PLZ HELP AS SOON AS POSSIBLE!!!!!!!!! WILL GIVE LIKE, 5 STARS, & BRAINLIEST; THE WHOLE SHEBANG;

Mathematics

2 answers:

serg [7]2 years ago

6 0

Answer:

a. Meg needs to find the part

b. Steve needs to find the percent.

Step-by-step explanation:

A. Does Meg need to find the part, the whole, or the percent.

Meg already knows the whole and the percent.

The whole is the entire number of dogs seen: 16 dogs.

The percent is 50%.

Hence, she needs to find the part: this is how many (what part) of the 16 dogs she saw were boxers.

And she can do with a simple operation:

Part = whole × percent = 16 × 50% = 8. This is 8 dogs of 16 were boxers.

B. Does Steve need to find the part, the whole, or the percent.

He already knows the part (7 of the 35 dogs were boxers) and the whole (35 dogs).

Then he needs to find the percent, which he can do using the next operation:

Percent = (part / whole) × 100 = (7/35)×100 = 20%

dimulka [17.4K]2 years ago

4 0

Step-by-step explanation:

Part 1:

Meg needs to find the whole as 50% of the dogs = 5 therefore 100% equals 10 dogs.

Part 2:

Steve saw 50% out of two dogs meaning that he only saw 1 dog that is a boxer.

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Triangle ABC has vertices ofA(–6, 7), B(4, –1), and C(–2, –9).Find the length of the median from

Svetradugi [14.3K]

We know that in geometry, a median of a triangle is a line segment joining a vertex to the midpoint of the opposing side. So, in a triangle there are three medians. We will find them.

As shown in the figure below, we have the medians:

P1C
P2A
P3B

We need to find P1, P2 and P3. The midpoint of the segment (x1, y1) to (x2, y2) is:

$(\frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2})$

Therefore:

For the segment AB:

$P_{1} = ( \frac{4-6}{2}, \frac{7-1}{2})$
$P_{1} = (-1,3)$

For the segment BC:
$P_{2} = ( \frac{4-2}{2}, \frac{-1-9}{2})$
$P_{2} = (1,-5)$

For the segment CA:
$P_{3} = ( \frac{-6-2}{2}, \frac{7-9}{2})$
$P_{3} = (-4,-1)$

We know that the distance d between two points P1(x1,y1) and P2(x2,y2) is given by the formula:

$d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}- y_{1})^{2} }$

Then of each median is:

Median P1C:

$d_{1} = \sqrt{(-9-3)^{2}+(-2-(-1))^{2}} = \sqrt{145}$

Median P2A:

$d_{2} = \sqrt{(7-(-5))^{2}+(-6-1)^{2}} = \sqrt{193}$

Median P3B:

$d_{3} = \sqrt{(-1-(-1))^{2}+(4-(-4))^{2} } = 8$