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Serjik [45]
3 years ago
12

Evalúa 9-b cuando b=8

Mathematics
1 answer:
Amiraneli [1.4K]3 years ago
8 0

Answer:

<h2><em>1</em></h2>

<em>Sol</em><em>ution</em><em>,</em>

<em>b</em><em>=</em><em>8</em>

<em>Now</em><em>,</em>

<em>9</em><em>-</em><em>b</em>

<em>=</em><em>9</em><em>-</em><em>8</em>

<em>=</em><em>1</em>

<em>Hope</em><em> </em><em>this </em><em>helps.</em><em>.</em><em>.</em>

<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em>

You might be interested in
Nonsense will be Reported!!​
USPshnik [31]

Step-by-step explanation:

1. when x is 0

y will be

4x+2y=10

4(0)+2y=10

0+2y=10

collect like terms

2y=10-0

2y=10

divide both sides by 2

2y/2=10/2

y=5

so when x =0 y will equal to 5

2. when x is 1

y will be

4x+2y=10

4(1)+2y=10

4+2y=10

collect like terms

2y=10-4

2y=6

divide both sides by two

2y/2=6/2

y=3

so when x is 1, y is 3

3. when x is 2

4x+2y=10

4(2)+2y=10

8+2y=10

(CLT) 2y=10-8

2y =2

y= 1

6 0
2 years ago
Read 2 more answers
Account (A) is a saving account , the interest earned after 1 year is $12 .If the interest rate is 3.2% for account (A). How muc
BaLLatris [955]

Answer:

The principal for the account is $375.

Step-by-step explanation:

This is a simple interest problem.

The simple interest formula is given by:

E = P*I*t

In which E is the amount of interest earned, P is the principal(the initial amount of money), I is the interest rate(yearly, as a decimal) and t is the time.

The interest earned after 1 year is $12 .If the interest rate is 3.2% for account (A).

This means, respectively, that t = 1, E = 12, I = 0.032

We want to find P.

E = P*I*t

12 = P*0.032*1

P = \frac{12}{0.032}

P = 375

The principal for the account is $375.

8 0
3 years ago
Suppose two dice are tossed and the numbers on the upper faces are observed. Let S denote the set of all possible pairs that can
Thepotemich [5.8K]

Answer:

▪A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)}

▪C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B bar = {(1,2) (1,4) (1,6) (3,2) (3,4) (3,6) (5,2) (5,4) (5,6)}

▪A bar∪B = {(1,1) (1,3) (1,5) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,3) (3,5) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,3) (5,5) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

▪A bar∩C = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)}

Step-by-step explanation:

S = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)} (second die is even)

B = {(1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5) (4,2) (4,4) (4,6) (5,1) (5,3) (5,5) (6,2) (6,4) (6,6)} (sum of the two numbers is even)

C = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,3) (2,5) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,3) (4,5) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,3) (6,5)} (at least one in the pair is odd i.e one of the pair is odd or both are odd)

A bar = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)} (the pairs that are not in A)

B bar = {(1,2) (1,4) (1,6) (2,1) (2,3) (2,5) (3,2) (3,4) (3,6) (4,1) (4,3) (4,5) (5,2) (5,4) (5,6) (6,1) (6,3) (6,5)} (the pairs that are not in B)

C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)} (the pairs that are not in C)

▪A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)}

▪C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)} (intersection: the pairs that are common to both A and B)

▪A∩B bar = {(1,2) (1,4) (1,6) (3,2) (3,4) (3,6) (5,2) (5,4) (5,6)} (intersection: the pairs that are common to both A and B bar)

▪A bar∪B = {(1,1) (1,3) (1,5) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,3) (3,5) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,3) (5,5) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)} (union: all the pairs in A bar and B )

▪A bar∩C = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)} (intersection: the pairs that are common to both A bar and C)

6 0
3 years ago
How do you find out area for the formzual for area would be what-
ser-zykov [4K]
This is simple:
area= Length x Width
7 0
3 years ago
Read 2 more answers
Anyone can help me solve this equation using cross multiplying
Natali5045456 [20]

9514 1404 393

Answer:

  x = 1 or 5

Step-by-step explanation:

The notion of "cross-multiplying" is the idea that the numerator on the left is multiplied by the denominator on the right, and the numerator on the right is multiplied by the denominator on the left. This looks like ...

  \displaystyle \frac{x-1}{7}=\frac{2x-2}{3x-1}\ \longrightarrow\ (x-1)(3x-1)=(7)(2x-2)

Then the solution proceeds by eliminating parentheses, and solving the resulting quadratic equation.

  3x^2-4x+1=14x-14\\\\3x^2-18x+15=0\qquad\text{subtract $14x-14$}\\\\x^2-6x+5=0 \qquad\text{divide by 3}\\\\(x-1)(x-5)=0\qquad\text{factor}\\\\x\in\{1,5\}

_____

<em>Comment on "cross multiply"</em>

Like a lot of instructions in Algebra courses, the idea of "cross multiply" describes <em>what the result looks like</em>. It doesn't adequately describe how you get there. The <em>one and only rule</em> in solving Algebra problems is "<em>whatever is done to one side of the equation must also be done to the other side of the equation</em>." If you multiply one side by one thing and the other side by a different thing, you are violating this rule.

What looks like "cross multiply" is really "<em>multiply by the product of the denominators</em> and cancel like terms from numerator and denominator." Here's what that looks like with the intermediate steps added.

  \displaystyle \frac{x-1}{7}=\frac{2x-2}{3x-1}\\\\\frac{x-1}{7}\times7(3x-1)=\frac{2x-2}{3x-1}\times7(3x-1)\\\\(x-1)(3x-1)=(2x-2)(7)\qquad\textit{looks like}\text{ cross multiply}

8 0
2 years ago
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