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2 years ago

Desperate please Hurry Question Down Below

Mathematics

2 answers:

Olenka [21]2 years ago

6 0

Answer:

(7+9)/4= 16/4= 4

Step-by-step explanation:

Nezavi [6.7K]2 years ago

5 0

Answer:

Step-by-step explanation:

7+9 = 16, so 16/4 u then see how many times 4 goes into 16, which is 4. because 4 x 4 = 16

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dolphi86 [110]

Answer:

x = 12/5

Step-by-step explanation:

Since the original is 10 and the second is 6, that means that the scale is 10:6 or 5:3. It can also be related as %60 of the length.

Step 1: Make an equation

10/6 = 4/x

Step 2: Cross Multiply

10x = 24

Step 3: Divide

10x / 10 = 24 / 10

24/10 = 12/5

x = 2.4

5 0

3 years ago

Read 2 more answers

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: ordinary differential equation ode integration separable variables initial value problem differential integral calculus

7 0

3 years ago

[[ PLEASE ANSWER ASAP (: ]]

Nady [450]

Angle A = 100 , Angle C = 100 , Angle D =80

6 0

3 years ago

Braden has already taken 1 page of notes on his own, and he will take 2 pages during each hour of class. Write an equation that

beks73 [17]

Answer:

2h=2p

1h=1p