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3 years ago

How do you write 56% as a fraction in simplest form?

Mathematics

1 answer:

Alecsey [184]3 years ago

3 0

56% can be represented with 56/100. That fraction can be reduced (or simplified) to 14/25.

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Please help if you answer correctly I will give brainliest thank you!!!!!!!!

ivolga24 [154]

Answer:

The Answer is C

Step-by-step explanation:

3x-4y=5

2x+4y=-3

This equation is the answer to this problem because just by looking at it, you can immediately cancel out the y value and solve for x.

The other equations are wrong because their is extra work needed to eliminate the values.

4 0

3 years ago

How do i do this question? someone help

lesya [120]

There are 360 degrees in the rhombus which means the equilateral triangles has 180 degrees (because it is half the rhombus and also it is the angle sum of a triangle). It is equilateral meaning all sides and angles are the same so they all must be 60 (180/3)
Now this must mean that angle a is equal to the opposing angle being 60

8 0

4 years ago

Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu

Alisiya [41]

Let $y=C_1x+C_2x^3=C_1y_1+C_2y_2$ . Then $y_1$ and $y_2$ are two fundamental, linearly independent solution that satisfy

$f(x,y_1,{y_1}',{y_1}'')=0$
$f(x,y_2,{y_2}',{y_2}'')=0$

Note that ${y_1}'=1$ , so that $x{y_1}'-y_1=0$ . Adding $y''$ doesn't change this, since ${y_1}''=0$ .

So if we suppose

$f(x,y,y',y'')=y''+xy'-y=0$

then substituting $y=y_2$ would give

$6x+x(3x^2)-x^3=6x+2x^3\neq0$

To make sure everything cancels out, multiply the second degree term by $-\dfrac{x^2}3$ , so that

$f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y$

Then if $y=y_1+y_2$ , we get

$-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0$

as desired. So one possible ODE would be

$-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0$

(See "Euler-Cauchy equation" for more info)

6 0

4 years ago

HELPPP ME PLSSSS!!!!!!!!!!!!!!!!!!!!!!!1 I BEGGGG

NeX [460]

The answer is A Ezra is right

5 0

3 years ago

Help with this proof

musickatia [10]

AD = DB

Angle ADE = Angle CDB

Angle DAE = Angle DBC (Alternate angles)

Angle DEA = Angle DCB (Alternate angles)

Since you have two congruent angles and one congruent side, triangle ADE is congruent to triangle CDB. This means that DE is congruent to DC implying D is the midpoint of CE.

6 0

3 years ago