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Alexxx [7]
3 years ago
13

A rectangle is 4 times as long as it is wide. If the length is increased by 4 inches and the width is decreased by 1 inch, the a

rea will be 60 square inches. What were the dimensions of the original rectangle?
Mathematics
2 answers:
Nataly [62]3 years ago
7 0

Let the Width of the Original Rectangle be : W

Given : Length of the Original Rectangle is 4 times as long as it's Width

⇒ Length of the Original Rectangle = 4 × W = 4W

Now Some Modifications are made, So that Original Rectangle becomes into a New Rectangle.

Given the Length of New Rectangle is 4 inches More than Length of Original Rectangle.

⇒ Length of New Rectangle = 4W + 4

Given the Width of New Rectangle is One Inch less than Width of Original Rectangle.

⇒ Width of New Rectangle = W - 1

We know that Area of a Rectangle is : Length × Width

Given the Area of New Rectangle is 60 inches²

⇒ (4W + 4)(W - 1) = 60

⇒ 4W² - 4W + 4W - 4 = 60

⇒ 4W² - 4 = 60

⇒ 4W² = 64

⇒ W² = 16

⇒ W = 4

<u>Dimensions of Original Rectangle :</u>

Length of Original Rectangle = 4W = 4 × 4 = 16 inches

Width of Original Rectangle = W = 4 inches

ruslelena [56]3 years ago
3 0

The equation can be formatted as (w-1)(l+4) = 60

There are many ways to solve this but I decided to just plug in numbers.

I established w is 7 and L was 6 initially.

The long way to solve this would be by completing the square which is literally impossible to diagram on this website.

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3х - 30 = у<br> 7y - 6 = 3х
mars1129 [50]

Answer:

Step-by-step explanation:

This one is simple substitution... at least, substitution is the easiest method. The first equation is 3<em>x</em> – 30 = <em>y</em>  and the second is 7<em>y</em> – 6 = 3<em>x</em>

As I look, I see 3 ways to use substitution to solve this:

  1. substitute 7<em>y</em> – 6 for 3<em>x</em>
  2. substitute 3<em>x</em> – 30 for <em>y</em>
  3. solve 3<em>x</em> – 30 = <em>y</em> for 3<em>x</em> and make it equal to 7<em>y</em> – 6

We're going to only use 1 method for the sake of time. Try the other two on your own. Assuming you don't make any mistakes, they will work.

<u>Method 1</u>:

3<em>x</em> – 30 = <em>y</em>

7<em>y</em> – 6 = 3<em>x</em>  — initial system of equations

7<em>y</em> – 6 – 30 = <em>y</em>  — substitute 7<em>y</em> – 6 for 3<em>x</em>

<u>7</u><u><em>y</em></u> – 6 – 30 = <u><em>y</em></u>  — marking like terms, bold for constants, <u>underlined</u> for variables

7<em>y</em> – 36 = <em>y</em>  — combining the constants and simplifying

Here, you could diverge into multiple paths: add 36 to both sides, subtract <em>y</em> from both sides, divide by 6 OR subtract 7<em>y</em> from both sides and divide by –6 . For the sake of time, I'm subtracting 7<em>y</em>, though I don't like dealing with negatives.

7<em>y</em> – 7<em>y</em> – 36 = <em>y</em> – 7<em>y</em>  — subtract 7<em>y</em> from both sides

–36 = –6<em>y</em>  — simplify

–36 ÷ –6 = –6<em>y</em> ÷ –6  — divide by –6 on both sides

<em>y</em> = 6  — simplify

Again, we can diverge here: substitute <em>y</em> into 3<em>x</em> – 30 = <em>y</em> or substitute <em>y</em> into 7<em>y</em> – 6 = 3<em>x</em>

I'm going to choose 3<em>x</em> – 30 = <em>y</em> but it will work either way, should you take the time (if you have it) to chase down every path this problem can take.

3<em>x</em> – 30 = <em>y</em>  — initial equation

3<em>x</em> – 30 = 6  — substitute 6 for <em>y</em>

3<em>x</em> – 30 + 30 = 6 + 30  — add 30 to both sides to isolate 3<em>x</em>

3<em>x</em> = 36  — simplify the expression

3<em>x</em> ÷ 3 = 36 ÷ 3  — divide both sides by 3 to isolate <em>x</em>

<em>x</em> = 12  — simplify

So, we have <em>x</em> = 12 and <em>y</em> = 6 . We know they work for 3<em>x</em> – 30 = <em>y</em>  but not if they work for 7<em>y</em> – 6 = 3<em>x</em> . Let's substitute those in to see if (12, 6) really is the solution point.

7<em>y</em> – 6 = 3<em>x</em>  — original equation

7(6) – 6 ≟ 3(12)  — substitute 6 for <em>y</em> and 12 for <em>x</em>

42 – 6 ≟ 36  — simplify by multiplying

36 = 36 ✔  — simplify by combining like terms on left side

Success! It works! We have found our solution!

I hope this helps increase your understanding of the concept. Have a great day!

8 0
3 years ago
If an 11year old child needs to take 1/5 amount of medication per day to heal their head. If a baby has to take 1/5 as much to g
MatroZZZ [7]

Answer:

Step-by-step explanation:

the child takes 1/5.....the baby needs to take 1/5 of what the child takes

so 1/5 of 1/5....." of " in math, means multiply

1/5 * 1/5 = 1/25 of the amount <==

6 0
3 years ago
Find the x-intercept of the line whose equation is 8x + 2y = 4.<br> 04<br> 02<br> 1/2
bezimeni [28]

Answer:

1/2

Step-by-step explanation:

The x-intercept is a point on the graph that is located on any point of x but must have a y-value of 0. To find the x-intercept, we must set y in the given equation equal to 0.

8x + 2y = 4

8x + 2(0) = 4

8x = 4

x = 4/8

x = 1/2

4 0
2 years ago
Is the answer to -8.8-3.4 positive or negative? I know. It is 12.2 but I don't know if it's -12.2 or +12.2
Marat540 [252]
It is negative.

When you subtract a negative a positive number from a negative number, the result is positive. The equation can be rewritten as -8.8-(+3.4).
6 0
3 years ago
The port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge, is the largest
Ksenya-84 [330]

Answer:

a) 0.7287

b) 0.9663

c) 0.237

d) 3.65 tons of cargo per week or more that will require the port to extend its operating hours.  

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  4.5 million tons of cargo per week

Standard Deviation, σ = 0 .82 million tons

We are given that the distribution of number of tons of cargo handled per week is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P( port handles less than 5 million tons of cargo per week)

P(x < 5)

P( x < 5) = P( z < \displaystyle\frac{5 - 4.5}{0.82}) = P(z < 0.609)

Calculation the value from standard normal z table, we have,  

P(x < 5) =0.7287= 72.87\%

b) P( port handles 3 or more million tons of cargo per week)

P(x \geq 3) = P(z \geq \displaystyle\frac{3-4.5}{0.82}) = P(z \geq −1.82926)\\\\P( z \geq −1.82926) = 1 - P(z < -1.829)

Calculating the value from the standard normal table we have,

1 - 0.0337 = 0.9663 = 96.63\%\\P( x \geq 3) = 96.63\%

c)P( port handles between 3 million and 4 million tons of cargo per week)

P(3 \leq x \leq 4) = P(\displaystyle\frac{3 - 4.5}{0.82} \leq z \leq \displaystyle\frac{4-4.5}{0.82}) = P(-1.829 \leq z \leq -0.609)\\\\= P(z \leq -0.609) - P(z < -1.829)\\= 0.271-0.034 = 0.237= 23.7\%

P(3 \leq x \leq 4) = 23.7\%

d) P(X=x) = 0.85

We have to find the value of x such that the probability is 0.85.

P(X > x)  

P( X > x) = P( z > \displaystyle\frac{x - 4.5}{0.82})=0.85  

= 1 -P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.85  

=P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.15  

Calculation the value from standard normal z table, we have,  

P( z \leq -1.036) = 0.15

\displaystyle\frac{x - 4.5}{0.82} = -1.036\\x = 3.65

Thus, 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

8 0
3 years ago
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