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4 years ago

Is it possible for two graphs that are parallel to share the same y-intercept

Mathematics

2 answers:

leonid [27]4 years ago

8 0

No they cant not share the same y intercept.

xxMikexx [17]4 years ago

7 0

No
lines that are paralell have the same slope
y=mx+b
m=slope
for them to be parlell, they cannot be the same line
that means that they have to have a different b
b is y intercpet, it tells how far up or down the line is
if they hit at the same value of y, then they would be the same line

no

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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 '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

Kelsie sold digital cameras on her website. She bought the cameras for 465 each and included a 60% markup to get the selling pri

melamori03 [73]

Each camera costs $651 dollars, assuming I understood the question format correctly.
Find 60% of 465, which is 186. Because it is markup, add 186 to the original price. 465+186 = 651.
Therefore, the price is $651.

5 0

4 years ago

What is the probability of tossing a standard coin and having it land on head three times in a row?

sergejj [24]

Answer:

the probability of a coin landing on heads 3 times in a row is 1/8 or 0.125 or 12.5%

Step-by-step explanation:

the chance of landing heads once is 1/2, for it to be 2 times in a row its 1/4, and 3 times in a row its 1/8, basically every time you want to do it in a row you multiply 1/2, so for example for this question you multiply 1/2 by however times you want the coin to land on heads in a row, (1/2) X(1/2)X(1/2)

4 0

3 years ago

X X 7

harina [27]

Answer:

x=7

Step-by-step explanation:

x/3 + x/6 = 7/2

Multiply each side by 6 to clear the fractions

6(x/3 + x/6) = 7/2 *6

Distribute

2x +x = 21

Combine like terms

3x = 21

Divide by 3

3x/3 = 21/3

x = 7

4 0

3 years ago

Read 2 more answers

I NEED HELP FAST PLEASE!!!

hammer [34]

So, we need to figure out how much times the body size of the flea it can jump. First, we need to figure out the size of the flea: 2^-4=0.0625

Now that we have the size of the flea, we need to figure out how high it can jump: 2^3=8

After we figure out the size of the flea, we need to divide 8 by .0625 because it is inverse operations to multiplication: 8/.0625=128

Answer: The flea can jump 128 times higher than it's body size.

3 0

4 years ago