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3 years ago

15

Model using integer tiles or a number line to show your thinking and

1 answer:

Neko [114]3 years ago

6 0

The answer is -2-(-3)=1

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What is 0.111 repeating as a fraction

Liono4ka [1.6K]

Answer: 1/9

Explanation: these types of repeating decimals are equivalent to the fraction that has the number doing the repeating in its numerator and the number 9 in its denominator.

Hope this helps!

7 0

3 years ago

The symmetry of a hyperbola with a center at (h, k) only occurs at y = k. True or False

Vikki [24]

A hyperbola with a center at (0, 0) can be defined as x²/a² − y²/b² = ±1.<span>
</span>The statement "<span>The symmetry of a hyperbola with a center at (h, k) only occurs at y = k" </span>is false, because a hyperbola have many different orientations.
It doesn't have to be symmetric about the lines y = k or x = h.

6 0

3 years ago

Which is true about the two triangles below?

Bezzdna [24]

Answer:

A

Step-by-step explanation:

Hope this helps!

4 0

4 years ago

Read 2 more answers

A punter kicks a football that follows the path of a parabola. The football reaches its maximum height in 3 seconds. At which ti

aksik [14]

cjggkxxfjxxk mv gjjgxgkxgixgjxugxjgxj

8 0

4 years ago

Solve the equation using the substitution u = y/x. When u = y/x is substituted into the equation, the equation becomes separable

bekas [8.4K]

Answer:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

Step-by-step explanation:

First step: I'm going to solve our substitution for y:

$u=\frac{y}{x}$

Multiply both sides by x:

$ux=y$

Second step: Differentiate the substitution:

$u'x+u=y'$

Third step: Plug in first and second step into the given equation dy/dx=f(x,y):

$u'x+u=\frac{x(ux)+(ux)^2}{3x^2+(ux)^2}$

$u'x+u=\frac{ux^2+u^2x^2}{3x^2+u^2x^2}$

We are going to simplify what we can.

Every term in the fraction on the right hand side of equation contains a factor of $x^2$ so I'm going to divide top and bottom by $x^2$ :

$u'x+u=\frac{u+u^2}{3+u^2}$

Now I have no idea what your left hand side is suppose to look like but I'm going to keep going here:

Subtract u on both sides:

$u'x=\frac{u+u^2}{3+u^2}-u$

Find a common denominator: Multiply second term on right hand side by $\frac{3+u^2}{3+u^2}$ :

$u'x=\frac{u+u^2}{3+u^2}-\frac{u(3+u^2)}{3+u^2}$

Combine fractions while also distributing u to terms in ( ):

$u'x=\frac{u+u^2-3u-u^3}{3+u^2}$

$u'x=\frac{-u^3+u^2-2u}{3+u^2}$

Third step: I'm going to separate the variables:

Multiply both sides by the reciprocal of the right hand side fraction.

$u' \frac{3+u^2}{-u^3+u^2-2u}x=1$

Divide both sides by x:

$\frac{3+u^2}{-u^3+u^2-2u}u'=\frac{1}{x}$

Reorder the top a little of left hand side using the commutative property for addition:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

The expression on left hand side almost matches your expression but not quite so something seems a little off.

5 0

3 years ago