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just olya [345]
2 years ago
9

Please help me possible

Mathematics
1 answer:
SIZIF [17.4K]2 years ago
6 0

hmm? Ok I don't know this or if this is old. 123

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Solve the system by adding. Drag and drop numbers into the coordinates to complete the ordered pair.
Mademuasel [1]

Answer:

(-3,5)

Step-by-step explanation:

4x + 3y = 3

x − 3y = −18

Solve by addition/Elimination

x = -3

y = 5

So,

The solution of the system is (-3,5).

HELPPP PLZZ (I got you)

<u><em>Ace </em></u>

8 0
2 years ago
Read 2 more answers
A:1/360<br> B:1/90<br> C:1/180<br> D:1/100
valkas [14]

the answer is 1/360.

\frac{1}{1}  = 360degrees  \\ x= 1degree

cross multiplication and u get the answer

5 0
3 years ago
The diameter of a cookie is 2 inches. If pico’s 3.14, what is the area of the cookie, rounded to the nearest 10th?
Amanda [17]

For this case we must find the area of the cookie, knowing that it has a circular shape. By definition, the area of a circle is given by:

A = \pi * r ^ 2

Where:

r: It is the radius of the circle

According to the data we have:

r = \frac {2} {2} = 1 \ in

Thus, the area of the cookie is:

A = \pi * (1) ^ 1

We have \pi = 3.14

A = 3.14 * 1\\A = 3.14

We round:

A = 3.1

Thus, the area of the cookie is 3.1 \ in ^ 2

Answer:

3.1 \ in ^ 2

6 0
2 years ago
The physical education instructor asked each student to do a total of 36 pull-ups and push-ups in 1 minute. The instructor wante
NeTakaya

Answer: Pull-ups = 4 and push up = 32

Step-by-step explanation:

Let x represents number of pull ups in one minute.

And, y represents number of push ups in one minute.

Therefore, According to the question,

Each student to do a total of 36 pull-ups and push-ups in 1 minute.

⇒ x+y= 36

But, again from equation, The instructor wanted students to do 8 times as many push-ups as pull-ups.

Therefore, 8x = y

Thus, required system of linear equations that represents this situation is,

x+y= 36 and 8x = y , where x is the number of pull ups and y is the number of push ups in one minute.

By solving these two equations we get, x=4 and y= 32

That is, In one minute 4 pull ups and 32 push ups are required.




8 0
3 years ago
Let C(x) be the statement "x has a cat," let D(x) be the statement "x has a dog," and let F(x) be the statement "x has a ferret.
jek_recluse [69]

Answer:

\mathbf{a)} \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)\\\mathbf{b)} \left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee \; F(x)\\\mathbf{c)} \left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)\\\mathbf{d)} \left( \forall x \in X\right) \; \neg C(x) \; \vee \; \neg D(x) \; \vee \; \neg F(x)\\\mathbf{e)} \left((\exists x\in X)C(x) \right) \wedge  \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)

Step-by-step explanation:

Let X be a set of all students in your class. The set X is the domain. Denote

                                        C(x) -  ' \text{$x $ has a cat}'\\D(x) -  ' \text{$x$ has a dog}'\\F(x) -  ' \text{$x$ has a ferret}'

\mathbf{a)}

Consider the statement '<em>A student in your class has a cat, a dog, and a ferret</em>'. This means that \exists x \in X so that all three statements C(x), D(x) and F(x) are true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

                         \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)

\mathbf{b)}

Consider the statement '<em>All students in your class have a cat, a dog, or a ferret.' </em>This means that \forall x \in X at least one of the statements C(x), D(x) and F(x) is true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

                        \left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee F(x)

\mathbf{c)}

Consider the statement '<em>Some student in your class has a cat and a ferret, but not a dog.' </em>This means that \exists x \in X so that the statements C(x), F(x) are true and the negation of the statement D(x) . We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

                      \left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)

\mathbf{d)}

Consider the statement '<em>No student in your class has a cat, a dog, and a ferret..' </em>This means that \forall x \in X none of  the statements C(x), D(x) and F(x) are true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as a negation of the statement in the part a), as follows

\neg \left( \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)\right) \iff \left( \forall x \in X\right) \; \neg C(x) \; \vee \; \neg D(x) \; \vee \; \neg F(x)

\mathbf{e)}

Consider the statement '<em> For each of the three animals, cats, dogs, and ferrets, there is a student in your class who has this animal as a pet.' </em>

This means that for each of the statements C, F and D there is an element from the domain X so that each statement holds true.

We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

           \left((\exists x\in X)C(x) \right) \wedge  \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)

5 0
3 years ago
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