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2 years ago

Hello could you help me please?

Mathematics

2 answers:

Ilya [14]2 years ago

3 0

Answer:

C. 6

Step-by-step explanation:

So it says read less thann 25 books --> that means the numbers are 13,14,14,16,19,21,21,22,23, and 24 --> a total of 10 numbers

Then it says read more than 35 books --> that means the numbers are 37,41,42,43 --> a total of 4 numbers

Now, it says how many more --> that means you subtract 4 by 10 = 10 - 4 = 6

Therefore the answer is 6 more students read less than 25 books than read more than 35 books.

MaRussiya [10]2 years ago

3 0

Yes ! what grade are you in? this is some difficult math

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Shaylee has 77 pennies. She will trade in the pennies to get nickels.

Ghella [55]

0.77 / 0.05 = 15.4 ...so we know it is 15 nickels
15 * 0.05 = 0.75.....0.77 - 0.75 = 0.02....with 2 pennies left over

ur answer is : 15 nickels and 2 pennies left over

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2 years ago

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2 years ago

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What is the result of 36 divided by 9? Use the number line to find the answer. A number line going from 0 to 39. The numbers 0,

Vladimir79 [104]

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Step-by-step explanation:

36/9=4

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3 years ago

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Please help with these math questions

Tanya [424]

1.
a^2+b^2=c^2
easy
a=a
b=13
c=21
a^2+13^2=21^2
a^2+169=441
minus 169 both sides
a^2=272
sqrt both sides
a=16.4924
D is answer

2. same thing
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18^2+26^2=c^2
324+676=c^2
1000=c^2
sqrt both sides
31.6228=c
C is answer

3.
one way is to plug them in
remember that hypotonuse is longest side
A. 7^2+24^2=25^2, is that true?, yes it is treu
answer is A

4.
legs are 6 and 4
a=6
b=4
c=x
4^2+6^2=x^2
16+36=x^2
52=x^2
sqrt both sides
7.2111=x
7.2=x

5.c=20
a=x
b=14

x^2+14^2=20^2
x^2+196=400
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1. D
2. C
3. A
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5. 14.3

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2 years ago

Prove A-(BnC) = (A-B)U(A-C), explain with an example

NikAS [45]

Answer:

Prove set equality by showing that for any element $x$ , $x \in (A \backslash (B \cap C))$ if and only if $x \in ((A \backslash B) \cup (A \backslash C))$ .

Example:

$A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace$ .

$B = \lbrace0,\, 1 \rbrace$ .

$C = \lbrace0,\, 2 \rbrace$ .

$\begin{aligned} & A \backslash (B \cap C) \\ =\; & \lbrace 0,\, 1,\, 2,\, 3 \rbrace \backslash \lbrace 0 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace \end{aligned}$ .

$\begin{aligned}& (A \backslash B) \cup (A \backslash C) \\ =\; & \lbrace 2,\, 3\rbrace \cup \lbrace 1,\, 3 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace\end{aligned}$ .

Step-by-step explanation:

Proof for $[x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))]$ for any element $x$ :

Assume that $x \in (A \backslash (B \cap C))$ . Thus, $x \in A$ and $x \not \in (B \cap C)$ .

Since $x \not \in (B \cap C)$ , either $x \not \in B$ or $x \not \in C$ (or both.)

If $x \not \in B$ , then combined with $x \in A$ , $x \in (A \backslash B)$ .
Similarly, if $x \not \in C$ , then combined with $x \in A$ , $x \in (A \backslash C)$ .

Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ as required.

Proof for $[x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]$ :

Assume that $x \in ((A \backslash B) \cup (A \backslash C))$ . Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

If $x \in (A \backslash B)$ , then $x \in A$ and $x \not \in B$ . Notice that $(x \not \in B) \implies (x \not \in (B \cap C))$ since the contrapositive of that statement, $(x \in (B \cap C)) \implies (x \in B)$ , is true. Therefore, $x \not \in (B \cap C)$ and thus $x \in A \backslash (B \cap C)$ .
Otherwise, if $x \in A \backslash C$ , then $x \in A$ and $x \not \in C$ . Similarly, $x \not \in C \!$ implies $x \not \in (B \cap C)$ . Therefore, $x \in A \backslash (B \cap C)$ .

Either way, $x \in A \backslash (B \cap C)$ .

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ implies $x \in A \backslash (B \cap C)$ , as required.

8 0

2 years ago