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3 years ago

Image is(-6,-4) what is the pre image

Mathematics

1 answer:

svlad2 [7]3 years ago

8 0

Answer:

In the photo, I want you to look at Graph 1. Look at the bottom left square.

Step-by-Step:

(-6,-4) this problem. For starters, you should know that the first number goes to the bottom. Since we are using negatives, the first cumber goes on top. When you look at the image and see graph 1 in the bottom left square. Notice how it actually shows the replica. Recommend to draw it out yourself. sorry, the image is blurry. <33

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A student club holds a meeting. The predicate M(x) denotes whether person x came to the meeting on time. The predicate O(x) refe

Novay_Z [31]

Answer:

a) $\exists \, x \in C : O(x) = 0$

b) $\{ x \in C : O(x) = 1 \} \subseteq \{ x \in C : M(x) = 1 \}$

c) $\{ x \in C: M(x) = 1 \} = C$

d) $\{ x \in C : D(x) = 1 \} \, \cup \, \{x \in C : M(x) = 1 \} = C$

e) $\exists \, x \in C : M(x) = 1 \, \wedge D(x) = 1$

f) $\exists \, x \in C : O(x) = 1 \, \wedge M(x) = 0$

Step-by-step explanation:

M(x) = 1 if the person x came to the meeting, and 0 otherwise.

O(x) = 1 if the person is an officer of the club and 0 otherwise.

D(x) = 1 if the person has paid hid/her club dues and 0 otherwise.

Lets also call C the set given by the members of the club. C is the domain of the functions M, O and D.

a) If someone is not an officer, the there should be at least one value x such that O(x) = 0. This can be expressed by logic expressions this way

$\exists \, x \in C : O(x) = 0$

b) If all the officers came on time to the meeting, then for a value x such that O(x) = 1, we also have that M(x) = 1. Thus, the set of officers of the Club is contained on the set of persons which came to the meeting on time, this can be written mathematically this way:

$\{ x \in C : O(x) = 1 \} \subseteq \{ x \in C : M(x) = 1 \}$

c) If everyone was in time for the meeting, then C is equal to the set of persons who came to the meeting on time, or, equivalently, the values x such that M(x) = 1. We can write that this way:

$\{ x \in C: M(x) = 1 \} = C$

d) If everyone paid their dues or came on time to the meeting, then if we take the set of persons who came to the meeting on time and the set of the persons who paid their dues, then the union of the two sets should be the entire domain C, because otherwise there should be a person that didnt pay nor was it on time. This can be expressed logically this way:

$\{ x \in C : D(x) = 1 \} \, \cup \, \{x \in C : M(x) = 1 \} = C$

e) If at least one person paid their dues on time and came on time to the meeting, then there should be a value x on C such that M(x) and D(x) are both equal to 1. Therefore

$\exists \, x \in C : M(x) = 1 \, \wedge D(x) = 1$

f) If there is an officer who did not come on time for the meeting, then there should be a value x in C such that O(x) = 1 (x is an officer), and M(x) = 0. As a result, we have

$\exists \, x \in C : O(x) = 1 \, \wedge M(x) = 0$

I hope that works for you!

7 0

4 years ago

Truer or false? subtracting 7 from the parent linear function results in a reflection in the x-axis

beks73 [17]

The parent linear function is y = x

A reflection in the x-axis is denoted by - f(x), so the answer is false

6 0

1 year ago

Given ABC find the values of x and y. In your final answer, include all of your calculations.

kow [346]

So with this, we will be using proportions. To solve for x, our proportion will be $\frac{x}{6} =\frac{6}{x-9}$

Firstly, cross-multiply: $x^2-9x=36$

Next, subtract 36 on both sides: $x^2-9x-36=0$

Next, replace -9x with 3x-12x $x^2+3x-12x-36=0$

Next, factor x^2+3x and -12x-36 separately: $x(x+3)-12(x+3)=0$ . After, rewrite it as $(x-12)(x+3)=0$

Next, solve for x separately in the two parentheses:

$x-12=0\\ x=12$

$x+3=0\\ x=-3$

Since we can't have a negative length, x = 12.

To solve for y, our proportion will be $\frac{y}{9} =\frac{12}{y}$

Firstly, cross-multiply: $y^2=108$

Next, just square root each side, and your answer will be (rounded to hundredths) $y=10.39$

In short, x = 12 and y = 10.39.

5 0

3 years ago

Can someone help me with number 7 the circled one?

Ivenika [448]

Your answer:
<span>Equation:
x(x+1) = 2(x+1)+77
-----

x^2+x = 2x+79
x^2-x-79 = 0
------
This quadratic does not have an integer solution.</span>

5 0

3 years ago

Drummers march in two patterns for one song. One pattern has

Dafna11 [192]

Answer:

Suppose we have a total of 16 drummers.

for the first pattern we have 2 drummers in each row, this means that we have a total of 16/2 = 8 rows.

Now, for the second pattern, we have 4 times as many in each row, so we have 4*2 = 8 drummers in each row, then we have 16/8 = 2 rows.

You can see that in each pattern we have the same number of drummers, but a different number of rows.

Now, we only can do this if the total number of drummers is a multiple of 8 (because in the second part we want 8 drummers in each row, so if we want that the numbers of rows to be a whole number, then we must have that the total number of drummers is a multiple of 8)

7 0

3 years ago