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4 years ago

Can someone help me please!!

Mathematics

2 answers:

podryga [215]4 years ago

8 0

I believe it is 3^2

CaHeK987 [17]4 years ago

3 0

Omg so easy !

i believe the answer would be 3^1
have a nice day :)

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Ms. Augustine is playing a game with her class. When a student tells her a number, she uses a rule to determine the new number.

Degger [83]

Answer:

When a student tells her a number, she uses a rule to determ

Step-by-step explanation:

Get the answers ... When a student tells her a number, she uses a rule to determine the new number. Ms. Augustine always uses the same rule. The table below shows some of the results. What would the number be when the student says 8? A. 58. B. 56. C. 54

7 0

3 years ago

What is greater and what is least 6.00 0.56 0.060 0.60 0.056 0.557781 0.059999

tangare [24]

Answer:

6.00, 0.56,0.557781,0.059999, 0.056

Step-by-step explanation:

Look at the place values

4 0

3 years ago

Jordan is cutting a 2 meter by 1 1/4 meter into two pieces by its diagonal line

Airida [17]

Question:

Jordan is cutting a 2 meter by $1\frac{1}{4}$ meter piece of rectangular paper into two pieces along its diagonal. what is the area of each of the pieces?

Answer:

Area of each piece of paper is $1 \frac{1}{4} \ m^2$ .

Solution:

Length of the paper = 2 m

Width of the paper = $1\frac{1}{4}$ m = $\frac{5}{4}$ m

<em>Area of the rectangle = length × width</em>

$$=2\times \frac{5}{4}$

$$= \frac{5}{2} \ m^2$

<em>Diagonal of a rectangle divides it into two equal parts.</em>

Area of each piece = Area of the rectangle ÷ 2

$$=\frac{ \frac{5}{2}}{2}$

$$=\frac{5}{4} \ m^2$

$$=1 \frac{1}{4} \ m^2$

Area of each piece of paper is $1 \frac{1}{4} \ m^2$ .

7 0

3 years ago

Please help quickly !!

anygoal [31]

Answer:

Step-by-step explanation:

4 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