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3 years ago

What is the value of x in 8/9x = 4/3

2 answers:

5 0

X= 3/2 because when you divide 4/3 by 8/9 it actually becomes 4/3 times 9/8 then you multiply the top and bottom and simplify

Talja [164]3 years ago

4 0

X= 3/2 you multiply the inverse (9/8) and then you probably know what to do

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segment M'N' has endpoints located at M' (−2, 0) and N' (2, 0). It was dilated at a scale factor of 2 from center (2, 0). Which

Anna11 [10]

Answer:

A. MN is located at M 0,0) and N (2, 0) is half the length of M'N'.

Step-by-step explanation:

MN was dilated with a factor 2 is is half the length of M'N'.

7 0

3 years ago

Read 2 more answers

PLEASE HELP, WILL MARK BRAINLIEST

KatRina [158]

Answer:

(y-0)=-3(x-2)

Step-by-step explanation:

im not sure if this os correct but i tried

4 0

3 years ago

Dy/dx = 2xy^2 and y(-1) = 2 find y(2)

Anarel [89]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2887301

—————

Solve the initial value problem:

dy
——— = 2xy², y = 2, when x = – 1.
dx

Separate the variables in the equation above:

$\mathsf{\dfrac{dy}{y^2}=2x\,dx}\\\\
\mathsf{y^{-2}\,dy=2x\,dx}$

Integrate both sides:

$\mathsf{\displaystyle\int\!y^{-2}\,dy=\int\!2x\,dx}\\\\\\
\mathsf{\dfrac{y^{-2+1}}{-2+1}=2\cdot \dfrac{x^{1+1}}{1+1}+C_1}\\\\\\
\mathsf{\dfrac{y^{-1}}{-1}=\diagup\hspace{-7}2\cdot \dfrac{x^2}{\diagup\hspace{-7}2}+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{y}=x^2+C_1}$

$\mathsf{\dfrac{1}{y}=-(x^2+C_1)}$

Take the reciprocal of both sides, and then you have

$\mathsf{y=-\,\dfrac{1}{x^2+C_1}\qquad\qquad where~C_1~is~a~constant\qquad (i)}$

In order to find the value of C₁ , just plug in the equation above those known values for x and y, then solve it for C₁:

y = 2, when x = – 1. So,

$\mathsf{2=-\,\dfrac{1}{1^2+C_1}}\\\\\\
\mathsf{2=-\,\dfrac{1}{1+C_1}}\\\\\\
\mathsf{-\,\dfrac{1}{2}=1+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-1=C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-\dfrac{2}{2}=C_1}$

$\mathsf{C_1=-\,\dfrac{3}{2}}$

Substitute that for C₁ into (i), and you have

$\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}}\\\\\\
\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}\cdot \dfrac{2}{2}}\\\\\\
\mathsf{y=-\,\dfrac{2}{2x^2-3}}$

So y(– 2) is

$\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot (-2)^2-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot 4-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{8-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{5}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: <em>ordinary differential equation ode integration separable variables initial value problem differential integral calculus</em>

7 0

3 years ago

Solve the following equation 5x-15=50

denis-greek [22]

X=13 I say.................

5 0

3 years ago

Read 2 more answers

the ratio of weight to mass is constant.which statement describes the ratio of weight to mass and the value of x in the table

anzhelika [568]

HERE IS THE FULL PROBLEM!!!!

the table below lists four names and their corresponding aproximant weights on earth.

________________________________

mass (kilograms) | weight (newtons)

20 | 196

50 | 490

x | 1078

130 | 1274

140 | 1372

the ratio weight to mass in contant. wich statemen describes the ratio of weight to mass and the value of x in the table?

A. 98/10; X =90

B. 98/10; =110

C. 10/98; X =90

D. 10/98; X= 110

6 0

3 years ago