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3 years ago

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1 answer:

Reika [66]3 years ago

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Hope this helps........

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The table below shows the proportional relationship between the weight of dog food, in ounces, and the number of bags of dog foo

luda_lava [24]

K = y/x
k = 30/2
k = 15 <=== constant of proportionality

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3 years ago

The following foreign adoptions (in the United States) occurred during these particular years. 2006 2010China 63403465Ethiopia 7

worty [1.4K]

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Step-by-step explanation:

Please find the attached file for the solution.

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3 years ago

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viktelen [127]

Answer: the answer is B

Step-by-step explanation:

The duty of the Vaishyas is sacrifice, giving gifts, agriculture, breeding, and trade. However, later the Sudras take over agriculture and breeding and the Vaishyas become traders, merchants, landowners, and money-lenders. ... The Vaishyas also focused on religious education, because they wanted to be “twice-born”

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3 years ago

Read 2 more answers

PLEASE HELP!!! A.S.A.P!!

Ksivusya [100]

Answer:

142m

Step-by-step explanation:

This problem can be solved by simply using the pythagorean theorem, as you stated at the beginning of the problem, which is: $a^{2} +b^{2} =c^{2}$

You are given the <em>a</em> side and <em>b</em> side that are needed for this equation, so it's all a matter of plugging in the information you have:

$110^{2} +90^{2} =c^{2}$

$110^{2} =12100$

$90^{2} =8100$

$12100+8100=c^{2}$

$20200=c^{2}$

Now, because the <em>c</em> is still squared, you must take the square root of 20200 in order to get the length of just side <em>c</em>:

$\sqrt{20200}$ ≈142m

7 0

3 years ago

Solve the following equation by factoring:9x^2-3x-2=0

olya-2409 [2.1K]

Answer:

The two roots of the quadratic equation are

$x_1= - \frac{1}{3} \text{ and } x_2= \frac{2}{3}$

Step-by-step explanation:

Original quadratic equation is $9x^{2}-3x-2=0$

Divide both sides by 9:

$x^{2} - \frac{x}{3} - \frac{2}{9}=0$

Add $\frac{2}{9}$ to both sides to get rid of the constant on the LHS

$x^{2} - \frac{x}{3} - \frac{2}{9}+\frac{2}{9}=\frac{2}{9}$ ==> $x^{2} - \frac{x}{3}=\frac{2}{9}$

Add $\frac{1}{36}$ to both sides

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{2}{9} +\frac{1}{36}$

This simplifies to

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{1}{4}$

Noting that (a + b)² = a² + 2ab + b²

If we set a = x and b = $\frac{1}{6}\right)$ we can see that

$\left(x - \frac{1}{6}\right)^2$ = $x^2 - 2.x. (-\frac{1}{6}) + \frac{1}{36} = x^{2} - \frac{x}{3}+\frac{1}{36}$

So

$\left(x - \frac{1}{6}\right)^2=\frac{1}{4}$

Taking square roots on both sides

$\left(x - \frac{1}{6}\right)^2= \pm\frac{1}{4}$

So the two roots or solutions of the equation are

$x - \frac{1}{6}=-\sqrt{\frac{1}{4}}$ and $x - \frac{1}{6}=\sqrt{\frac{1}{4}}$

$\sqrt{\frac{1}{4}} = \frac{1}{2}$

So the two roots are

$x_1=\frac{1}{6} - \frac{1}{2} = -\frac{1}{3}$

and

$x_2=\frac{1}{6} + \frac{1}{2} = \frac{2}{3}$

7 0

2 years ago