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3 years ago

Teach me how to add this problem

Mathematics

1 answer:

Arada [10]3 years ago

7 0

start with the ones and every time it goes over ten take the first number and add it to the tens for example 1+8+4+8=21 take the one and put it in the one spot below. Then take the 2 and add it to the six in the tens spot so it would be Instead of 6+8+3+2 it would be 8+8+3+2 which would equal 21 then put the 1 u=in the ten spot below add the 2 to the hundreds spot making is 5+5+7=4=14 put the 4 Below and add the 1 the the thousands making it 3+4+5+1=13 put the one on the ten thousands and put the three below the whole problem would be 13,411.

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Pls help I’m really confused How are vertical

skad [1K]

Vertical angles:
Vertical Angles are the angles opposite each other when two lines cross.
https://www.mathsisfun.com/geometry/vertical-angles.html
The link has a visual of vertical angles.

Adjacent Angles:
Adjacent angles are two angles that share a common vertex point and a common side. They basically sit right next to each other.
https://www.mathsisfun.com/geometry/adjacent-angles.html
The link has a visual of adjacent angles.

Complementary Angles:
Two angles whose measures add to 90º.
Basically, together they make a right angle.

Supplementary Angles:
Two angles whose measures add to 180º.
Basically, together they form a straight line.

7 0

2 years ago

Add the following fractions. 3/8 + 5/12 = <br> a 8/20 <br> b 19/24 <br> c 8/24 <br> d 1/3

loris [4]

3/8 + 5/12 =

B. 19/24

8 0

3 years ago

Read 2 more answers

Construct a polynomial function with the following properties: fifth degree, 4 is a zero of multiplicity 3, −4 is the only other

Nonamiya [84]

Answer:

Step-by-step explanation:

Hello,

degree 5

4 is a zero of multiplicity 3 -> (x-4)^3 is a factor

-4 is the only other zero, so the multiplicity is 5-3=2 -> (x+4)^2 is a factor

leading coefficient is 4 so we can write

$\boxed{4(x-4)^3(x+4)^2}$

If there is something that you do not understand or you are blocked somewhere let us know what / where.

Thank you.

8 0

3 years ago

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

What is the value of k in the function ƒ(x) = 112 - kx if ƒ(-3) = 121?

DedPeter [7]

f(x) = 112 - kx
f(-3) = 112 - k(-3)
f(-3) = 112 - (-3k)
f(-3) = 112 + 3k
121 = 112 + 3k
- 112 - 112
      9 = 3k
      3    3
      3 = k

6 0

2 years ago