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3 years ago

What is the length and the width of a rectangle with perimeter of 48 inches if its length is 7 inches longer than its width.

Mathematics

2 answers:

sattari [20]3 years ago

4 0

Answer:

Length and the width of the rectangle is 15.5 inches and 8.5 inches respectively.

Step-by-step explanation:

Given:

Length = 7 inches + Width

=> P = 48 inches

=> L = (7 + W) inches............. (1)

Perimeter 'P' is the sum of all sides of the rectangle 2 (L + W)

∴ 48 inches = 2 (L + W) .................(2)

Substitute for L in equation (2)

∴ 48 inches = 2 [(7 + W) + W]

48 = 2[7 + 2W]

48 = 14 +4W

48 - 14 = 4W

34 = 4W

W = 8.5 inches

Recall that: Length, L = (7 + W)

∴ L = (7 + 8.5)

L = 15.5 inches

liq [111]3 years ago

3 0

Answer:

Width = 8.5 inches

Length = 15.5 inches

Step-by-step explanation:

Width = W

Length = 7 + W

Perimeter = 2(Length) + 2(Width)

Substitute

48 = 2(7 + W) + 2(W)

Multiply

48 = 14 + 2W + 2W

Add

48 = 14 + 4W

Subtract 14 from both sides of the equation

34 = 4W

Divide both sides of the equation by 4

W = 8.5

Width = 8.5 inches

Length = 7 + Width = 7 + 8.5 = 15.5 inches

Hope this helps :)

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3 years ago

Find the limit (enter 'DNE' if the limit does not exist)

vaieri [72.5K]

Answer:

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\sqrt{-2x^2-6y^2+1}+1\right)=2$

Step-by-step explanation:

We need to first simplify the expression using rationalization(i.e. if a square root term exists in the denominator, then multiply and divide the whole expression by the denominator(but the change the sign of its middle term))

here, we need to find:

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\dfrac{-2x^2-6y^2}{\sqrt{-2x^2-6y^2+1}-1}\right)$

first we'll rationalize our expression:

$\dfrac{-2x^2-6y^2}{\sqrt{-2x^2-6y^2+1}-1}\left(\dfrac{\sqrt{-2x^2-6y^2+1}+1}{\sqrt{-2x^2-6y^2+1}+1}\right)$

$\dfrac{-(2x^2+6y^2)(\sqrt{-2x^2-6y^2+1}+1)}{(\sqrt{-2x^2-6y^2+1}+1)^2-(1)^2}$

$\dfrac{-(2x^2+6y^2)(\sqrt{-2x^2-6y^2+1}+1)}{-2x^2-6y^2+1-1}$

$\dfrac{-(2x^2+6y^2)(\sqrt{-2x^2-6y^2+1}+1)}{-(2x^2+6y^2)}$

$\sqrt{-2x^2-6y^2+1}+1$

this is our simplified expression, now we can apply our limits:

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\sqrt{-2x^2-6y^2+1}+1\right)$

$\sqrt{-2(0)^2-6(0)^2+1}+1$

$1+1$

$2$

the limit does exists and it is 2.

5 0

3 years ago

What is 9 and 19/18 simplified

Anastasy [175]

10 1/18...........................................

...................................................

6 0

3 years ago

Read 2 more answers

I need help quick please help me thnak you

quester [9]

Answer:

28 numbers

Step-by-step explanation:

There is a distance of 81 between -0.5 and 80.5. Divide this number by 3, since you are only counting every third number and you will get 27, but you need to count the first number, which has been excluded. Add one more to get the answer, 28.

3 0

3 years ago

A worn, poorly set-up machine is observed to produce components whose length X follows a normal distribution with mean 14 centim

Akimi4 [234]

Answer:

74.86% probability that a component is at least 12 centimeters long.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 14$