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3 years ago

HELPPPPPPPPPPPPPPPPPPPPPP

Mathematics

1 answer:

RideAnS [48]3 years ago

4 0

Answer:omg I gotchu I just learned these for the letters for example would be q/1 times 2/7 because if you have one letter or number you ALWAYS put a one under it then u multiply another easy trick don’t butterfly multiply go straight across and example 8/1 (all ways add the one) times 3/4. Ok how to multiply that always go across so it would be 8x3 and 1x4 your answer should be 24/4 and simplify if needed

Step-by-step explanation:

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PLEASSSEEEE HELPPPPPP !!!!!

Angelina_Jolie [31]

Marking the questions as 12345.
1 is G, 2 is E, 3 is D, 4 is C, and 5 is A. But I might be misunderstanding this too.

5 0

3 years ago

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What is the equation of a line in slope intercept form that is perpendicular to y=2/3x+2 and passes through the points (-2,2)? H

JulijaS [17]

Answer:

$y = -\frac{3}{2}x -1$

Step-by-step explanation:

Given

Perpendicular to $y = \frac{2}{3}x + 2$

Pass through $(-2,2)$

Required

Determine the line equation

First, we need to determine the slope of $y = \frac{2}{3}x + 2$

An equation is of the form:

$y = mx + b$

Where

$m = slope$

In this case:

$m = \frac{2}{3}$

Next, we determine the slope of the second line.

Since both lines are perpendicular, the second line has a slope of:

$m_1 = \frac{-1}{m}$

$m_1 = \frac{-1}{2/3}$

$m_1 = -\frac{3}{2}$

Since this line passes through (-2,2); The equation is calculated as thus:

$y - y_1 = m_1(x - x_1)$

Where

$(x_1,y_1) = (-2,2)$

This gives:

$y - 2 = -\frac{3}{2}(x - (-2))$

$y - 2 = -\frac{3}{2}(x +2)$

$y - 2 = -\frac{3}{2}x -3$

Add 2 to both sides

$y - 2+2 = -\frac{3}{2}x -3 + 2$

$y = -\frac{3}{2}x -1$

8 0

3 years ago

Is h(x)=x+2 linear or nonlinear?

ANTONII [103]

the answer to your question is linear I think

4 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

4 years ago

Use the distributive property to simplify the following expression 4x(x-6)

katrin [286]

We need to distribute the term 4x to x - 6. Basically, we need to multiply 4x by x, which gives us 4x², and 4x by -6, which gives us -24x. Putting our two derived terms together, we get 4x²- 24, which is your answer. Hope this helps and have a great day!

3 0

4 years ago

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