How do i factor 10x-10y?

The length of the hypotenuse of a right triangle is 157 units. The length of one leg of the triangle is 132. Lara wrote the foll

ASHA 777 [7]

the correct answer for this is the last choice:

It is incorrect because the length of the unknown side is the square root of 7,225.

4 0

3 years ago

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The measures of ∠1, ∠2, and ∠3 are 40%, 12.5%, and 25% of the sum of the angle measures of the quadrilateral. Find the value of

sweet [91]

The value of x is 81

Step-by-step explanation:

The sum of the interior angles of any quadrilateral is 360°

The measure of ∠1 is 40% of the sum of the angle measures of the quadrilateral
The measure of ∠2 is 12.5% of the sum of the angle measures of the quadrilateral
The measure of ∠3 is 25% of the sum of the angle measures of the quadrilateral
We need to find the value of x

∵ The figure have 4 sides

∴ The figure is a quadrilateral

∵ The sum of the measures of the interior angles of a

quadrilateral is 360°

- Add the four angles and equate the sum by 360

∴ m∠1 + m∠2 + m∠3 + x = 360

∵ m∠1 = 40% of the sum of the angle measures of the quadrilateral

∴ m∠1 = 40% × 360 = $\frac{40}{100}$ × 360 = 144°

∵ m∠2 = 12.5% of the sum of the angle measures of the quadrilateral

∴ m∠2 = 12.5% × 360 = $\frac{12.5}{100}$ × 360 = 45°

∵ m∠3 = 25% of the sum of the angle measures of the quadrilateral

∴ m∠3 = 25% × 360 = $\frac{25}{100}$ × 360 = 90°

- Substitute these values in the equation above

∴ 144 + 45 + 90 + x = 360

- Add the like terms in the left hand side

∴ 279 + x = 360

- Subtract 279 from both sides

∴ x = 81°

The value of x is 81

Learn more:

You can learn more about the polygons in brainly.com/question/6281564

#LearnwithBrainly

5 0

2 years ago

Why might you want to use the commutative property to change the order of the integers in the following sum before adding

Delicious77 [7]

<span>Commutative Property is the property in which you can move around numbers in numerical operations like, addition and multiplication while retaining their result. In contrast to subtraction and division in which position is an important factor for every result, here it is regardless. </span>Why might you want to use this property?<span>Well, most importantly it suits the operation of addition and hence, to ensure the arrangement of the number is in symmetric proportion to its counterpart such as 3 + 2=2 + 3. Or rather, understanding that the equations in both sides are but the same and equal in sum. Thus, this is much more usable or will make more sense if used in a larger scale of complex equations and integers.<span>
</span></span>

4 0

2 years ago

Which of these is the absolute value parent function?

kotykmax [81]

Answer:

Option d. is correct answer.

Step-by-step explanation:

<em>f(x) = [x]</em><em> </em>is the absolute value parent function.

7 0

3 years ago

An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem

Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

Case 1: both balls are white.

At the beginning we have $b+w$ balls. We want to pick a white one, so we have a probability of $\frac{w}{b+w}$ of picking a white one.

If this happens, we're left with $w-1$ white balls and still $b$ black balls, for a total of $b+w-1$ balls. So, now, the probability of picking a white ball is

$\dfrac{w-1}{b+w-1}$

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

$\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}$

Case 2: both balls are black

The exact same logic leads to a probability of

$\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}$

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

$\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of $\frac{w}{b+w}$ of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability $\frac{b}{b+w}$ of picking a black ball with both picks.

This leads to an overall probability of

$\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}$

Of picking two balls of the same colour.

We want to prove that

$\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Expading all squares and products, this translates to

$\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}$

As you can see, this inequality comes in the form

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k}$

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y$

And this is our case, because in our case we have

$x=b^2+w^2$
$y=b^2+w^2+2bw$ so, y has an extra piece and it is larger
$k=b+w$ which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2

4 0

2 years ago