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kobusy [5.1K]
2 years ago
6

I NEED HELP RIGHT AWAY!!!! I'LL GIVE YOU 30 POINTS AND TRY TO MARK YOU AS BRAINLIST.Q: What do you need to do to the equations s

o you can put them into the Graph tool? Put the equations into the Graph tool. To create the graph, select the correct relationship and then enter the values for the variables. Paste a screenshot of your graph in the space provided. Do you get the same solution as when you solved it algebraically?
Mathematics
1 answer:
MAVERICK [17]2 years ago
5 0

The variables are the alphabets which can have any value depending on the relationship with the other numbers and variables in the equations.

To create the graph, select the correct relationship and then enter the values for the variables.

The graph is the representation of the equation in the form of lines or shapes.

The correct method to create a graph is to find the relationship of the variables and put their values to get the exact points.

The points are joined together to draw the required line or shape formed.

The best option of the given question is:

To create the graph, select the correct relationship and then enter the values for the variables.

For example in the given graph we see that the rate is increasing as the mass is decreasing.

The mass is plotted along x-axis and the <em>rate</em> along the <u><em>y-axis.</em></u>

Their relationship that they are <em><u>inversely proportional</u></em> i.e.  one increases as the other decreases can be seen in the given graph.

brainly.com/question/13282648

brainly.com/question/14961829

brainly.com/question/17174762

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1. Check the drawing of the rhombus ABCD in the picture attached.

2. m(CDA)=60°, and AC and BD be the diagonals and let their intersection point be O.

3. The diagonals:
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     ii) are perpendicular to each other, so m(DOC)=90°

4. In a right triangle, the length of the side opposite to the 30° angle is half of the hypothenuse, so OC=3 in.

5. By the pythagorean theorem, DO= \sqrt{ DC^{2}- OC^{2} }=  \sqrt{ 6^{2}- 3^{2} }=\sqrt{ 36- 9 }=\sqrt{ 27 }= \sqrt{9*3}= 
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6. The 4 triangles formed by the diagonal are congruent, so the area of the rhombus ABCD = 4 Area (triangle DOC)=4*\frac{DO*OC}{2}=4 \frac{3 \sqrt{3} *3}{2}==2*9 \sqrt{3}=18 \sqrt{3} (in^{2})

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