Quick i dont have time

Describe and correct the error in solving the equation

Free_Kalibri [48]

Answer:

2(s + 5) = 2s + 10, not 2s - 10

2s - 10 = 2s + 10

$0 \neq 20$

3 0

3 years ago

Trenton is building a skateboarding ramp in the shape of a triangular prism. According to his plan, the ramp would have a base o

Vikki [24]

Answer:

The answer is 21 square feet of plywood.

Step-by-step explanation:

8 0

4 years ago

Read 2 more answers

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

4 years ago

This is a problem that I have to solve with matrixes, Please show me all of the work. I will award brainliest.

Elina [12.6K]

Answer:

The total points by Washington High School = 162 points

The total points by Johnson High School = 159 points

The total points by Roosevelt High School = 108 points

The total points by Lewis High School = 196 points

Washington High School has WON the meet.

Johnson High School came second.

The difference in the points= 2 points

Step-by-step explanation:

Points awarded for first place = 10 points

Points awarded for second place = 8 points

Points awarded for third place = 7 points

About Washington High School

10 first places = 10 x ( 10) = 100 points

6 second places = 6 x ( 8) = 48 points

2 first places = 2 x ( 7) = 14 points

So, the total points by Washington High School = Sum of all points

= 100 + 48 + 14 = 162 points .............. (1)

About Johnson High School

8 first places = 8 x ( 10) = 80 points

9 second places = 9 x ( 8) = 72 points

1 first places = 1 x ( 7) = 7 points

So, the total points by Johnson High School = Sum of all points

= 80 + 72 + 7 = 159 points .............. (2)

About Roosevelt High School

4 first places = 4 x ( 10) = 40 points

5 second places = 5 x ( 8) = 40 points

4 first places = 4 x ( 7) = 28 points

So, the total points by Roosevelt High School = Sum of all points

= 40 + 40 + 28 = 108 points .............. (3)

About Lewis High School

3 first places = 3 x ( 10) = 30 points

3 second places = 3 x ( 8) = 24 points

6 first places = 6 x ( 7) = 42 points

So, the total points by Lewis High School = Sum of all points

= 30 + 42 + 24 = 96 points .............. (4)

Now, comparing all four equation, we get that:

Washington High School has WON the meet.

Johnson High School came second.

The difference in the points = 162 - 159 = 2 points

6 0

3 years ago

Which expression can be expanded using the binomial theorem?

kirill [66]

Answer:

The third choice: (x² - 1)³

Step-by-step explanation:

That is the only expression that is a binomial. Binomial means "two numbers". For this expression, x² is one number, and -1 is the other. Choice 1 is a monomial (one number) and choices 2 and 4 are trinomials (three numbers)

7 0

3 years ago