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Anuta_ua [19.1K]
3 years ago
10

13. P = a + b + c (perimeter of a triangle);P= 12, a = 3, c = 5​

Mathematics
1 answer:
Kamila [148]3 years ago
5 0

Answer:

if you’re looking for b, then its 4

Step-by-step explanation:

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Find solution to the system of linear equations. 3x1 + 6x2 = 2 , 9x1 + 12x2 = 1
iogann1982 [59]

Answer:

x₁ = -1

x₂ = 5/6

Step-by-step explanation:

The equations of the given linear system are:

3x_1 +6x_2=2\\9x_1 +12x_2 = 1

Multiply the first equation by -2 and add it to the second one to solve for x₁:

-6x_1 -12x_2=-4\\9x_1 +12x_2 +(-6x_1 -12x_2)= 1-4\\3x_1=-3\\x_1=-1

Since we already know the value of x₁, simply replace it in the first equation  to solve for x₂:

3x_1 +6x_2=2\\x_2=\frac{2+3}{6} =\frac{5}{6}

The solution for the system is x₁ = -1 and x₂ = 5/6

6 0
3 years ago
Will choose brainliest and 12 points
sveta [45]
The answer is equation 2 as it expands to 0=0
7 0
3 years ago
A bag contains 5 blue marbles, 2 green marbles, and 3 yellow marbles. What is the probability of choosing one green marble and t
Vesna [10]

Answer:

Step-by-step explanation:

Total number of marbles = 5 + 2 + 3 = 10

Probability of choosing 1 green marble = 2/10

Probability of choosing 1 yellow marble = 3/10

Notice (and this is important) that the denominator didn't change. Why?

Because you replaced the first marble into the bag. That word replacement is critical in a problem of this nature. There is the term non replacement which means that the second draw would have a denominator of 9.

So what is the probability of P(green then yellow)?

P(green then yellow) = 3/10 * 2/10 = 6/100

Answer: P(green then yellow) = 3/50 because 6 / 100 reduces.

4 0
3 years ago
Which is the solution set of the inequality 15y-9<36
m_a_m_a [10]
15y - 9 <36.
Add 9 to each side.

15y- 9 (+9) < 36 + 9.
15y < 45.
y < 3.
Hope it helps! :)
7 0
3 years ago
Read 2 more answers
Help please!! How do I solve this
egoroff_w [7]
This is an exponential equation. We will solve in the following way. I do not have special symbols, functions and factors, so I work in this way
 2 on (2x) - 5 2 on x + 4=0 =>. (2 on x)2 - 5 2 on x + 4=0  We will replace expression ( 2 on x) with variable t => 2 on x=t  =. t2-5t+4=0 => This is quadratic equation and I solve this in the folowing way => t2-4t-t+4=0 =>     t(t-4) - (t-4)=0 => (t-4) (t-1)=0 => we conclude t-4=0 or t-1=0 => t'=4 and t"=1 now we will return t' => 2 on x' = 4 => 2 on x' = 2 on 2 => x'=2 we do the same with t" => 2 on x" = 1 => 2 on x' = 2 on 0 => x" = 0 ( we know that every number on 0 gives 1). Check 1: 2 on (2*2)-5*2 on 2 +4=0 =>                   2 on 4 - 5 * 4+4=0 => 16-20+4=0 =. 0=0 Identity proving solution. 
Check 2:   2 on (2*0) - 5* 2 on 0 + 4=0 => 2 on 0 - 5 * 1 + 4=0 => 
1-5+4=0 => 0=0  Identity provin solution.
5 0
3 years ago
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