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dimaraw [331]
3 years ago
8

Diego cut 7 smaller boards of equal length from a board that is 9 1/3 feet long. How long is each of the 7 smaller boards? Write

your answer as an improper fraction using the / symbol for the fraction.
Mathematics
1 answer:
Temka [501]3 years ago
4 0
It would be 9 1/3 divided by 7/1 or 28/3 * 1/7

It would give you 28/21 which would simplify to 4/3 and still leave you with an improper fraction
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A cyclist has rode his bike 174 miles in 3 hours. If he continues at this rate, how long would it take him to complete another 4
In-s [12.5K]

Answer:

7 hours

Step-by-step explanation:

174/3 is 58 miles per hour so 406/58 is 7 hours

6 0
3 years ago
One of the factors of the polynomial x3 + 5x2 + 6x is (x2 + 3x). What is the other factor?
butalik [34]

Answer:

x+2

Step-by-step explanation:

Dividing the poly by the known factor:

x3+5x2+6x    <u>| x2+3x</u>

-x3-3x2             x + 2  <- the answer is here

      2x2+6x

      -2x2-6x

To prove that, let's multiply the two factors and find the pylonomial back on:

(x2+3x).(x+2) = x3 + 2x2 + 3x2 + 6x = x3 + 5x2 + 6x

Tks your attention!

8 0
3 years ago
Which expression is the same as 10÷4?
prisoha [69]
10/4 is not the simplified version that it can be. So 5/2 would be the same expression.
3 0
3 years ago
Read 2 more answers
Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam
MariettaO [177]
  • X-(A\cup B)=(X-A)\cap(X-B)

I'll assume the usual definition of set difference, X-A=\{x\in X,x\not\in A\}.

Let x\in X-(A\cup B). Then x\in X and x\not\in(A\cup B). If x\not\in(A\cup B), then x\not\in A and x\not\in B. This means x\in X,x\not\in A and x\in X,x\not\in B, so it follows that x\in(X-A)\cap(X-B). Hence X-(A\cup B)\subset(X-A)\cap(X-B).

Now let x\in(X-A)\cap(X-B). Then x\in X-A and x\in X-B. By definition of set difference, x\in X,x\not\in A and x\in X,x\not\in B. Since x\not A,x\not\in B, we have x\not\in(A\cup B), and so x\in X-(A\cup B). Hence (X-A)\cap(X-B)\subset X-(A\cup B).

The two sets are subsets of one another, so they must be equal.

  • X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices i\in I.

Proof of one direction for example:

Let x\in X-\left(\bigcup\limits_{i\in I}A_i\right). Then x\in X and x\not\in\bigcup\limits_{i\in I}A_i, which in turn means x\not\in A_i for all i\in I. This means x\in X,x\not\in A_{i_1}, and x\in X,x\not\in A_{i_2}, and so on, where \{i_1,i_2,\ldots\}\subset I, for all i\in I. This means x\in X-A_{i_1}, and x\in X-A_{i_2}, and so on, so x\in\bigcap\limits_{i\in I}(X-A_i). Hence X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i).

4 0
3 years ago
Teri has $6 less than does her sister, Jennie, who has x dollars. Teri
elixir [45]

The expression for the amount of money, in dollars, Teri has is x-2.

Since Jennie has x dollars while Teri has $6 less than does Jennie, this means that Teri will have: x - 6

Since Teri does not spend any money and earns $4, the the total amount that Teri has will be:

= x - 6 + 4

= x - 2

Therefore, Teri has x-2.

Read related link on:

brainly.com/question/24499674

4 0
3 years ago
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