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3 years ago

One number exceeds another by 2 the sum of the number is 48 what are the numbers

1 answer:

3 0

$x-\ first\ number\\y-\ second\ number\\\\
 \left \{ {{x-y=2} \atop {x+y=48}} \right. \\\\ \left \{ {{x=2+y} \atop {x+y=48}} \right. \\\\Substution\ method\\\\
2+y+y=48\\\\
2+2y=48\\\\
2y=46\ \ \ |:2\\\\
\boxed{y=23}$

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Consider the ODE, dy dx = y 2 1 + x (2) subject to condition y = 1 when x = 0, use your Euler code from class (modified if neces

makkiz [27]

Answer:

Computation.

Step-by-step explanation:

I'm not really sure if that's the analytical solution of the inital value problem,

because y(0)=11-ln(1-0)(3)=11. Howevwer, let us procede with the given values...

Let us assume that we are going to use euler with n=2 (two steps) and h=0.2(the size of each step)

The update rules of the Euler Methode are

$X_i = X_{i-1}+h=X_0+ih$

$m_i=\dfrac{dY}{dX} \biggr \rvert_{x_i} \\\\Y_{i+1}=m_i\cdot h+y_i$

Since the initial value problem tells us that Y=1 when X=0, we know that

$X_0=0$ and that $Y_0=1$ . Then, we have

$X_0=0\\\\X_1=0.2\\\\X_2=0.4$

and

$Y_0=1\\\\m_0=21 \cdot Y_0 + X_0 \cdot 2=21 \cdot 1 + 0=21\\\\Y_1=0.2 \cdot 21 + 1 =5.2\\\\m_1=21 \cdot Y_1 + X_1\cdot 2=21 \cdot 5.2 + 0.2 \cdot 2=109.6\\ \\Y_2=m_1 \cdot h + Y_1 = 109.6 \cdot 0.2 + 5.2= 27.12$

which gives us the points (0,1), (0.2, 5.2) and (0.4, 27.12).

Now, since we want to compare the analyticaland the Euler result, we first compute the value of y=11-ln(1-x)(3) for the values x=0, 0.2 and 0.4. We get that

$y(0)=11-\ln(1-0)(3)=11-ln(1)(3)=11\\\\Y(1)=11-\ln(1-0.2)(3)=11.67\\\\Y(2)=11-\ln(1-0.4)(3)=12.53$

and we compute $Y(i)-Y_i$ for each i.

It holds

$Y(0)-Y_0=11-1=10\\\\Y(1)-Y_1=11.67-5.2=6.47\\\\Y(2)-Y_2=12.53-27.12=-14.59$

which tells us that we have a really bad approximation, as I already stated there must be a mistake in the analytical solution since the intial values don't coincide. Also note that the curve that we get using the euler methose is growing faster than the analitical solution.

4 0

3 years ago

Solve the following problems. Remember that all reasoning must be explained, and all steps of math work must be shown!

iris [78.8K]

Answer to Question 1:

<OGE (also m˚) = 19˚

Step-by-step explanation:

We know that all the angles in a triangle add up to 180˚.

Given:

<EGO = 71˚

<OEG = 90˚ (the angle forms a right angle because line EG is perpendicular to point O)

To Solve:

⇒ Since we know two angles, we should know the third; and they all add up to 180˚. This can be written as:

71˚ + 90˚ + <OGE = 180˚

⇒ Solve for <OGE by simplifying:

161˚ + <OGE = 180˚

⇒ Subtract 161˚ from both sides to get:

161˚ − 161˚ + <OGE = 180˚ − 161˚

⇒ Isolate <OGE:

<OGE = 19˚

Answer: <OGE (also m˚) = 19˚

Answer:

<OAC (also x˚) = 50˚

<OCB (also y˚) = 90˚

Step-by-step explanation:

–How to solve for <OCB (also y˚)

Just by looking, you can already tell that <OCB equals 90˚ because it forms a right angle.

Answer: <OCB (also y˚) = 90˚

–How to solve for <OAC (also x˚)

Given (and what we can already know):

<AOC = 40˚

<OCA = 90˚

⇒ Reason: Line AB is a straight line and a straight line equals 180˚. We already know on side of the line because <OCB is 90˚, and so the other side, <OCA, should also be 90˚, since the both sides that make up the straight line equal 180˚ (90˚ + 90˚ = 180˚). You can also tell that <OCA forms a right angle in the triangle because line AB is perpendicular to point O, and right angles are 90˚.

To Solve:

⇒ Since we know two angles in the triangle, we should know the third; and they all add up to 180˚. This can be written as:

40˚ + 90˚ + <OAC = 180˚

⇒ Solve for <OAC by simplifying:

130˚ + <OAC = 180˚

⇒ Subtract 130˚ from both sides to get:

130˚ − 130˚ + <OAC = 180˚ − 130˚

⇒ Isolate <OAC:

<OAC = 50˚

Answer: <OAC (also x˚) = 50˚

I hope you understand and that this helps with your question! :)

3 0