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3 years ago

9

A veterinarian knows that a 50-pound dog gets 0.5 milligrams of certain medicine and that the number of milligrams, m, varies di

rectly with the weight of the dog, w. The vet uses these steps to find the amount of medicine to give a 10-pound dog.
In which step did the veterinarian make the first error?

2 answers:

ioda3 years ago

6 0

The veterinary should give the dog 0.1 m if medicine

erastovalidia [21]3 years ago

4 0

Answer:

step 3

Step-by-step explanation:

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23.49 is 27% of what number?

Digiron [165]

Answer: 100

Step-by-step explanation: 100 divided by 27= 0.27, 23.49x0.27= 6.34.

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Yoda soda is the intergalacatic party drink .You are throwing a party and you need 5 liters of Yoda soda for 12 guest. If you ha

aksik [14]

Answer:

15 liters

Step-by-step explanation:

Given:

We need 5 liters of Yoda soda for 12 guest.

To find:

If we have 36 guest how many liters of Yoda soda we need.

Solution:

<u>By unitary method:</u>

For 12 guests, we need liters of Yoda soda = 5

For 1 guest, we need liters of Yoda soda = $\frac{5}{12}$

For 36 guests, we need liters of Yoda soda = $\frac{5}{12}\times36=\frac{180}{12} =15\ liters$

Therefore, we need 15 liters of Yoda soda for 36 guests.

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3 years ago

A metalworker has a metal alloy that is 15% copper and another alloy that is 75% copper. How many kilograms of each alloy shou

lisov135 [29]

48 kilograms of 15 % copper is combined with 72 kilograms of 75 % copper to create 120 kg of a 51% copper alloy

<em><u>Solution:</u></em>

Let "x" be the kilograms of 15 % copper

Then, (120 - x) be the kilograms of 75 % copper

Then, according to question,

"x" kilograms of 15 % copper is combined with (120 - x) kilograms of 75 % copper to create 120 kg of a 51% copper alloy

Thus we frame a equation as:

15 % of x + 75 % of (120 - x) = 51 % of 120

<em><u>Solve the equation for "x"</u></em>

$15 \% \times x + 75 \% \times (120-x) = 51 \% \times 120\\\\\frac{15}{100} \times x + \frac{75}{100} \times (120-x) = \frac{51}{100} \times 120\\\\0.15x + 0.75(120-x) = 0.51 \times 120\\\\0.15x + 90 - 0.75x = 61.2\\\\0.6x = 90 - 61.2\\\\0.6x = 28.8\\\\Divide\ both\ sides\ by\ 0.6\\\\x = 48$

Thus 48 kilograms of 15 % copper used

Then, (120 - x) = (120 - 48) = 72

Thus, 72 kilograms of 75 % copper is used

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3 years ago

Please help! I'll mark as brainliest

Rudik [331]

Answer:

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3 years ago

Prove A-(BnC) = (A-B)U(A-C), explain with an example

NikAS [45]

Answer:

Prove set equality by showing that for any element $x$ , $x \in (A \backslash (B \cap C))$ if and only if $x \in ((A \backslash B) \cup (A \backslash C))$ .

Example:

$A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace$ .

$B = \lbrace0,\, 1 \rbrace$ .

$C = \lbrace0,\, 2 \rbrace$ .

$\begin{aligned} & A \backslash (B \cap C) \\ =\; & \lbrace 0,\, 1,\, 2,\, 3 \rbrace \backslash \lbrace 0 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace \end{aligned}$ .

$\begin{aligned}& (A \backslash B) \cup (A \backslash C) \\ =\; & \lbrace 2,\, 3\rbrace \cup \lbrace 1,\, 3 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace\end{aligned}$ .

Step-by-step explanation:

Proof for $[x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))]$ for any element $x$ :

Assume that $x \in (A \backslash (B \cap C))$ . Thus, $x \in A$ and $x \not \in (B \cap C)$ .

Since $x \not \in (B \cap C)$ , either $x \not \in B$ or $x \not \in C$ (or both.)

If $x \not \in B$ , then combined with $x \in A$ , $x \in (A \backslash B)$ .
Similarly, if $x \not \in C$ , then combined with $x \in A$ , $x \in (A \backslash C)$ .

Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ as required.

Proof for $[x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]$ :

Assume that $x \in ((A \backslash B) \cup (A \backslash C))$ . Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

If $x \in (A \backslash B)$ , then $x \in A$ and $x \not \in B$ . Notice that $(x \not \in B) \implies (x \not \in (B \cap C))$ since the contrapositive of that statement, $(x \in (B \cap C)) \implies (x \in B)$ , is true. Therefore, $x \not \in (B \cap C)$ and thus $x \in A \backslash (B \cap C)$ .
Otherwise, if $x \in A \backslash C$ , then $x \in A$ and $x \not \in C$ . Similarly, $x \not \in C \!$ implies $x \not \in (B \cap C)$ . Therefore, $x \in A \backslash (B \cap C)$ .

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

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ implies $x \in A \backslash (B \cap C)$ , as required.

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3 years ago