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4 years ago

I need to find out x y and z and am very lost and confused please help!

Mathematics

2 answers:

Law Incorporation [45]4 years ago

4 0

I dont know the answer becouse im not there but i can show you how to get the answer..

you know the rulers the are half way cut?use that and measure each angle

(i hope i helped!)

zhenek [66]4 years ago

4 0

X^2 = 6^2 + 17^2
x^2 = 36 + 289
x^2 = 325
x = 18.0277563773 

1/2 Vertex angle = arc tan (6/17)
1/2 Vertex angle = arc tan (0.3529411765 )
1/2 Vertex angle = 19.44 Degrees

So, the other "half of the vertex angle = 90 - 19.44 degrees = 70.56

cosine (70.56) = 17 / z
z = 17 / cosine (70.56)
z = 17 / 0.33282
z = 51.0786611382 

half of y^2 = 51.0786611382^2 - 17^2
half of y^2 = 2609.0296236711 -289
half of y^2 = 2,320.03 
half of y = 48.1666858282 
y = 48.1666858282 + 6
y = 48.1666858282

wow - lots of work!!

You might be interested in

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

4. A local DVD rental machine rents

Mrrafil [7]

It takes 16 movie rentals for both options to be the same price

Solution:

Given that, A local DVD rental machine rents movies out to customers for $5.75 per movie.

A digital movie company allows you to rent movies for $3.25 but requires a membership fee of $40.

We have to find how many movie rentals will it take for both options to be the same price?

Now, let the number of movies rented be "n"

Then, according to the given information,

$\begin{array}{l}{5.75 \times n=3.25 \times n+40} \\\\ {5.75 n=3.25 n+40} \\\\ {5.75 n-3.25 n=40} \\\\ {2.50 n=40} \\\\ {n=\frac{40}{2.5}=16}\end{array}$