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4 years ago

13

Tirzah wants to put a fence around her garden. She has 22 yards Of fence material. Does she have enough to go all the way around

the garden? 4 8/12 6 9/12

1 answer:

CaHeK987 [17]4 years ago

3 0

No. She needs at least 31 yards of material.

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A machine can produce a maximum of 600 parts per day. Yesterday, it produced 75 % of its maximum capacity. How many parts did th

gizmo_the_mogwai [7]

Answer:

450

Step-by-step explanation:

here is a fast way:

1- 75% = 75/100 = 3/4

2- 600 × (3/4) = 450

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3 years ago

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Mark wants to order a pizza. Which is the better deal? Explain.

tangare [24]

10/12= $.83 per inch
20/18= $1.11 per inch

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3 years ago

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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

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

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4 years ago

Translate the following statement to an inequality. Then, find the solution.

bogdanovich [222]

Answer:

3(n+5) ≥ -6

n ≥ -7

Step-by-step explanation:

3(n+5) ≥ -6

3n+15 ≥ -6

3n ≥ -21

n ≥ -7

4 0

3 years ago

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A 16\dfrac1216 2 1 ? 16, start fraction, 1, divided by, 2, end fraction kilometer stretch of road needs repairs. Workers can rep

alexdok [17]

Answer: It will take $7\dfrac{1}{3}$ weeks to repair this stretch of road.

Step-by-step explanation:

Since we have given that

Part of road needs repair = $16\dfrac{1}{2}=\dfrac{33}{2}$

Part of road worker can repair per week = $2\dfrac{1}{4}=\dfrac{9}{4}$

We need to find the number of weeks that it will take to repair this stretch of road.

So, Number of weeks is given by

$\dfrac{33}{2}\div \dfrac{9}{4}\\\\=\dfrac{33}{2}\times \dfrac{4}{9}\\\\=\dfrac{66}{9}\\\\=\dfrac{22}{3}\\\\=7\dfrac{1}{3}$

Hence, it will take $7\dfrac{1}{3}$ weeks to repair this stretch of road.

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3 years ago

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