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2 years ago

How many parts are there in the format of a two-column proof?

Mathematics

2 answers:

stich3 [128]2 years ago

5 0

6 parts in 2 collums........

iogann1982 [59]2 years ago

4 0

There is 6 parts in a two column format.

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A triangle has sides with lengths of 7 yards, 12 yards, and 13 yards. Is it a right triangle?

sergey [27]

Answer:

Step-by-step explanation:

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3 years ago

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If Jasper installs a floor himself, it will take him 7 hours. Working with Yolanda, it will take them 3 hours.

Svetllana [295]

Answer: 5 $\frac{1}{4}$ hrs

Explanation:

Jasper: $\frac{1}{7}$ per hr

Yolanda: $\frac{1}{x}$ per hr

Together: $\frac{1}{3}$ per hr

Jasper + Yolanda = Together

$\frac{1}{7}$ + $\frac{1}{x}$ = $\frac{1}{3}$

$(21x)\frac{1}{7}$ + $(21x)\frac{1}{x}$ = $(21x)\frac{1}{3}$

3x + 21 = 7x

21 = 4x

$\frac{21}{4}$ = x

$5\frac{1}{4}$

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3 years ago

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A sheet of construction paper is 9 inches long and 11 inches wide. How many 1- inch squares of paper can cut out one sheet of pa

Serggg [28]

Answer:

Step-by-step explanation: 9*11=99

8 0

3 years ago

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What is the answer = 8w- 8 - 6w = 4w - 7 w= ??

ololo11 [35]

Answer:

- 1/2 (or - 0.5)

Step-by-step explanation:

so first, simplify by combining like terms.

so 8w-6w = 2w

so now we have

2w - 8 = 4w - 7

now we have to get the w's to one side of the equation so we can solve for it. so that mean we have to subtract 2w from both sides. 2w - 2w is 0. and then 4w - 2w = 2w. so now we have

-8 = 2w - 7

then to slowly start isolating w, add 7 to both sides of th equation. so now its

-1 = 2w

finally to get w by itself divide 2 from both sides.

and the answer is

- $\frac{1}{2}$ = w

5 0

2 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