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4 years ago

9

Choose what the expressions below best represent within the context of the word problem.The length of a rectangle is 12 inches m

ore than its width. What is the width of the rectangle that if the perimeter is 42 inches?

1 answer:

Virty [35]4 years ago

8 0

Answer:

L=16.5 inches

Step-by-step explanation:

L=W+12

P= 2L+2W

P=2(W+12)+2W

42=4W+24

4W=42-24

W=18/4=4.5

L=12+4.5=16.5 inches

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How can the statement c(1.5)+9 be interpreted

nordsb [41]

The answer is 10.5
Hope I Helped!

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4 years ago

A store sells televisions for $360 and DVD burners for $270. The entire stock is worth $52,920 and there are a total of 164 tele

bogdanovich [222]

Answer: 96 Televisions, 68 DVD burners

<u>Step-by-step explanation:</u>

Let T represent the quantity of televisions

Let D represent the quantity of DVD burners

<u>Given:</u>

value of television (T) = 360

value of DVD burners (D) = 270

Total value = 52,920

<u>Equations:</u>

Value: 360T + 270D = 52,920

Quantity: T + D = 164

We can solve the system of equations using the Elimination Method:

-1(360T + 270D = 52,920) → -360T - 270D = -52,920

360(T + D = 164) → <u>360T + 360D</u> = <u>59,040 </u>

90D = 6,120

<u> ÷90 </u> <u> ÷90 </u>

D = 68

Input the value of D into one of the equations to solve for T

T + D = 164

T + 68 = 164

T = 96

6 0

3 years ago

The measure of c is ___

jonny [76]

Answer:

3 $\sqrt{13}$

Step-by-step explanation:

i'm not really sure if i'm correct

7 0

3 years ago

Read 2 more answers

If a game consists of picking a number from a bag of beans numbered 2, 2, 3, 4, and 5 and flipping a coin, which probabilities a

Semenov [28]

The true statement is (d) none of the above

<h3>Missing information</h3>

I. P(3 and tail)

II. P(even and head)

III .P(odd and head)

<h3>How to determine the equivalent probabilities?</h3>

We start by calculating each probability

<u>I. P(3 and tail) </u>

There is only one 3 in the 5 numbers.

So:

P(3) = 1/5

The probability of a tail in a coin is:

P(Coin) = 1/2

So, we have:

P(3 and tail) = 1/5 * 1/2

P(3 and tail) = 1/10

<u>II. P(even and head) </u>

There are three even numbers in the 5 numbers.

So:

P(even) = 3/5

The probability of a head in a coin is:

P(head) = 1/2

So, we have:

P(even and head) = 3/5 * 1/2

P(even and head) = 3/10

<u>III .P(odd and head)</u>

There are two odd numbers in the 5 numbers.

So:

P(odd) = 2/5

The probability of a head in a coin is:

P(head) = 1/2

So, we have:

P(odd and head) = 2/5 * 1/2

P(even and head) = 1/5

By comparing the results, none of the probabilities are equal.

Hence, the true statement is (d) none of the above

Read more about probability at:

brainly.com/question/25870256

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6 0

2 years ago

Find the sum of a finite geometric series. The Smiths spent $2,000 on clothes in a particular year. If they increase the amount

boyakko [2]

$\bf \stackrel{\textit{first year}}{2000}~~,~~\stackrel{\textit{second year}}{2000+\stackrel{\textit{10\% of 2000}}{\frac{2000}{10}}}\implies 2200$

now, if we take 2000 to be the 100%, what is 2200? well, 2200 is just 100% + 10%, namely 110%, and if we change that percent format to a decimal, we simply divide it by 100, thus $\bf 110\%\implies \cfrac{110}{100}\implies 1.1$ .

so, 1.1 is the decimal number we multiply a term to get the next term, namely 1.1 is the common ratio.

$\bf \qquad \qquad \textit{sum of a finite geometric sequence}\\\\S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\r=\textit{common ratio}\\----------\\a_1=2000\\r=1.1\\n=4\end{cases}\\\\\\S_4=2000\left[ \cfrac{1-(1.1)^4}{1-1.1} \right]\implies S_4=2000\left(\cfrac{-0.4641}{-0.1} \right)\\\\\\S_4=2000(4.641)\implies S_4=9282$

7 0

3 years ago