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2 years ago

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Mathematics

1 answer:

tatuchka [14]2 years ago

6 0

The additional set of values which could be included in the table is; (10,2).

<h3>Which additional set of values can be included in the table?</h3>

As evident in the task content, it follows that for a 15 unit increase in x, depth, there is a 3 unit increase in y, pressure.

On this note, it follows that with each 5 unit increase in x corresponds to 1 unit increase in y.

And ultimately, at x = 10, y = 2 × 1 = 2.

Read more on proportional relationship;

brainly.com/question/2143065

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Find an equation of the curve that satisfies dy/dx = 102yx^16 and whose y-intercept is 6. y(x) =

otez555 [7]

Answer:

The solution of the given differential equation is:

$y(x)=6e^{6x^{17}}$

Step-by-step explanation:

The given differential equation is in variable separable form and is solved as shown under

$\frac{dy}{dx}=102\times x^{16}\times y\\\\\frac{dy}{y}=102x^{16}dx$

Integrating bot sides we get

$\int \frac{dy}{y}=\int 102x^{16}dx\\ln(y)=102\times \frac{x^{17}}{17}+c\\\\ln(y)=6x^{17}+c\\\\y=e^{6x^{17}+c}\\\\y=e^{c}\cdot e^{6x^{17}}\\\\\therefore y=ke^{6x^{17}}$

where 'k' is a constant whose value can be found by the given condition that

y(0) = 6

$y(0)=ke^{6\times 0^{17}}\\\\\therefore k=6$

The final solution is

$y(x)=6e^{6x^{17}}$

8 0

3 years ago

Write an expression for the calculation the product of 4 and 4 added to double the number 5

Gnesinka [82]

4*(4+(2*5))
4*(4+10)
4*14
56 is the answer.

4 0

3 years ago

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Please help me please i really need help please

givi [52]

It is the last choice. 4 teams with a total of 24 players

8 0

3 years ago

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What breaks yet never falls, and what falls yet never breaks?

guajiro [1.7K]

Answer:

day and night

Step-by-step explanation:

that is the answer

5 0

3 years ago

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Which of the following statements is true for all sets A,B and C? Give a proof

ELEN [110]

Answer:

(a) and (b) are not true in general. Refer to the explanations below for counterexamples.

It can be shown that (c) is indeed true.

Step-by-step explanation:

This explanation will use a lot of empty sets $\phi$ just to keep the counterexamples simple.

Note that $A \cap B$ can well be smaller than $A$ . It should be alarming that the question is claiming $A\!$ to be a subset of something that can be smaller than $\! A$ . Here's a counterexample that dramatize this observation:

Consider:

$A = \left\lbrace 1 \right\rbrace$ .
$B = \phi$ (an empty set, same as $\left\lbrace \right\rbrace$ .)
$C = \phi$ (another empty set.)

The intersection of an empty set with another set should still be an empty set:

$A \cap B = \left\lbrace 1\right\rbrace \cap \left\lbrace\right\rbrace = \left\lbrace\right\rbrace$ .

The union of two empty sets should also be an empty set:

$((A \cap B) \cup C) = \left\lbrace\right\rbrace \cup \left\lbrace\right\rbrace = \left\lbrace\right\rbrace$ .

Apparently, the one-element set $A = \left\lbrace 1 \right\rbrace$ isn't a subset of an empty set. $A \not \subseteq ((A\cap B) \cup C)$ . Contradiction.

Consider the same counterexample

$A = \left\lbrace 1 \right\rbrace$ .
$B = \phi$ (an empty set, same as $\left\lbrace \right\rbrace$ .)
$C = \left\lbrace 2 \right\rbrace$ (another empty set.)

Left-hand side:

$(A \cup B) \cap C = \left(\left\lbrace 1 \right\rbrace \cup \left\lbrace \right\rbrace\right) \cap \left\lbrace 2 \right\rbrace\right = \left\lbrace 1 \right\rbrace \cap \left\lbrace 2 \right\rbrace = \left\lbrace \right\rbrace$ .

Right-hand side:

$(A \cap B) \cup C = \left(\left\lbrace 1 \right\rbrace \cap \left\lbrace \right\rbrace\right) \cup \left\lbrace 2 \right\rbrace\right = \left\lbrace \right\rbrace \cup \left\lbrace 2 \right\rbrace = \left\lbrace 2 \right\rbrace$ .

Apparently, the empty set on the left-hand side $\left\lbrace \right\rbrace$ is not the same as the $\left\lbrace 2 \right\rbrace$ on the right-hand side. Contradiction.

Part one: show that left-hand side is a subset of the right-hand side.

Let $x$ be a member of the set on the left-hand side.

$x \in (A \backslash B) \cap C$ .

$\implies x\in A \backslash B$ and $x \in C$ (the right arrow here reads "implies".)

$\implies x \in A$ and $x \not\in B$ and $x \in C$ .

$\implies (x \in A\cap C)$ and $x \not\in B \cap C$ .

$\implies x \in (A \cap C) \backslash (B \cap C)$ .

Note that $x \in (A \backslash B) \cap C$ (set on the left-hand side) implies that $x \in (A \cap C) \backslash (B \cap C)$ (set on the right-hand side.)

Therefore:

$(A \backslash B) \cap C \subseteq (A \cap C) \backslash (B \cap C)$ .

Part two: show that the right-hand side is a subset of the left-hand side. This part is slightly more involved than the first part.

Let $x$ be a member of the set on the right-hand side.

$x \in (A \cap C) \backslash (B \cap C)$ .

$\implies x \in A \cap C$ and $x \not\in B \cap C$ .

Note that $x \not\in B \cap C$ is equivalent to:

$x \not \in B$ , OR
$x \not\in C$ , OR
both $x \not\in B$ AND $x \not \in C$ .

However, $x \in A \cap C$ implies that $x \in A$ AND $x \in C$ .

The fact that $x \in C$ means that the only possibility that $x \not\in B \cap C$ is $x \not \in B$ .

To reiterate: if $x \not \in C$ , then the assumption that $x \in A \cap C$ would not be true any more. Therefore, the only possibility is that $x \not \in B$ .

Therefore, $x \in (A \backslash B)\cap C$ .

In other words, $x \in (A \cap C) \backslash (B \cap C) \implies x \in (A \backslash B)\cap C$ .

$(A \cap C) \backslash (B \cap C) \subseteq (A \backslash B)\cap C$ .

Combine these two parts to obtain: $(A \backslash B) \cap C = (A \cap C) \backslash (B \cap C)$ .

8 0

4 years ago