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2 years ago

Explain how to estimate a percent equivalent of the ratio 344/513

Mathematics

2 answers:

N76 [4]2 years ago

5 0

Just simplify and from there you will get it

Temka [501]2 years ago

3 0

Simplify, then see what its close to.
7/8 is close to 0 or 1, 1!

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3s+2p=85.50 4s+2p=123

Gala2k [10]

Answer:s=10.50 and p=27

Step-by-step explanation:

3 0

3 years ago

emily drives home from work in 30 minutes driving 60 miles per hour. If Emily need to get home 10 minutes earlier, how fast does

AlladinOne [14]

The correct answer is 90 miles per hour

Explanation:

The first step to know how fast Emily needs to drive to get 10 minutes earlier is to determine the distance from her work to her home. This can be calculated by using the information provided (speed and time). The process is shown below:

speed = distance ÷ time

distance = speed x time

distance = 60 miles per hour x 0.5 (30 minutes represent 0.5 hours)

distance= 30 miles

Now, using the same formula let's calculate the speed for 20 minutes (30 minutes - 10 minutes earlier = 20 minutes)

speed = distance ÷ time

speed = 30 miles ÷ 0.333 (20 minutes represents 0.333 hours as 20 minutes is 1/3 of an hour)

speed= 90 miles per hour

8 0

3 years ago

A lighthouse operator sights a sailboat at an angle of depression of 12°. If the sailboat is 80 m away, how tall is the lighthou

Brilliant_brown [7]

Answer:

‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎ ‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎ ‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎ ‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎ ‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎ ‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎ ‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎ ‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎ ‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎ ‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎ ‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎ ‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎ ‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎ ‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎ ‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎ ‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎ ‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎ ‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎ ‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎ ‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎‎ ‏‏‎

5 0

2 years ago

Your friend has w blueberries. You have 13 blueberries. You and your friend

murzikaleks [220]

Answer:

x = 27. The friend has 27 bluberries

Step-by-step explanation:

x + 13 = 40

-13 -13

x = 27

Hope this helped! Have a nice rest of ur day! Plz mark as brainliest!!! :D

-Lil G

4 0

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.

8 0

2 years ago