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Molodets [167]
3 years ago
12

Exterior of scalene triangle sides are 25 and 15 and x. solve x.

Mathematics
2 answers:
Lyrx [107]3 years ago
5 0

-- All three sides of a scalene triangle have different lengths.
So 'x' can't be 15 and it can't be 25.

-- 'x' must be 10 or more in order to reach between the ends
of the 25 and the 15.

-- 'x' must be less than 40 in order for the 25 and the 15 to reach
between its ends. 

So the value of 'x' must satisfy these conditions:

<em>0 < x < 15</em>
<em>15 < x < 25</em>
<em>25 < x < 40</em>

Any number that satisfies these conditions is an acceptable value for 'x'.


arlik [135]3 years ago
5 0
Let's say we have a triangle with three side lengths.
This triangle is really flat. So much so that you could put it on a number line.
It literally cannot get any flatter, and the largest side literally cannot get any bigger.
Well, the smaller two sides would <em>add <u />up</em> to equal the third.

From this we have realized something: The largest side of any triangle cannot be larger than the sum of the other two.

Let's think back to our problem.
What are the possibilities for x?

If x was the biggest side, it couldn't be any bigger than 40.

If x was a smaller side, then that would make 25 the bigger side.
15 and x together can't be any bigger than 25, so x has to be less than 10.

Since it's a scalene triangle, x cannot be 15 or 25 either, so add that too.

\{x|10\ \textless \ x\ \textless \ 40,\ x\neq15, x\neq25\}
<em />
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Svet_ta [14]
#2- yes because both can be divided or multiplied for example 3 and 9, u can multiply 3x3 to get 9 since the dilation is 3 then yes it is similar triangles
4 0
3 years ago
Question 5^ Need help ... ASAP ... your help is really appreciated ...
goblinko [34]

Answer:

See below

Step-by-step explanation:

I think we had a question similar to this before. Again, let's figure out the vertical and horizontal distances figured out. The distance from C at x=8 to D at x=-5 is 13 units while the distance from C at y=-2 to D at y=9 is 11 units. (8+5=13 and 2+9=11, even though some numbers are negative, we're looking at their value in those calculations)

Next, we have to divide each distance by 4 so we can apply it to the ratio. 13/4=\frac{13}{4} and 11/4=\frac{11}{4}. Next, we need to read the question carefully. It's asking us to place the point in the ratio <em>3</em> to <em>1</em> from <em>C</em> to <em>D</em>. The point has to be closer to endpoint D because of this. Let's take each of our fractions, multiply them by 3, then add them towards the direction of endpoint D to get our answer (sorry if that sounds confusing):

\frac{13}{4}*3 = \frac{39}{4}\\ 8-\frac{39}{4} =  \frac{-7}{4}\\\frac{11}{4}*3 = \frac{33}{4}  \\-2+\frac{33}{4} = \frac{25}{4}

Therefore, our point that partitions CD into a 3:1 ratio is (\frac{-7}{4}, \frac{25}{4}).

I'm not sure if there was more to #5 judging by how part B was cut off. From what I can understand of part B, however, I believe that Beatriz started from endpoint D and moved towards C, the wrong direction. She found the coordinates for a 1:3 ratio point.

Also, for #6, since a square is a 2-dimensional object, the answer needs to be written showing that. The answer for #6 is 9 units^2.

7 0
3 years ago
Someone help me please
kherson [118]

Answer:

I will just choosé the second one

Because it is possible to compare sizes In one triangle rather than comparing with another one which we don't reàlly know of its sizes.

4 0
3 years ago
A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa
gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4  | red dice) = \dfrac{1}{6}

P (4 | green dice) = \dfrac{3}{6} =\dfrac{1}{2}

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = \dfrac{1}{2}

The probability of two 1's and two 4's in the first dice can be calculated as:

= \begin {pmatrix}  \left \begin{array}{c}4\\2\\ \end{array} \right  \end {pmatrix} \times  \begin {pmatrix} \dfrac{1}{6}  \end {pmatrix}  ^4

= \dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4

= \dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4

= 6 \times ( \dfrac{1}{6})^4

= (\dfrac{1}{6})^3

= \dfrac{1}{216}

The probability of two 1's and two 4's in the second  dice can be calculated as:

= \begin {pmatrix}  \left \begin{array}{c}4\\2\\ \end{array} \right  \end {pmatrix} \times  \begin {pmatrix} \dfrac{1}{6}  \end {pmatrix}  ^2  \times  \begin {pmatrix} \dfrac{3}{6}  \end {pmatrix}  ^2

= \dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times  ( \dfrac{3}{6})^2

= 6 \times ( \dfrac{1}{6})^2 \times  ( \dfrac{3}{6})^2

= ( \dfrac{1}{6}) \times  ( \dfrac{3}{6})^2

= \dfrac{9}{216}

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = \dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}

The probability of two 1's and two 4's in both die = \dfrac{1}{432}  + \dfrac{1}{48}

The probability of two 1's and two 4's in both die = \dfrac{5}{216}

By applying  Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's )  = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's )  = \dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}

P(second die (green) | two 1's and two 4's )  = \dfrac{0.5 \times 0.04166666667}{0.02314814815}

P(second die (green) | two 1's and two 4's )  = 0.9

Thus; the probability the die chosen was green is 0.9

8 0
3 years ago
Which category would have the lowest conditional relative frequency?
lianna [129]
I think it might be B.. but i’m not sure bc I would think it would be amen who prefer dance
6 0
3 years ago
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