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rosijanka [135]
4 years ago
6

Simplify the expression: (8+-9)(9+-8)

Mathematics
2 answers:
Alexus [3.1K]4 years ago
8 0

Answer:

-1

Step-by-step explanation:

[8 + −9][9 + −8] = [8 - 9][9 - 8] = [1][-1] = -1

I am joyous to assist you anytime.

Paha777 [63]4 years ago
5 0

Answer:

-1

Step-by-step explanation:

(8 + -9)(9 + -8)

A negative times a positive is a negative.

8 + -9

+- = -

8 - 9

8 - 9 = -1

(-1)(9 + -8)

9 + -8

+- = -

9 - 8

9 - 8 = 1

(-1)(1)

Now, we multiply

Remember, a negative times a positive equals a negative.

-1 * 1 = -1

Our answer is -1

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Shelby is starting a business selling handmade necklaces. She has decided to invest an initial amount of $172 for advertising, a
8_murik_8 [283]

Answer:

-43 necklaces.

-total expenses=Sales=$473

Step-by-step explanation:

The breakeven point is the point where the sales revenue equals the expenditure.

-The fixed expenses is $172

-Let x be the number of necklaces sold, the breakeven point is expressed as:

Sales= Expenses\\\\11x=7x+172\\\\4x=172\\\\x=43

Hence, Shelby has to sell 43 necklaces to breakeven.

#The expenses at the breakeven point is:

E=7x+172\\\\=7\times43+172\\\\=473

Hence, expenses at breakevent point is $473

#Sales is equal to expenses at this point, hence, sales of $473

4 0
3 years ago
Esther the Clown does face painting at the city carnival. She paints 7 faces every 21 minutes and spends the same amount of time
Lena [83]

Answer:   M=3f     hope this helps guys

7 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
3 years ago
tim has a box of 15 markers. He gives 3 markers each to 4 friends..what expression can show the numbers of markers tim has left?
SOVA2 [1]
15-(3•4) = m

m=how many markers Tim has left.
8 0
3 years ago
2. (10.03)
faust18 [17]

This question is incomplete because it was not written properly

Complete Question

A teacher gave his class two quizzes. 80% of the class passed the first quiz, but only 60% of the class passed both quizzes. What percent of those who passed the first one passed the second quiz? (2 points)

a) 20%

b) 40%

c) 60%

d) 75%

Answer:

d) 75%

Step-by-step explanation:

We would be solving this question using conditional probability.

Let us represent the percentage of those who passed the first quiz as A = 80%

and

Those who passed the first quiz as B = unknown

Those who passed the first and second quiz as A and B = 60%

The formula for conditional probability is given as

P(B|A) = P(A and B) / P(A)

Where,

P(B|A) = the percent of those who passed the first one passed the second

Hence,

P(B|A) = 60/80

= 0.75

In percent form, 0.75 × 100 = 75%

Therefore, from the calculations above, 75% of those who passed the first quiz to also passed the second quiz.

3 0
3 years ago
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