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agasfer [191]
3 years ago
14

Help me please I'm a bit stuck ❤

Mathematics
1 answer:
dedylja [7]3 years ago
6 0
There we go!
X + 3x-10 + 70-x = 180
3x = 120
X = 40

The size of the largest angle is 3x - 10, because it is the biggest side.

Hope it helps, have a great day!
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Part A.
g100num [7]

Answer:

for points sorry nsm km dj

7 0
3 years ago
3. ADD:<br> 35 degrees 25 minutes 14 seconds<br> 50 degrees 38 minutes 55 seconds
ValentinkaMS [17]
1. 924.849971 rad s
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7 0
3 years ago
Select the correct answer from each drop-down menu. If u = 5 − 2i and v = 2 − 5i, u + v = and u− v = .
Alex Ar [27]

Answer:

u+v=7-7i

u-v=3+3i

Step-by-step explanation:

u+v

(5-2i)+(2-5i)

Put like terms together:

(5+2)+(-2i-5i)

Simplify:

7+-7i

7-7i

u-v

(5-2i)-(2-5i)

Distribute:

(5-2i)+(-2+5i)

Put like terms together:

(5-2)+(-2i+5i)

3+3i

6 0
3 years ago
Help on section A :)
N76 [4]
Hey there!

To start, the domain of a function is the set of x values a function has, in other words, the values the function has on the x-axis.

In this function, the graph appears to have a function going on infinitely in both directions (left and right). Because of this, your domain in interval notation would be (-∞,∞) 

**Your interval notation would be in parenthesis because you cannot include negative or positive infinity as one of your values since while your graph is approaching these values, it never really reaches it.

Hope this helps and have a wonderful day! :)
5 0
3 years ago
A stick is broken at a point, chosen at random, along its length. Find the probability that the ratio, R, of the length of the s
Oksanka [162]

Solution :

Let the distance of the stick from one break be X

And let us assume that $X \leq l/2$.

Here, l = length of stick

Therefore, $P(X

We know that, $R=\frac{X}{l-X}$  ,  so by definition we get

$X =\frac{lR}{1+R}$

The cumulative distribution function for R is

$P(R

When it starts at zero, then r =0. It ends at one when the r has a maximum

value of one.

The probability density function is given by

\frac{d}{dr}(\frac{2r}{1+r})= \frac{2}{(r+1)^2}

Now integrating, we find E(R) and $E(R)^2$ gives :

$E(R) =\int\limits^1_0 \frac{2r}{(1+r)^2} \, dr = 2 \ln 2-1 $

$E(R)^2= 3 - 4\ln 2$

Therefore, Var(R)= $2-4(\ln \ 2)^2$

3 0
3 years ago
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