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2 years ago

2+2/4+7-9x(-5). Please help me

Mathematics

1 answer:

Cerrena [4.2K]2 years ago

3 0

Answer:

109/2 is the answer of your question . Use PEMDAS rule

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Jan age is 3 less than the twice triss agethe sum of thier ages is 30

tresset_1 [31]

Jan is 19, Triss is 11.

3 0

3 years ago

9. Todd makes $9.25 per hour plus a $50 bonus. Todd made $235 this week.

Daniel [21]

Answer:

20 hours

Step-by-step explanation:

use this linear equation

y = 9.25x + 50

plug in 235 for y and solve for x

235 = 9.25x + 50

185 = 9.25x

20 = x

so he worked for 20 hours

hope this helped <3

8 0

3 years ago

Ion understand this one

Dafna1 [17]

Answer:

114

Step-by-step explanation:

\+_+/

3 0

2 years ago

Read 2 more answers

Answer and how to do it

ollegr [7]

Answer:

$(x - 9)^2 + y^2 = 36$

Step-by-step explanation:

The equation of a circle in standard form is

$(x - h)^2 + (y - k)^2 = r^2$

You are given

$x^2 + y^2 - 18x + 45 = 0$

In order to put the equation in standard from, we need to complete the square. Since there is no y term, the y part is simply y^2, and there is no need to complete the square for y. For x, we do have an x term, so we must complete the square in x.

Start by grouping the x terms and subtracting 45 from both sides.

$x^2 - 18x + y^2 = -45$

Now we need to complete the square for x.

$x^2 - 18x ~~~~~~+ y^2 = -45$

The number that completes the square will go in the blank above, and it will also be added to the right side of the equation.

To find the number you need to add to complete the square, take the coefficient of the x term. It is -18. Divide it by 2. You get -9. Now square -9 to get 81. The number that completes the square in x is 81. Now you add it to both sides of the equation.

$x^2 - 18x + 81 + y^2 = -45 + 81$

$(x - 9)^2 + y^2 = 36$

Answer: $(x - 9)^2 + y^2 = 36$

6 0

3 years ago

a. Determine all bijections from the {1,2,3} into {a,b,c}. b. Determine all bijections from {1, 2, 3} into {a,b,c,d}.

avanturin [10]

Part A

There are 6 bijections from {1,2,3} to {a,b,c}. This is effectively the same as asking the question "how many ways are there to arrange {a,b,c} where order matters?" We use a factorial to answer this question.

3 factorial = 3! = 3*2*1 = 6

You can also use a permutation, which is composed of factorials, to get the same answer.

======================================================

Part B

There are no bijections from {1,2,3} to {a,b,c,d}. Why is this? Because a bijection has two properties: it must be one-to-one, and it must be onto. The term "onto" in mathematics means "every value in the range is targeted". In the case of the range {a,b,c,d} it is not possible for each value to show up. This is because there are only three items in the domain {1,2,3}. You'll always be one letter short.

As you can probably guess, a bijection is only possible if and only if n(D) = n(R), where D and R are the domain and range respectively. The notation n(D) represents the count or number of items in set D.

3 0

3 years ago