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fenix001 [56]
3 years ago
8

How do you do proofs

Mathematics
1 answer:
AnnyKZ [126]3 years ago
8 0

Answer:

you can watch YT videos or search online. search: How to solve proofs. Also search up proof theorems. This will help you even more towards solving more complex proof problems.

Step-by-step explanation:

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(c+4)^2(4-c)^2 i need help
Furkat [3]

Answer

c^4-32c^2+256 (Simplified)

8 0
2 years ago
Whoever answers question 19 correctly with both parts of the questions answered
diamong [38]
Let's focus on 0.7*0.8 for now. 

Start by drawing a large square. Cut this figure into 10 rows and 10 columns. So this means you'll have 10*10 = 100 little squares. 

Now highlight the first 7 rows. Shade in all 70 squares (7*10 = 70)
Starting on the left side, highlight the first 8 columns. You'll shade in 80 squares (8*10 = 80)

Use different colors for your highlighting or somehow indicate different shading styles. This way you can see the overlapping region. The overlapping region consists of 56 squares (7 rows, 8 columns ---> 7*8 = 56 little squares)

Each little square represents 0.01, so having 56 of them means we have 0.56

This shows that 0.7*0.8 = 0.56

-----------------------------------------------------------

How is this different if we had 1.7*0.8? Well we can break 1.7 into 1+0.7 to have

1.7*0.8 = (1+0.7)*0.8
1.7*0.8 = (1)*0.8+(0.7)*0.8
1.7*0.8 = (1*0.8)+(0.7*0.8)

The portion (0.7*0.8) was done earlier. That result was 0.56. So we just need to compute (1*0.8), which is simply 0.8; recall that 1 times any number is that number itself.

Now simply add 0.8 to 0.56 to get 1.36

So, 1.7*0.8 = 1.36


5 0
3 years ago
Find two integers whose sum is -18 and product is 77
dmitriy555 [2]

Answer:

-11 and -7

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Can you help me solve an infinite geometric sequence
zaharov [31]

A geometric series like

\displaystyle \sum_{n=1}^\infty \dfrac{1}{\alpha^n}

Converges if and only if |\alpha|>1. If this is the case, the sum equals

\displaystyle \sum_{n=1}^\infty \dfrac{1}{\alpha^n} = \dfrac{1}{\alpha-1}

So, in your case, you have convergence if and only if

|2+a|>1 \iff 2+a>1 \lor 2+a-1 \lor a

And if this is the case, the sum equals

\displaystyle \sum_{n=1}^\infty \dfrac{1}{(2+a)^n} = \dfrac{1}{2+a-1} = \dfrac{1}{a+1}

8 0
3 years ago
The length of the minute hand on a clock is 5in
GenaCL600 [577]

Answer:

25 Minutes

Step-by-step explanation:

Each minute sign represents 5 minutes gone by. When hand is on the 5, it is 25 minutes.

6 0
2 years ago
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