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2 years ago

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)

Ace

8 0

2 years ago

Read 2 more answers

A:1/360 B:1/90 C:1/180 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 'A student in your class has a cat, a dog, and a ferret'. 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 'All students in your class have a cat, a dog, or a ferret.' 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 'Some student in your class has a cat and a ferret, but not a dog.' 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 'No student in your class has a cat, a dog, and a ferret..' 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 ' For each of the three animals, cats, dogs, and ferrets, there is a student in your class who has this animal as a pet.'

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