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never [62]
10 months ago
10

Which of the following are solutions to the equation below? Check all that apply. (3x - 5)2 = 19 O A. x= 19+ olm B. x= 119 +5 I

C. x= - 19 + D. x= _V14 3 O E. x= -19 +5 3 O F. X= 14 3
Mathematics
1 answer:
asambeis [7]10 months ago
8 0

Solution

(3x-5)^2=19

For this case we can take square root in both sides and we have:

3x-5=\pm\sqrt[]{19}

And solving for x we got:

x=\frac{5\pm\sqrt[]{19}}{3}

then the solutions for this case are:

B and E

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A coin is tossed then a number 1-10 is chosen at random what is the probability of getting heads then a number less than 4
Darya [45]
The probability of getting heads on a coin toss is 0.5
The probability of choosing a number <u>less</u> than four is 3/10 (1,2,3)

Therefore, the probability of getting heads <u>and</u> choosing a number less than four is:
=  \frac{1}{2} (\frac{3}{10}) = \frac{3}{20} = 0.15

Hope this helps!
3 0
3 years ago
Help!!! please what's is the figure​
S_A_V [24]

Hello from MrBillDoesMath!

Answer:

108 cm^2

Discussion:

The area of the logo  =

area of Trapezoid on top (I)  + area of triangle on bottom (II)

I  = (1/2) h ( b1 + b2)            h = altitude, b1, b2 = bases

  = (1/2)  6  ( 6  + 12)

  = (1/2) 6 ( 18)

  = 54

II  = (1/2) b h

   = (1/2) (12) ( 15 - 6)               => altitude of triangle = 15 - 6 NOT (1/2) 15

   =  54

I + II= 54 + 54 = 108

Thank you,

MrB

3 0
2 years ago
Read 2 more answers
What is 5.6 = p - 8.3
vredina [299]

Answer:

p - 8.3 = 5.6

p = 5.6+8.3= 13.9

7 0
3 years ago
bob and mark talk about their families. bob says he has 3 kids, the product of their ages is 72. Mark points out that there is s
Sonbull [250]
Given:
3 kids
72 is the product of their ages.
There is a youngest child.

This means that all three are of different ages. We need to factor 72 into 3 integers.
Let us use prime factorization:

72 ÷ 2 = 36
36 ÷ 2 = 18
18 ÷ 2 = 9
  9 ÷ 3 = 3

Prime factors are: 2 x 2 x 2 x 3 x 3    OR 2³ x 3²

3 ages.

eldest      = 3³ = 9
middle     = 2² = 4
youngest  = 2¹ = 2

2 x 4 = 8
8 x 9 = 72
8 0
3 years ago
Show that if the first 10 positive integers 1,2,3,···,10 are placed around a circle, in any order,there exists three integers in
zysi [14]

Answer:

Let A1=a1+a2+a3, A2=a2+a3+a4, and so on, A10=a10+a1+a2. Then A1+A2+⋯+A10=3(a1+a2+⋯+a10)=(3)(55)=165, so some Ai≥165/10=16.5, so some Ai≥17.

Step-by-step explanation:

7 0
3 years ago
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