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3 years ago

14

a pet store has 2 kittens. one kitten weighs 15/16 pounds. the other kitten weighs 12/16 pounds. what is the difference of the w

eights of the two kittens?

1 answer:

Lostsunrise [7]3 years ago

5 0

Subtract 15/16 and 12/16 to get 3/16

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Cual es la diferencia entre divisor y divisible

Anna [14]

La diferencia entre divisor y divisible es que los números compuestos (número que posee más de dos divisores) son los divisores y los números divisible son los números enteros como el 10,15,20,25 y 30.

espero que esto ayude

7 0

3 years ago

What is the answer for this question?

drek231 [11]

Answer: B

Step-by-step explanation:

(-3, -1) and (4, 5) are the points given

Slope formula: $\frac{Y2-Y1}{X2-X1}$

$\frac{5--1}{4--3} = \frac{5+1}{4+3} \\$

$\frac{5+1}{4+3} = \frac{6}{7}$

8 0

3 years ago

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

Find the x intercepts of the parabola with vertex (-1,-16) and y intercept (0,-15)

Arlecino [84]

<span> first, write the equation of the parabola in the required form: </span>
<span>(y - k) = a·(x - h)² </span>

<span>Here, (h, k) is given as (-1, -16). </span>
<span>So you have: </span>
<span>(y + 16) = a · (x + 1)² </span>

<span>Unfortunately, a is not given. However, you do know one additional point on the parabola: (0, -15): </span>

<span>-15 + 16 = a· (0 + 1)² </span>
<span>.·. a = 1 </span>

<span>.·. the equation of the parabola in vertex form is </span>
<span>y + 16 = (x + 1)² </span>

<span>The x-intercepts are the values of x that make y = 0. So, let y = 0: </span>

<span>0 + 16 = (x + 1)² </span>
<span>16 = (x + 1)² </span>

<span>We are trying to solve for x, so take the square root of both sides - but be CAREFUL! </span>

<span>± 4 = x + 1 ...... remember both the positive and negative roots of 16...... </span>

<span>Solving for x: </span>
<span>x = -1 + 4, x = -1 - 4 </span>
<span>x = 3, x = -5. </span>

<span>Or, if you prefer, (3, 0), (-5, 0). </span>

8 0

3 years ago

Let mZA = 40°. If zB is a complement of ZA, and ZC is a supplement of ZB, find these measures.

seraphim [82]

Angle b= 50

Angle c= 130

Complementary angles are two angles that add to equal 90

Supplementary angles are two angles that add to equal 180

To find the measurement of angle b, subtract 40 from 90 (50)

To find the measurement of angle c, subtract 50 from 180 (130)

7 0

4 years ago