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1 year ago

Factorise 15x+25thank you for any help

Mathematics

2 answers:

leva [86]1 year ago

6 0

The answer is this:
5(3x+5)

scoray [572]1 year ago

4 0

Answer: 5(3x+5)

Step-by-step explanation:

To factor this expression we need to pull out the greatest common factor of the 2 numbers, 15 and 25. This factor is 5.

Now divide both numbers by 5 and put a 5 on the outside of the new expression to look like: $5(3x+5)$ and this is as far as it can be factored.

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If LN=54 and LM=31, find MN

kirza4 [7]

In this item, it is unfortunate that a figure, drawing, or illustration is not given. To be able to answer this, it is assumed that these segments are collinear. Points L, M, and N are collinear, and that L lies between MN.

The length of the whole segment MN is the sum of the length of the subsegments, LN and LM. This can be mathematically expressed,
LN + LM = MN

We are given with the lengths of the smalller segments and substituting the known values,
MN = 54 + 31
MN = 85

<em>ANSWER: MN = 85</em>

7 0

1 year ago

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The cross section of rectangular prism A measures 3 units by 2 units. The cross section of triangular prism B has a base that me

mamaluj [8]

Answer:

Volume A= one third

Step-by-step explanation:

Use volume B

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2 years ago

Randy presses RAND on his calculator twice to obtain two random numbers between 0 and 1. Let $p$ be the probability that these t

anygoal [31]

Answer:

$\frac{\pi}{4}$

Step-by-step explanation:

Lets call x,y the numbers we obtain from the calculator. x and y are independent random variables of uniform[0,1] distribution.

Lets note that, since both x and y are between 0 and 1, then 1 is the biggest side of the triangle.

Lets first make a geometric interpretation. If the triangle were to be rectangle, then the side of lenght 1 should be its hypotenuse, and therefore x and y should satisfy this property:

x²+y² = 1

Remember that in this case we are supposing the triangle to be rectangle. But the exercise asks us to obtain an obtuse triangle. For that we will need to increase the angle obtained by the sides of lenght x and y. We can do that by 'expanding' the triangle, but if we do that preserving the values of x and y, then the side of lenght 1 should increase its lenght, which we dont want to. Thus, if we expand the triangle then we should also reduce the value of x and/or y so that the side of lenght 1 could preserve its lenght. With this intuition we could deduce that

x²+y² < 1

Now lets do a more mathematical approach. According to the Cosine theorem, a triangle of three sides of lenght a,b,c satisfies

a² = b²+c² - 2bc* cos(α), where α is the angle between the sides of lenght b and c.

Aplying this formula to our triangle, we have that

$1^2 = x^2 + y^2 - 2bc* cos(\alpha)$ , where $\alpha$ is the angle between the sides of lenght x and y.

Since the triangle is obtuse, then $\pi/2 < \alpha < \pi$ , and for those values $cos(\alpha)$ is negative , hence we also obtain

1 > x² + y²

Thus, we need to calculate P(x²+y² <1). This probability can be calculated throught integration. We need to use polar coordinates.

(x, y) = (r*cosФ,r*senФ)

Where r is between 0 and 1, and Ф is between 0 and $\pi /2$ (that way, the numbers are positive).

The jacobian matrix has determinant r, therefore,

${\int\int}\limits_{x^2+y^2 < 1} \, dxdy = \int\limits^1_0\int\limits^{\frac{\pi}{2}}_0 {r} \, d\phi dr = \frac{\pi}{2} * \int\limits^1_0 {r} \, dr = \frac{\pi}{2} * (\frac{r^2}{2} |^1_0) = \frac{\pi}{4}$