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4 years ago

5

In Cathy’s garden, 5/6 of the area is planted with flowers. Of the flowers, 3/10 of them are red. What fraction of Cathy’s garde

n is planted with red flowers?

1 answer:

Leno4ka [110]4 years ago

7 0

50/33 you have to add then mutlyply

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Which of the following is the graph of the quadratic function? y=x^2+4-12

tiny-mole [99]

Answer:

Graph B

Step-by-step explanation:

The solutions to a quadratic equation are the points on which we have the graph of the curve touching the x-axis

Now the first thing we will do here is to solve the quadratic equation graph;

x^2 + 4x -12 = 0

x^2 +6x - 2x -12 = 0

x (x + 6) -2(x + 6) = 0

(x-2)(x + 6) = 0

x = 2 or -6

So the graph that touches the x-axis at the points x = 2 and x = -6 is the solution to the quadratic equation

Graph B is the closest to what we have as answer

5 0

4 years ago

Which of the following relations are not functions? Select all that apply

alekssr [168]

A,b,c, d that’s the answer

5 0

3 years ago

56 points for whoever answers correctly !

vladimir2022 [97]

We need to use what we know about rectangles to get:

1)

Total length = 2*W + 8ft
Total width = W + 8ft

2) area = 2*W^2 + 24ft*W + 64ft^2

<h3>Working with rectangles:</h3>

We know that rectangles are defined by two measures, width W and length L.

Here we do know that the length of the pool is twice the width, and the width is W, then the length of the pool is:

L = 2*W

And we also have a sidewalk of 4ft all around the pool, now we want to get:

1) The total length and the total width.

This will be equal to the length/width of the pool <u>plus twice the width of the sidewalk</u> (we add it twice because is in both ends) then we have:

Total length = L + 2*4ft = 2*W + 8ft
Total width = W + 2*4ft = W + 8ft

2) Now we want to get an expression for the total area of the pool.

Remember that for a rectangle the area is just the product between the width and the length, so to get the area of the pool with the sidewalk we just take:

area = (total length)*(total width)

area = (2*W + 8ft)*(W + 8ft) = 2*W^2 + 3*W*8ft + 64ft^2

area = 2*W^2 + 24ft*W + 64ft^2

This is the equation that gives the total area as a function of W, the width of the pool.

If you want to learn more about rectangles, you can read:

brainly.com/question/17297081

6 0

3 years ago

Which product is positive?<br> ( 13 ( 31 )<br> ( 31 )<br> (3) ( 13 (1)

Anastaziya [24]

The first one because it’s (13(31)

8 0

4 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