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3 years ago

Select the best answer for the question.

Mathematics

2 answers:

Natalka [10]3 years ago

8 0

A is the right answer to this question

Mamont248 [21]3 years ago

3 0

<h3>Answer: Linear</h3>

Explanation: The population increases by the same amount each time, so the rate is constant, meaning we have linear growth. The slope of this line is 30.

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Make a frequency distribution and find the relative frequencies for the following number set. Round the relative frequency to th

Mandarinka [93]

Number......frequency.....relative frequency
10 2 2/24 = 8.3%
20 2 2/24 = 8.3%
30 4 4/24 = 16.7%
40 2 2/24 = 8.3%
50 3 3/24 = 12.5%
60 5 5/24 = 20.8%
70 3 3/24 = 12.5%
80 1 1/24 = 4.2%
90 2 2/24 = 8.3%
total : 24

7 0

3 years ago

*Brainliest*<br> I need help with 4 problems like this...KHAN ACADEMY.

Sedaia [141]

9514 1404 393

Answer:

(x +1)² = 4
(D) x = -1 ±2

Step-by-step explanation:

Start by getting the x-terms on one side of the equal sign. We can do that by subtracting x from both sides.

3 = x² +2x

Now, add the square of half the x-coefficient.

3 +1 = x² +2x +1

(x +1)² = 4 . . . . . . . . write as a square in the desired form

To solve, take the square root, and add the opposite of the left-side constant.

x +1 = ±2

x = -1 ±2 . . . . . . matches choice (D)

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3 years ago

BRAINLIEST!! What’s 36.3 square root ?

Marat540 [252]

Answer:

6.02

Step-by-step explanation:

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3 years ago

Did I get this right? If I did t could someone explain it to me

Kisachek [45]

Answer:

Yes, you are correct.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Please help?

Mazyrski [523]

Answer:

$23\sqrt{3}\ un^2$

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

$A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\$

2. Note that

$A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2$

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

$A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2$

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

7 0

3 years ago