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adelina 88 [10]
2 years ago
9

Help this is for a big part of my grade :'(

Mathematics
2 answers:
Mars2501 [29]2 years ago
5 0

Answer:

BCD are wrong

Step-by-step explanation:

Lerok [7]2 years ago
3 0
It’s the last one because a expression doesnt just needs numbers;)
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Gina wants to ride the waves at flagship pier in Galveston so she rents a surfboard for $5 per hour plus an initial or beginning
anastassius [24]
Fourteen hours. The equation is 5x+20=90. Solve for X.
5 0
2 years ago
A newspaper carrier has $5.40 in change. He has six more quarters than dimes but five times as many nickles as quarters. How man
BigorU [14]
X= number of dimes
x+6= number of quarters
5(x+6)= number of nickels
$5.40= 540 cents

Add the number of each coin and set it equal to the total in cents.


x + (x + 6) + 5(x + 6)= $5.40
multiply 5 by parentheses; change dollar total to cents

x + (x + 6) + (5 * x) + (5 * 6)= 540

x + (x + 6) + 5x + 30= 540
combine like terms

7x + 36= 540
subtract 36 from both sides

7x= 504
divide both sides by 7

x= 72 dimes

x + 6= 78 quarters

5(x + 6)= 390 nickels


CHECK
72 + 78 + 390= 540
540= 540


ANSWER: There are 72 dimes, 78 quarters and 390 nickels.

Hope this helps! :)
5 0
2 years ago
Six bags of Takis costs $13.50. Five bags of Takis costs $10. Which has
viktelen [127]

Answer:

the correct answer is 5 bags

4 0
2 years ago
Read 2 more answers
An angle measures 54° less than the measure of a supplementary angle. What is the measure of each angle?
Katen [24]
X+x+54=180
2x+54=180
2x=126
x=63
63+54=117
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8 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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