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allsm [11]
3 years ago
7

How do i solve this pls​

Mathematics
1 answer:
miskamm [114]3 years ago
7 0
First we would solve both x’s
3x^2 - 5x
But 3 has the thingy thing (forgot the name haha)
Which means you have to multiply it by itself that many times which would be
3x3=9
So now it would be
9x- 5x = 4x
Which leaves 7 and 4x

Now you may subtract those two

7-4x= 3x
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Answer:

-2x^2-20x-46

Step-by-step explanation:

-2(x^2)-20x-46

-2(x^2) -2(10x)-46

-2(x^2)-2(10x)-2(23)

-2(x^2+10x)-2(23)

-2x^2-20x-46

7 0
3 years ago
The line plot shows how long you read during each reading session last week. What is the total time you read during these sessio
Diano4ka-milaya [45]

Answer: 15 1/2

Step-by-step explanation: There are two 1/2 s so 1/2 + 1/2 is 2/2 or 1 whole. Next is 1. There are 4 ones. 1+1+1+1 = 4. There is one 1 1/2. There are  two 2. 2+2 is 4. There are two 2 1/2 so 2 1/2 + 2 1/2 is 4 2/2 which is 5. Now you have to add all the answers together. 1 +4+ 1 1/2 +4+5  which is 15 and 1/2. 15 1/2 is the answer.

5 0
4 years ago
Read 2 more answers
A student has test scores of 68%, 75%, and 79% in a government class. What must she score on the last exam to earn a B (80% or b
FinnZ [79.3K]
68% +75 % +79 % +x ÷4≥ 80%
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She has to get at least 98% to get an B.
3 0
3 years ago
A car dealership has seven cars in the lot. Unfortunately, the keys to the cars have been mixed up. The manager randomly grabs a
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8 0
3 years ago
Use mathematical induction to prove the statement is true for all positive integers n. 1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = (n(2n-
Charra [1.4K]

Answer:

The statement is true is for any n\in \mathbb{N}.

Step-by-step explanation:

First, we check the identity for n = 1:

(2\cdot 1 - 1)^{2} = \frac{2\cdot (2\cdot 1 - 1)\cdot (2\cdot 1 + 1)}{3}

1 = \frac{1\cdot 1\cdot 3}{3}

1 = 1

The statement is true for n = 1.

Then, we have to check that identity is true for n = k+1, under the assumption that n = k is true:

(1^{2}+2^{2}+3^{2}+...+k^{2}) + [2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}

\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)}{3} +[2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}

\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot [2\cdot (k+1)-1]^{2}}{3} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}

k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot (2\cdot k +1)^{2} = (k+1)\cdot (2\cdot k +1)\cdot (2\cdot k +3)

(2\cdot k +1)\cdot [k\cdot (2\cdot k -1)+3\cdot (2\cdot k +1)] = (k+1) \cdot (2\cdot k +1)\cdot (2\cdot k +3)

k\cdot (2\cdot k - 1)+3\cdot (2\cdot k +1) = (k + 1)\cdot (2\cdot k +3)

2\cdot k^{2}+5\cdot k +3 = (k+1)\cdot (2\cdot k + 3)

(k+1)\cdot (2\cdot k + 3) = (k+1)\cdot (2\cdot k + 3)

Therefore, the statement is true for any n\in \mathbb{N}.

4 0
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