All categories

4 years ago

Write five fractions that each simplify to one-ninth

1 answer:

3 0

2/18
3/27
4/36
5/45
6/54

Take any number let's say 9
Then 9/9 = 1
Or 9/9 = 1/1
Then if you multiply the bottom half of those by 9, you get
9/9 becomes 9/81
and
1/1 becomes 1/9

So to generate any fraction equivalent to 1/9
take any number and write it as one, i.e. 5/5 = 1 (or 1/1)
and then multiply the bottom part by 9 = 5/45

You might be interested in

I need help !! It’s geometry!!!

murzikaleks [220]

Answer:

x = 13, y = 18

Step-by-step explanation:

Since it has a 90 degree angle and a 45 degree angle, it is an isosceles triangle, so x = 13.

Use Pyth Theorem to find y:

13^2 + 13^2 = c^2

169 + 169 = c^2

338 = c^2

18.384 = c

6 0

2 years ago

Please help me with this geometry question:((

Fudgin [204]

Answer:

x = 13

RQ = 160°

Step-by-step explanation:

Inscribed angles are half the intercepted arc angle.

5x + 15 = ½ (RQ)

RQ = 10x + 30

Arc angles add up to 360°.

10x + 30 + 9x + 17 + 6x − 12 = 360

25x + 35 = 360

25x = 325

x = 13

Therefore, RQ = 160°.

7 0

3 years ago

Read 2 more answers

Find an explicit solution to the Bernoulli equation. y'-1/3 y = 1/3 xe^xln(x)y^-2

NNADVOKAT [17]

$y'-\dfrac13y=\dfrac13xe^x\ln x\,y^{-2}$

Divide both sides by $\dfrac13y^{-2}(x)$ :

$3y^2y'-y^3=xe^x\ln x$

Substitute $v(x)=y(x)^3$ , so that $v'(x)=3y(x)^2y'(x)$ .

$v'-v=xe^x\ln x$

Multiply both sides by $e^{-x}$ :

$e^{-x}v'-e^{-x}v=x\ln x$

The left side can be condensed into the derivative of a product.

$(e^{-x}v)'=x\ln x$

Integrate both sides to get

$e^{-x}v=\dfrac12x^2\ln x-\dfrac14x^2+C$

Solve for $v(x)$ :

$v=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x$

Solve for $y(x)$ :

$y^3=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x$

$\implies\boxed{y(x)=\sqrt[3]{\dfrac14x^2e^x(2\ln x-1)+Ce^x}}$

4 0

3 years ago

Evaluate cot 120° without using a calculator by using ratios in a reference triangle.

Ksenya-84 [330]

The reference angle for $120\degree$ is $60\degree$ .

Recall that $cot(120\degree) = \frac{1}{tan(120\degree)}$

This implies that $cot(120\degree) = \frac{1}{tan(60\degree)}$

Recall also from $30\degree - 60\degreee - 90\degree$ triangle that

$tan(60\degree)=\sqrt(3)$ and also since $120\degree$ is the in the second quadrant, the tangent ratio is negative.

Putting all together we have

$cot(120\degree) = \frac{1}{-\sqrt(3)}$

Rationalizing the denominator gives

$cot(120\degree) = \frac{-\sqrt(3)}{3}$

6 0

3 years ago

Read 2 more answers

2.8 m<br> 3 m<br> What is the area of this triangle

vlada-n [284]

I think the answer is 7.6m

4 0

3 years ago