All categories

4 years ago

2(x+1)+3=7 how do I solve this? Is this correct? 2(x+1)+3=7 2x + 2 +3=7 2x+2 +3=7 2x+5=7 -5 7 2x= 2 2 2 x=1

2 answers:

7 0

Yes, that's correct.

Here's another way to approach it:

2 (x + 1) + 3 = 7

Subtract 3 from each side: 2 (x + 1) = 4

Divide each side by 2 : x + 1 = 2

Subtract 1 from each side: x = 1

kaheart [24]4 years ago

7 0

You betcha! Your work is correct! Here is a little different way to solve it. So you can choose which one is simpler for you!

2(x+1)+3=7

Simplify both sides of the equation.

2(x+1)+3=7

Simplify

2x+5=7

Subtract 5 from both sides.

2x+5−5=7−5

2x=2

Divide both sides by 2.

2x / 2=2 / 2

x=1

You might be interested in

What is the value of x in simplest radical form?

GalinKa [24]

Answer:

Step-by-step explanation:

3 0

3 years ago

Daphne rolls two 6-sided number cubes. What is the probability that she rolls

Alex Ar [27]

The answer is A. 2/36 or 1/18

7 0

3 years ago

Read 2 more answers

A circular hedge surrounds a sculpture on a square base. The radius of the hedge is 6x. The side length of the square sculpture

andrey2020 [161]

Hi, thank you for posting your question here at Brainly.

Since the hedge is surrounding the square base, the area of the base can be determined by subtracting the area of the base from the hedge.

A = pi*r^2 - (4x)^2
A = pi*(6x)^2 - (4x)^2
A = 2x^2*(3pi - 2)

I hope I was able to answer your question. Have a good day.

6 0

3 years ago

This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 120 customer orders to f

Aliun [14]

Answer:

a. P(X = 0) = 0.02586

b. $\mathbf{P(X \leq 2 ) =0.2879}$

c. $\mathbf{P(X \leq 5 ) =0.8387}$

Step-by-step explanation:

From the given information:

a. If the manufacturer stocks 120 components, what is the probability that the 120 orders can be filled without reordering components?

$P(X = 0)=(^{120}_{0}) (0.03)^0 (1-0.03)^{n-0}$

$P(X = 0)=\dfrac{120!}{0!(120-0)!} (0.03)^0 (1-0.03)^{n-0}$

P(X = 0) = 1 × 1 ( 0.97)¹²⁰ ⁻ ⁰

P(X = 0) = 0.02586

b. ) If the manufacturer stocks 122 components, what is the probability that the 120 orders can be filled without reordering components?

$P(X \leq 2 ) = [ P(X=0) + P(X =1) + P(X = 2) ]$

$P(X \leq 2 ) = [(^{122}_{0})(0.03)^0 (0.97)^{122-0}+(^{122}_{1})(0.03)^1 (0.97)^{122-1}+(^{122}_{2})(0.03)^2 (0.97)^{122-2}]$ $P(X \leq 2 ) = [\dfrac{122!}{0!(122-0)! } \times 1 \times (0.97)^{122}+\dfrac{122!}{1!(122-1)! } \times (0.03) (0.97)^{121}+\dfrac{122!}{2!(122-2)! } \times 9 \times 10^{-4} \times (0.97)^{120}]$

$P(X \leq 2 ) = [(1 \times 1 \times 0.02433 )+(122 \times (0.03) \times 0.025083)+(7381 \times 9 \times 10^{-4} \times 0.02586)]$

$\mathbf{P(X \leq 2 ) =0.2879}$

(c) If the manufacturer stocks 125 components, what is the probability that the 120 orders can be filled without reordering components?

$P(X \leq 5 ) = [ P(X=0) + P(X =1) + P(X = 2) +P(X = 3)+P(X = 4)+ P(X = 5) ]$

$P(X \leq 5 ) = [(^{122}_{0})(0.03)^0 (0.97)^{122-0}+(^{122}_{1})(0.03)^1 (0.97)^{122-1}+(^{122}_{2})(0.03)^2 (0.97)^{122-2} + (^{122}_{3})(0.03)^3 (0.97)^{122-3} + (^{122}_{4})(0.03)^4 (0.97)^{122-4}+ (^{122}_{5})(0.03)^5 (0.97)^{122-5}]$ $P(X \leq 5 ) = [\dfrac{122!}{0!(122-0)! } \times 1 \times (0.97)^{122}+\dfrac{122!}{1!(122-1)! } \times (0.03) (0.97)^{121}+\dfrac{122!}{2!(122-2)! } \times 9 \times 10^{-4} \times (0.97)^{120} + \dfrac{122!}{3!(122-3)! }*(0.03)^3(0.97)^{122-3)}+ \dfrac{122!}{4!(122-4)! }*(0.03)^4(0.97)^{122-4)} +\dfrac{122!}{5!(122-5)! } *(0.03)^5(0.97)^{122-5)}]$

$\mathbf{P(X \leq 5 ) =0.8387}$

5 0

3 years ago

Solve for −5(x+1)=−3(2x−2)

Nitella [24]

Answer:

x=11

Step-by-step explanation:

7 0

3 years ago