\ 4 & 1\end{bmatrix}\times \begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A.A^{-1}=\begin{pmatrix}\left(-7\right)\cdot \:1+\left(-2\right)\left(-4\right)&\left(-7\right)\cdot \:2+\left(-2\right)\left(-7\right)\\ 4\cdot \:1+1\cdot \left(-4\right)&4\cdot \:2+1\cdot \left(-7\right)\end{pmatrix}\\\Rightarrow A.A^{-1}=\begin{pmatrix}1&0\\ 0&1\end{pmatrix}$

Hence, confirmed

6 0

3 years ago

Mr. Jones took a survey of college students and found that 60 out of 65 students are liberal arts majors. If a college has 8,943

Anit [1.1K]

Should be 5812.95 Students

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3 years ago