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notsponge [240]
3 years ago
15

Simplify: √(4√3-√15)^2 I WILL AWARD BRAINLIEST TO THE CORRECT ANSWER!

Mathematics
2 answers:
disa [49]3 years ago
6 0

Answer:

3.05521988407

Step-by-step explanation:

used a calculator lol

krok68 [10]3 years ago
6 0
I believe the answer is 3.05

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Is (4,2) a solution of the system y=x-2 and y=3x+4
babunello [35]

Answer:

False

Step-by-step explanation:

y=x-2  y=3x+4

Substitute the point into both equations and see if it is true

2=4-2

2=2  true

and

2=3*4+4

2 = 12+4

2 =16

False

It is not a solution

4 0
3 years ago
Read 2 more answers
Subtract-a^2-5ab+3ab^2 from 3a^2-2ab+3b2
sergejj [24]

The correct answer is 4a^2 + 3ab

In order to find this, we can write the equation out as described and then follow the order of operations.

3a^2 - 2ab + 3b^2 - (-a^2 - 5ab + 3b^2) ----> Distribute the negative

3a^2 - 2ab + 3b^2 + a^2 + 5ab - 3b^2 ----> Combine like terms

4a^2 + 3ab

5 0
3 years ago
Read 2 more answers
AMSWER ASAP PLEASE PLEASE PLEASE
Marysya12 [62]

Answer:

(2,14)

Step-by-step explanation:

so i solved this with substitution but there are other methods to solve this!

equation 1 : y = -x + 16
equation 2 : y = x + 12

solve for y in equation 1 !!
y = -x + 16
(add x on both sides)
y + x = 16
(subtract y on both sides)
x = -y + 16

equation 1 now equals : x = -y + 16

substitute equation 1 into equation 2 :)

y = (-y + 16) + 12
(add 16 + 12)
y = -y + 28
(add y on both sides)

2y = 28
(divide 2 on both sides)
y = 14


now to solve for x you can insert y !
you can use equation 1 or equation 2 to solve for x.

i'm using equation 2 :
y = x + 12
(14) = x + 12
(subtract 12 on both sides)
x = 2

therefore, the answer is (2, 14).

hope this helps :p

3 0
2 years ago
In a large school, it was found that 77% of students are taking a math class, 74% of student are taking an English class, and 70
Iteru [2.4K]

Answer:

0.81 = 81% probability that a randomly selected student is taking a math class or an English class.

0.19 = 19% probability that a randomly selected student is taking neither a math class nor an English class

Step-by-step explanation:

We solve this question working with the probabilities as Venn sets.

I am going to say that:

Event A: Taking a math class.

Event B: Taking an English class.

77% of students are taking a math class

This means that P(A) = 0.77

74% of student are taking an English class

This means that P(B) = 0.74

70% of students are taking both

This means that P(A \cap B) = 0.7

Find the probability that a randomly selected student is taking a math class or an English class.

This is P(A \cup B), which is given by:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

So

P(A \cup B) = 0.77 + 0.74 - 0.7 = 0.81

0.81 = 81% probability that a randomly selected student is taking a math class or an English class.

Find the probability that a randomly selected student is taking neither a math class nor an English class.

This is

1 - P(A \cup B) = 1 - 0.81 = 0.19

0.19 = 19% probability that a randomly selected student is taking neither a math class nor an English class

6 0
3 years ago
(D)Using the model formula: P= -55/4 t+340, predict the park’s owl population in 2022.
Anastaziya [24]

Answer:

D: 216 E: 2038

Step-by-step explanation:

For D:

P= - 55/4 × t + 340

Substitute in for t (2022-2013=9)

-55/4 × 9 = - 123.75

-123.75+340= 216.25

Round down to 216

For E:

The owl population =0

0=-55/4×t+340

-340=-55/4 × t

-340÷-55/4=t

T=24.7272

Round to 25

25+2013 =2038

4 0
3 years ago
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