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uranmaximum [27]
3 years ago
8

Can u please help me do this question​

Mathematics
1 answer:
Evgesh-ka [11]3 years ago
8 0

Answer:

a. 20x > 10x + 80

b.  •20x > 10x + 80

• Subtract 10x from both sides: 20x -10x > 10x + 80 - 10x

• Simplify: 10x > 80

• Divide both sides by 10:   10x/10 > 80/10

• Simplify: x > 8

Step-by-step explanation:

10x + 80 = number of seashells Pilar has.

20x = number of seashells Marisol has.  

answer for a:  20x > 10x + 80

answer for b: 20x > 10x + 80

• Subtract 10x from both sides: 20x -10x > 10x + 80 - 10x

• Simplify: 10x > 80

• Divide both sides by 10:   10x/10 > 80/10

• Simplify: x > 8

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In a class of 120 student 70 student passes in english 80 student pass in hindi and 40 passes in both english and hindi how many
expeople1 [14]

Answer:

Total students that failed in both subjects would be 10

Step-by-step explanation:

Students that passed English : 70 - 40 = 30

Students that passed Hindi : 80 - 40 = 40

Total students that passed both subjects 110 (40 + 40 + 30)

120 - 110 = 10 students failed in both subjects

The 30 & 40 come from the students that passed each subject above, then the extra 40 comes from the amount of students that passed both.

3 0
3 years ago
Tasha invests in an account that pays 1.5% compound interest annually. She uses the expression P(1+r)t to find the total value o
zhenek [66]
\bf ~~~~~~ \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}\to &\$3000\\
r=rate\to 1.5\%\to \frac{1.5}{100}\to &0.015\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &5
\end{cases}
\\\\\\
A=3000\left(1+\frac{0.015}{1}\right)^{1\cdot 5}\implies A=3000(1.015)^5\implies A\approx 3231.852
5 0
3 years ago
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Write the sum of the numbers as the product of their GCF and another sum <br> 56+64
olga2289 [7]
This is a question about finding the GCF or the greatest common factor for these two numbers.  The real question there is what is the largest number that both 56 and 64 can be divided by and still give out integers (meaning, whole regular counting numbers).  This number turns out to be 8.  56/8 = 7.  64/8 = 8.  The GCF gets pulled to the outside of the parenthesis and the divided numbers remain on the inside being added together.  It looks like this, 8(7+8).
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3 years ago
6 + [-25 = (-3 + 2. 1)]
antiseptic1488 [7]

Answer:

6+[25=(-0.9)]

6+[-22.5]

-16.5

Step-by-step explanation:

the answe of this question will be -16.5

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3 years ago
The temperature at 6pm is
Ostrovityanka [42]

Answer:

12

Step-by-step explanation:

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3 years ago
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