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3 years ago

C an anyone solve this 3(2b + 3)2 = 36 and if so what is it

2 answers:

6 0

Assuming it's linear:

3(2b+3)2 = 36

6(2b+3) = 36
12b + 18 = 36
12b = 18
b = 1.5

Assuming you meant to put the 2 as an exponent:
3(2b+3)² = 36

First do the exponent:
3(4b² + 12b + 9) = 36

Multiply it out:
12b² + 36b + 27 = 36
12b² + 36b - 9 = 0

Factor:
3(4b² + 12b - 3) = 0

Apply Quadratic Formula:

x = (-12 +- √(144 + 48)) / 8
x = (-12+√192) / 8 and x = (-12 - √192) / 8

This makes it so that the variable you're solving for = .232 and -3.232.

Natali [406]3 years ago

5 0

$3(2b + 3)^2 = 36$

Divide by 3 on both sides:
$(2b + 3)^2 = 12$

Square root both sides:
$2b + 3= \pm \sqrt{12}$

Take away 3 from both sides:
$2b= \pm \sqrt{12} - 3$

Divide by 2 on both sides:
$b= \dfrac{ \sqrt{12} -3 }{2} \ or \ b= \dfrac{ -\sqrt{12} -3 }{2}$

Answer:
$b= 0.23 \ or \ -3.23$

You might be interested in

-1/3+(-3/5)=

Illusion [34]

$\frac{-1}{3} + \frac{-3}{5}$
You need to use the equation for adding fractions, which is $\frac{a}{b} + \frac{c}{d} =(ad+bc)/bd$

In this case, a=-1, b=3, c=-3, d=5.

$
 \frac{(-1*5)+(3*-3)}{(3*5)}$

simplify

answer: $\frac{-14}{15}$

3 0

3 years ago

Read 2 more answers

On the Algebra test the number of students who got a grade A was three more than twice the number of students who got lower grad

svlad2 [7]

X= # students who got lower grades
2x+3= # students with grade A

STEP 1:
Add the two sets above equal to 24.

x + (2x + 3)= 24
combine like terms

3x + 3= 24
subtract 3 from both sides

3x= 21
divide both sides by 3

x= 7 (# students lower than A)

STEP 2:
substitute x value from step 1 to find # grade A students

=2x + 3
=2(7) + 3
=14 + 3
=17 (# grade A students)

CHECK:
x + (2x + 3)= 24
7 + 2(7) + 3= 24
7 + 14 + 3= 24
24= 24

ANSWER: There were 17 students with grade A.

Hope this helps! :)

4 0

3 years ago

Solve each of the following systems of equations by substitution

Delvig [45]

Answer:

1) The solution to the given equations is (3,6)

2) The solution to the given equations is (7,-1)

3) The solution to the given equations is (6,5)

Step-by-step explanation:

1) Given equations are $x+y=5\hfill (1)$

and $x=1+y\hfill (2)$

To solve the given equations by substitution method :

From equation (2) we have the value x=1+y

Substitute the value of x=1+y in equation (1) we get

(1+y)+y=5

1+y+y=5

1+2y=5

2y=5-1

2y=4

$y=\frac{4}{2}$

Therefore y=2

Now substitute the value of y=2 in equation (2) we get

x=1+y

x=1+2

Therefore x=3

The solution is (3,6)

2) Given equations are $2x+3y=11\hfill (1)$

and $x+y=6\hfill (2)$

To solve the given equations by substitution method :

From equation (2) we have the value x=6-y

Substitute the value of x=6-y in equation (1) we get

2(6-y)+3y=11

12-2y+3y=11

y=11-12

Therefore y=-1

Now substitute the value of y=-1 in equation (2) we get

x+(-1)=6

x=6+1

Therefore x=7

The solution is (7,-1)

3) Given equations are $x-y=1\hfill (1)$

and $6x-7y=1\hfill (2)$

To solve the given equations by substitution method :

From equation (1) we have the value x=1+y

Substitute the value of x=1+y in equation (2) we get

6(1+y)-7y=1

6+6y-7y=1

-y=1-6

Therefore y=5

Now substitute the value of y=5 in equation (1) we get

x-5=1

x=1+5

Therefore x=6

The solution is (6,5)

8 0

3 years ago

Gwen used elimination with multiplication to solve the system:

il63 [147K]

The correct answer for this question is this one:

Gwen used elimination with multiplication to solve the system:
2x + 6y = 3
x − 3y = −5

Her work to find x is shown. Complete the explanation of her error. Then solve the system. If necessary, enter your answer as a reduced fraction.
2(x − 3y) = −5 --> 2(x-3y) = 2(-5)
2x − 6y = −5 --> 2x - 6y = -10
2x + 6y = 3 --> 2x + 6y = 3
2x − 6y = −5 --> 2x - 6y = -10
4x + 0y = −2 --> 4x = -7
x = −2/4 --> x = -7/4

x - 3y = -5
4(-7/4 - 3y = -5)
-7 - 12y = -20
-12y = -13
y = 12/13

Gwen forgot to multiply the right side by 2.
What is the solution as an ordered pair

8 0

3 years ago

In a start-up company which has 20 computers, some of the computers are infected with virus. The probability that a computer is

Alex777 [14]

Answer:

(1) The probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.

(2) The probability at least 5 computers are infected is 0.949.

Step-by-step explanation:

The probability that a computer is defective is, p = 0.40.

(1)

Let X = number of computers to be tested before the 1st defect is found.

Then the random variable $X\sim Geo(p)$ .

The probability function of a Geometric distribution for k failures before the 1st success is:

$P (X = k)=(1-p)^{k}p;\ k=0, 1, 2, 3,...$

Compute the probability that the technician tests at least 5 computers before the 1st defective computer is found as follows:

P (X ≥ 5) = 1 - P (X < 5)

= 1 - [P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)]

$=1 -[(1-0.40)^{0}0.40+(1-0.40)^{1}0.40+(1-0.40)^{2}0.40\\+(1-0.40)^{3}0.40+(1-0.40)^{4}0.40]\\=1-[0.40+0.24+0.144+0.0864+0.05184]\\=0.07776\\\approx0.078$

Thus, the probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.

(2)

Let Y = number of computers infected.

The number of computers in the company is, n = 20.

Then the random variable $Y\sim Bin(20,0.40)$ .

The probability function of a binomial distribution is:

$P(Y=y)={n\choose y}p^{y}(1-p)^{n-y};\ y=0,1,2,...$

Compute the probability at least 5 computers are infected as follows:

P (Y ≥ 5) = 1 - P (Y < 5)

= 1 - [P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3) + P (Y = 4)] $=1-[{20\choose 0}(0.40)^{0}(1-0.40)^{20-0}+{20\choose 1}(0.40)^{1}(1-0.40)^{20-1}\\+{20\choose 2}(0.40)^{2}(1-0.40)^{20-2}+{20\choose 3}(0.40)^{3}(1-0.40)^{20-3}\\+{20\choose 4}(0.40)^{4}(1-0.40)^{20-4}]\\=1-[0.00004+0.00049+0.00309+0.01235+0.03499]\\=1-0.05096\\=0.94904$

Thus, the probability at least 5 computers are infected is 0.949.

7 0

3 years ago