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3 years ago

14

2.–/1 pointsMy Notes

1 answer:

lesya692 [45]3 years ago

7 0

Tan(A) = a/b
a = b*tan(A) = 10.58*tan(36.15°) = 7.729 . . . yd

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Factories 54m3n + 81m4n2 b) 15x2y3z + 25x3y2z + 35x2y2z

Oliga [24]

<u>Part a)</u>

Given the expression

$54m^3n\:+\:81m^4n^2$

Apply exponent rule: $a^{b+c}=a^ba^c$

$54m^3n\:+\:81m^4n^2=54m^3n+81m^3mnn$ ∵ $m^4n^2=m^3mnn$

Rewrite 81 as 3 · 27

Rewrite 54 as 2 · 27

$=2\cdot \:27m^3n+3\cdot \:27m^3mnn$

Factor out the common term: 27m³n

$=27m^3n\left(2+3mn\right)$

Therefore,

$54m^3n\:+\:81m^4n^2=54m^3n+81m^3mnn=27m^3n\left(2+3mn\right)$

<u>Part B)</u>

Given the expression

$15x^2y^3z+25x^3y^2z+35x^2y^2z$

Apply exponent rule: $a^{b+c}=a^ba^c$

$15x^2y^3z+25x^3y^2z+35x^2y^2z=15x^2y^2yz+25x^2xy^2z+35x^2y^2z$

Rewrite as

$=3\cdot \:5y^2x^2zy+5\cdot \:5y^2x^2zx+7\cdot \:5y^2x^2z$

Factor out common term 5y²x²z

$=5y^2x^2z\left(3y+5x+7\right)$

Therefore,

$15x^2y^3z+25x^3y^2z+35x^2y^2z=5y^2x^2z\left(3y+5x+7\right)$

8 0

3 years ago

Need this answer please

marshall27 [118]

Answer:

line s

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Can anyone do this?

elixir [45]

The answer is Tuesday there will be 15 teachers for 179 students and 12 times 15 is 180

8 0

3 years ago

Read 2 more answers

the first group of students consists of 10 and their average age was 13 years old the next group consisted of 30 students and th

Kazeer [188]

Answer: The average was 9 years old find the average age of both groups is 10 years old.

Step-by-step explanation:

Formula foe average : $\text{Average} =\dfrac{\text{Sum of all observations}}{\text{Number of observations}}$

Given : The first group of students consists of 10 and their average age was 13 years old.

i.e. $13=\dfrac{\text{Sum of ages of students of first group}}{10}\\\\\Rightarrow\ \text{Sum of ages of students of first group}=13\times10=130$ (1)

The next group consisted of 30 students and their average was 9 years old.

i.e. $9=\dfrac{\text{Sum of ages of students of next group}}{30}\\\\\Rightarrow\ \text{Sum of ages of students of nextgroup}=9\times30=270$ (2)

Then from (1) and (2) , the sum of both groups (first group and next group )students = 130+270 =400

Combined students of both groups (first and next group )= 10+30=40

Now , the average of both groups = $\dfrac{\text{Sum of combined students}}{\text{combined students}}$

$=\dfrac{400}{40}=10$

Hence, the average was 9 years old find the average age of both groups is 10 years old.

4 0

3 years ago

How many real does this 2x^2-5x+3=0 equation have

Pavlova-9 [17]

2 $x^{2}$ -5x+3=0

x= $\frac{-(-5)+ \sqrt{(-5) ^{2}-4*2*3 } }{2*2}$

x=3/2;x=1

3 0

2 years ago