Using proportions, it is found that it takes 886 more mini-bears than regular-bears to have the same weight as one super-bear.
<h3>What is a proportion?</h3>
A proportion is a fraction of a total amount, and the measures are related using a rule of three. Due to this, relations between variables, either direct(when both increase or both decrease) or inverse proportional(when one increases and the other decreases, or vice versa), can be built to find the desired measures in the problem, or equations to find these measures.
10 mini-bears weights to 12.1 grams, hence the weight of a mini-bear is of:
12.1/10 = 1.21 grams.
10 regular bears weights to 23.1 grams, hence the weight of a regular bear is of:
23.1/10 = 2.31 grams.
1 super bear weights to 2250 grams, hence the proportion between the <u>weight of a super bear and the weight of a mini-bear</u> is:
2250/1.21 = 1860.
The proportion between the <u>weight of a super bear and the weight of a regular bear</u> is:
2250/2.31 = 974.
The difference of proportions is given by:
1860 - 974 = 886.
It takes 886 more mini-bears than regular-bears to have the same weight as one super-bear.
More can be learned about proportions at brainly.com/question/24372153
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Answer:
-3x+6
Step-by-step explanation:
-3x-6(-1)
First, multiplying two negatives equals a positive: (-)×(-)=(+). Then,
multiply the numbers -3×+6×1 = -3x+6. Then, you got the answer and the answer is -3x+6
Answer:
![\left[\begin{array}{cc}3&9\\5&-2\end{array}\right] +\left[\begin{array}{cc}6&0\\-8&4\end{array}\right]=\left[\begin{array}{cc}9&9\\-3&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%269%5C%5C5%26-2%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%260%5C%5C-8%264%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D9%269%5C%5C-3%262%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
To add matrices, we add the corresponding components.
The given matrices is
![\left[\begin{array}{cc}3&9\\5&-2\end{array}\right] +\left[\begin{array}{cc}6&0\\-8&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%269%5C%5C5%26-2%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%260%5C%5C-8%264%5Cend%7Barray%7D%5Cright%5D)
We add the corresponding components to get;
![\left[\begin{array}{cc}3&9\\5&-2\end{array}\right] +\left[\begin{array}{cc}6&0\\-8&4\end{array}\right]=\left[\begin{array}{cc}3+6&9+0\\5+-8&-2+4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%269%5C%5C5%26-2%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%260%5C%5C-8%264%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%2B6%269%2B0%5C%5C5%2B-8%26-2%2B4%5Cend%7Barray%7D%5Cright%5D)
We simplify to get:
Answer:
Check it below
Step-by-step explanation:
Hello!
By definition, the intercept of the 3 bisectors, within a triangle is the Incenter. So let's get started, by tracing three bisectors.
In the triangle PQR::
- Use a compass, open its legs with a little more than half of
. Place the needle point in the vertex of P. Mark! - Do the same, placing the needle point in Q, following
direction. Mark - Repeat it with the other triangle's line segment, namely:
4.Trace with a ruler, from each intersection point a line. You'll trace three lines then.
5. Place the compass' needle point in this intersection point, (inside the triangle) with a hinge up to the triangle side, draw the circle inscribed.
A is the correct answer. When put into calculator, the decimals ends.
