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Snezhnost [94]
2 years ago
13

Simplify the problem 6(x-2)+7

Mathematics
2 answers:
Alika [10]2 years ago
5 0

Answer:

step 1: open bracket

6x - 12 + 7

= 6x -5

Maurinko [17]2 years ago
4 0

Answer:

6x - 5

Step-by-step explanation:

distribute

6x - 12 + 7

combine like terms

6x - 5  

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If you add the digits in a two-digit number and multiply the sum by 7, you get the original number. If you reverse the digits in
kolbaska11 [484]
The answer is A. 42

Solution:
Let x= ones digit, y=tens digit

1st condition (original number) : 7(x+y)=10y + x
2nd condition (new number by reversing the digits): 18+x+y=10x+y

simplifying:
1st condition: 6x=3y
2nd condition: x=2
substituting x=2 to 6x=3y
<span>y=4</span>
3 0
3 years ago
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Lorico [155]
8 is 6.2
9 is 7
The pattern here is +0.8.
4.6 - 3.8=0.8
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6 0
3 years ago
How do write 85 less than the product of 23 and an unknown number is 191.
guapka [62]
23(x)-85
Substitute 12 for x
23(12)-85
276-85
191<span />
8 0
3 years ago
Laura takes very good care of her vehicles. She owns a blue van and a red truck. Although she bought them both new, she has owne
MrRissso [65]

The age of the van owned = 12 years

The age of the truck owned  = 29 years

Step-by-step explanation:

Here, the given question is INCOMPLETE.

Laura takes very good care of her vehicles. She owns a blue van and a red truck. Although she bought them both new, she has owned the truck for 17 years longer than she has owned the van. If the sum of the ages of the vehicles is 41 years, how old is the van and how old is the truck?

Let us assume the number of years Laura has owned the van = S years

So, according to the question:

The number of years she has owned truck = S + 17 years

Now, Sum of the age of ( Truck + Van)  = 41 years

⇒ (S + 17 years) + (S years)  = 41 years

or, 2 S +  17 = 41

or, 2 S  = 4 1 -1 7 = 24

or, S  = 24/2 = 12 years

or, S  = 12 years

Hence the age of the van owned = 12 years

The age of the truck owned  = S  + 17  = 12 + 17 = 29 years

6 0
3 years ago
Businesses deposit large sums of money into bank accountsImagine an account with $10 million dollars in it.
adoni [48]

again, let's assume daily compounding means 365 days per year.

~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$10000000\\ r=rate\to 2.12\%\to \frac{2.12}{100}\dotfill &0.0212\\ t=years\dotfill &1 \end{cases} \\\\\\ I = (10000000)(0.0212)(1)\implies \boxed{I=212000}

~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$10000000\\ r=rate\to 2.12\%\to \frac{2.12}{100}\dotfill &0.0212\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{daily, thus 365} \end{array}\dotfill &365\\ t=years\dotfill &1 \end{cases}

A=10000000\left(1+\frac{0.0212}{365}\right)^{365\cdot 1}\implies A\approx 10214256.88 \\\\\\ \underset{\textit{earned interest amount}}{10214256.88~~ - ~~10000000 ~~ \approx ~~ \boxed{214256.88}}

what's their difference?  well

\stackrel{\textit{compounded daily}}{\approx 214256.88}~~ - ~~\stackrel{\textit{simple interest}}{212000}\implies \boxed{2256.88}

7 0
2 years ago
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