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12345 [234]
4 years ago
14

I need to create my own area word problem for extra credit (algebra 1)

Mathematics
1 answer:
vichka [17]4 years ago
3 0
Bob has a garden
the area is 75 square feet
the legnth is 3 times the width
solve
75=lw
l=3w
subsitute
75=3w times w=3w^2
75=3w^2
sdivide by 3
25=w^2
square root
5=width
subsitute
3w=l
3 times 5=l
l=15

width=5
legnth=15
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What is the change in each termof the sequence? 5, -2, -9, -16, -23​
nordsb [41]

Answer:

-7

Step-by-step explanation:

To find the common difference, take the second term and subtract the first term

-2 - 5 = -7

Check by subtracting the second term from the third term

-9 - (-2) = -9+2 = -7

The common difference is -7

7 0
4 years ago
You select a card at random from the cards that make up the word "replacement". Without replacing the card, you choose a second
lutik1710 [3]

Answer:

The probability is 14/55

Step-by-step explanation:

The word replacement contains 11 letters.

In which 4 letters are vowel = e,a,e,e

And 7 letters are not vowel = r,p,l,c,m,n,t

The probability of getting the vowel in the first try = 4/11

The probability of getting something that is not a vowel = 7/10.

The denominator is one less because the vowel card is not replaced, meaning there is one less card to choose from.

Multiply the values:

=4/11 * 7/10

=28/110

=14/55

The probability is 14/55....

5 0
3 years ago
PLEASE ANSWER FAST!!!
Ilya [14]
S= 11/12
15%-p=x
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5 0
4 years ago
Read 2 more answers
How to solve logarithmic equations as such
Serga [27]

\bf \textit{exponential form of a logarithm} \\\\ \log_a b=y \implies a^y= b\qquad\qquad a^y= b\implies \log_a b=y \\\\\\ \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} ~\hspace{7em} \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \log_2(x-1)=\log_8(x^3-2x^2-2x+5) \\\\\\ \log_2(x-1)=\log_{2^3}(x^3-2x^2-2x+5) \\\\\\ \log_{2^3}(x^3-2x^2-2x+5)=\log_2(x-1) \\\\\\ \stackrel{\textit{writing this in exponential notation}}{(2^3)^{\log_2(x-1)}=x^3-2x^2-2x+5}\implies (2)^{3\log_2(x-1)}=x^3-2x^2-2x+5

\bf (2)^{\log_2[(x-1)^3]}=x^3-2x^2-2x+5\implies \stackrel{\textit{using the cancellation rule}}{(x-1)^3=x^3-2x^2-2x+5} \\\\\\ \stackrel{\textit{expanding the left-side}}{x^3-3x^2+3x-1}=x^3-2x^2-2x+5\implies 0=x^2-5x+6 \\\\\\ 0=(x-3)(x-2)\implies x= \begin{cases} 3\\ 2 \end{cases}

5 0
3 years ago
Ignore that bottom but help pleaseee
NNADVOKAT [17]
The answer is 49 because it is
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