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Alexxx [7]
3 years ago
5

12x - 5 = 8x - x + 50 Help please show work and check the solution

Mathematics
1 answer:
Inga [223]3 years ago
3 0

Answer:

x = 11

Step-by-step explanation:

12x - 5 = 8x - x + 50 \\ 12x - 5 = 7x + 50 \\ 12x - 5 - 7x = 7x + 50 - 7x \\ 12x - 7x - 5 = 50  \\ 5x - 5 = 50 \\ 5x - 5 + 5 = 50 + 5 \\ 5x = 55 \\ x = 11

and check:

12(11) - 5 = 8(11) - (11) + 50 \\ 132 - 5 = 77 + 50 \\ 127 = 127

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Wittaler [7]

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5 0
2 years ago
Fabian harvests 10 pounds of tomatoes from his garden. He needs 225 pounds to make a batch of soup. If he sets aside 2.8 pounds
nadya68 [22]
He can make 3 batches of soup.

Starting with 10 pounds of tomatoes and taking out the number of pounds he sets aside for spaghetti sauce, 2.8, we have:
10-2.8 = 7.2 pounds left.

0.2 is read as "two tenths," so 7.2 = 7 2/10

We divide this remaining number of pounds by 2 2/5 for each batch of soup:

7 2/10 ÷ 2 2/5

Convert each to an improper fraction:
72/10 ÷ 12/5

Multiply by the reciprocal:
72/10 × 5/12 = 360/120 = 3


5 0
3 years ago
NEED CORRECT ANSWER...
Misha Larkins [42]

39.50-23.76-2.57-1.49=11.68

take the sales and subtract the expenses

6 0
3 years ago
It's a question from real and complex numbers which I can't solve. so someone PLZ HeLp​
Semmy [17]

Answer:

\frac{1}{5}

Step-by-step explanation:

Using the rules of exponents

a^{m} × a^{n} = a^{(m+n)}, \frac{a^{m} }{a^{n} } = a^{(m-n)}, (a^m)^{n} = a^{mn}

Simplifying the product of the first 2 terms

\frac{a^{p^2+pq} }{a^{pq+q^2} } × \frac{a^{q^2+qr} }{a^{qr+r^2} }

= a^{p^2-q^2} × a^{q^2-r^2}

= a^{p^2-r^2}

Simplifying the third term

5((a^p+r)^{p-r}

= 5a^{(p+r)(p-r)} = 5a^{(p^2-r^2)}

Performing the division, that is

\frac{a^{(p^2-r^2)} }{5a^{(p^2-r^2)} } ← cancel a^{(p^2-r^2)} on numerator/ denominator leaves

= \frac{1}{5}

4 0
3 years ago
Can somebody prove this mathmatical induction?
Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

\sum \limits_{j=1}^k2^j=2(2^k-1)

3 step:

Check for n=k+1 whether the statement

\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)

is true.

Start with the left side:

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)

According to the 2nd step,

\sum \limits_{j=1}^k2^j=2(2^k-1)

Substitute it into the \ast

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)

So, you have proved the initial statement

4 0
3 years ago
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