Ask question

Ask question

All categories

Ksenya-84 [330]

3 years ago

7

The width of a rectangle is 7 inches less than its length. The area of the rectangle is 120 square inches. Solve for the dimensi

ons of the rectangle. Length: inches Width: inches

2 answers:

Helen [10]3 years ago

8 0

area = L x W

W=L-7

120 = L x L-7

120 = L^2-7L

L^2-7L+120 =0

(L-15) (L+8)

L=15, L=-8. It can't be a negative number so L=15

W=15-7 = 8

15*8 =120

length = 15 inches

width = 8 inches

Paul [167]3 years ago

8 0

Answer:

The dimensions of the rectangle. <em><u>Length= 15 inches and Width= 8 inches</u></em>

Step-by-step explanation:

Let width of rectangle be W and length be L then

L=W+7 ---- (A)

Also given that area of rectangle = 120 square inches

=> WxL=120 -----(B)

From equation (A) and (B)

Wx(W+7) = 120

=> $W^{2}+7W-120=0$

=> $W^{2}+15W-8W-120=0 =>(W+15)(W-8)=0$

=> $W=-15 or 8$

Since the width can not be a negative quantity , so W= 8 inches

=> L= W+7= (8+7) inches = 15 inches

Thus the dimensions of the rectangle. <em><u>Length= 15 inches and Width= 8 inches</u></em>

You might be interested in

An accelerated life test on a large number of type-D alkaline batteries revealed that the mean life for a particular use before

joja [24]

Answer:

option (d) 16.6 and 21.4

Step-by-step explanation:

Data provided in the question:

The mean life for a particular use before they failed = 19.0 hours

The distribution of the lives approximated a normal distribution

The standard deviation of the distribution = 1.2 hours

To find:

The values between which 95.44 percent of the batteries failed

Now,

In Normal distribution, the approximately 95% ( ≈ 95.44% of all values ) falls within 2 standard deviations of the mean

Therefore,

Upper limit = Mean + 2 × standard deviation

⇒ Upper limit = 19.0 + 2 × 1.2 = 21.4

Lower limit = Mean - 2 × standard deviation

⇒ Lower limit = 19 - 2 × 1.2 = 16.6

Hence,

the answer is option (d) 16.6 and 21.4

8 0

3 years ago

Raphael is traveling in a straight line at a constant speed from point A to point B. His distance from point B in miles is -20t+

Rudik [331]

Answer:

Speed = 20 mi/h

Explanation:

There are two points A and B

Distance between both the points are define with function -20t + 45 miles

where t represent the number of hours of travelling.

Now firstly we find the initial position. At starting time is 0, so we put t=0 in given function of t.

At t=0;

d₀ = -20(0) + 45 = 0 + 45 = 45 miles

Now we find the distance travell after starting first hour.

Than we put t = 1 in the given function

At t = 1;

d₁ = -20(1) + 45 = -20 + 45 = 25 miles

Difference between d₁ & d₀ is

45 - 25 = 20 miles

We see that in one hour, total distace is covered 20 miles

Now we use time, speed, distance relation

Speed = distance/time

Speed= $\frac{20}{1}$

Speed = 20 miles/hour

That's the final answer.

I hope it will help you.

7 0

3 years ago

Two urns both contain green balls and red balls. Urn I contains 6 green balls and 4 red balls and Urn II contains 8 green balls

34kurt

Presumably, whatever is drawn from Urn I is independent of what is drawn from Urn II. This means

$P(\text{red from Urn 1 AND red from Urn 2})=P(\text{red from Urn 1})\cdot P(\text{red from Urn 2})$

We have

$P(\text{red from Urn 1})=\dfrac4{10}=\dfrac25$

$P(\text{red from Urn 1})=\dfrac7{15}$

so the probability of drawing red balls from both urns is $\dfrac25\cdot\dfrac7{15}=\dfrac{14}{75}$

3 0

3 years ago

Can you guys please help me out

marta [7]

C , it’s become more adapted to and more modern

5 0

3 years ago

Find+the+positive+value+for+α+if+the+radius+of+the+circle+3x^2+3y-6αx+12y-3α=0+is+4

Alik [6]

Answer:

$\alpha=3$

Step-by-step explanation:

<u>Equation of a Circle</u>

A circle of radius r and centered on the point (h,k) can be expressed by the equation

$(x-h)^2+(y-k)^2=r^2$

We are given the equation of a circle as

$3x^2+3y^2-6\alpha x+12y-3\alpha=0$

Note we have corrected it by adding the square to the y. Simplify by 3

$x^2+y^2-2\alpha x+4y-\alpha=0$

Complete squares and rearrange:

$x^2-2\alpha x+y^2+4y=\alpha$

$x^2-2\alpha x+\alpha^2+y^2+4y+4=\alpha+\alpha^2+4$

$(x-\alpha)^2+(y+2)^2=r^2$

We can see that, if r=4, then

$\alpha+\alpha^2+4=16$

Or, equivalently

$\alpha^2+\alpha-12=0$

There are two solutions for $\alpha$ :

$\alpha=-4,\ \alpha=3$

Keeping the positive solution, as required:

$\boxed{\alpha=3}$

8 0

3 years ago