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irina [24]
3 years ago
8

Ten-ninths of the sum of 3x, 12y and -6z

Mathematics
1 answer:
jenyasd209 [6]3 years ago
8 0
9/10(3x+12y+-6z) is the expression, simplified it's:
2.7x+10.8y-5.4z 

Hope this helped!
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The area of a rectangle is 3 2/3square meters. If the rectangle is 2 3/4meters long, find the width of the rectangle?
NNADVOKAT [17]
This is your answer...

7 0
3 years ago
One of the legs of a right triangle is twice as long as the other, and the perimeter of the triangle is 35. Find the
Aloiza [94]

Answer:

6.68, 13.37, 14.95

Step-by-step explanation:

One of the legs is twice as long as the other.

b = 2a

The perimeter is 35.

35 = a + b + c

The triangle is a right triangle.

c² = a² + b²

Three equations, three variables.  Start by plugging the first equation into the second and solving for c.

35 = a + 2a + c

c = 35 − 3a

Now plug this and the first equation into the Pythagorean theorem:

(35 − 3a)² = a² + (2a)²

1225 − 210a + 9a² = a² + 4a²

1225 − 210a + 4a² = 0

Solve with quadratic formula:

a = [ -(-210) ± √((-210)² − 4(4)(1225)) ] / 2(4)

a = (210 ± √24500) / 8

a ≈ 6.68 or 45.82

Since the perimeter is 35, a = 6.68.  Therefore, the other sides are:

b ≈ 13.37

c ≈ 14.95

8 0
3 years ago
During a flu epidemic, 35% of the school's students have the flu. Of those with the flu, 90% have high
frutty [35]

Answer: 0.8015

Step-by-step explanation:

Let F= event that a person has flu

H= event that person has a high temperature.

As per given,

P(F) =0.35

Then P(F')= 1- 0.35= 0.65               [Total probability= 1]

P(H | F) = 0.90

P(H|F') = 0.12

By Bayes theorem, we have

P(F|H)=\dfrac{P(F)\timesP(H|F)}{P(F)\timesP(H|F)+P(F')\timesP(H|F')}\\\\=\dfrac{0.35\times0.90}{0.35\times0.90+0.65\times0.12}\\\\=\dfrac{0.315}{0.315+0.078}\approx0.8015

Required probability = 0.8015

8 0
3 years ago
Halla dos pares de enteros cuya suma sea -6​
SCORPION-xisa [38]

Answer:

0 y -6 o 1 y -7

Step-by-step explanation:

7 0
3 years ago
(3x+2)∧2=9 how do i solve this in the square root property
Irina18 [472]
(3x+2)^2=9 \\\\9x^2+12x+4-9=0\\\\9x^2+12x-5=0\\\\a=9,\ \ b=12, \ \ c=-5

x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-12-\sqrt{12^2-4 \cdot9 \cdot (-5)}}{2 \cdot 9}=\frac{-12-\sqrt{144+180}}{18}=\\\\=\frac{-12-\sqrt{324}}{18}=\frac{-12-18}{18}=\frac{-30}{18}=-\frac{5}{3}\\\\x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{-12+\sqrt{12^2-4 \cdot9 \cdot (-5)}}{2 \cdot 9}=\frac{-12+18}{18}=\frac{6}{18}=\frac{1}{3} \\\\Answer: \ x=-\frac{5}{3}\ \ and \ \ x=\frac{1}{3}
 

6 0
3 years ago
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