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inn [45]
3 years ago
9

Find the slope using the following two points: (2,-7) and (-1,6)

Mathematics
1 answer:
Advocard [28]3 years ago
7 0

Answer:

13/-3

Step-by-step explanation:

Y2-Y1/X2-X1

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We can describe 15×-10 as an expression. we would describe 6×-2< 35 as an...
algol [13]

Answer: Inequality

Step-by-step explanation:

From the question, if we can describe 15×-10 as an expression, then we would describe 6×-2< 35 as an inequality. An

inequality is used to compare two values, and shows if one is less than, or greater than, or maybe not equal to the other value.

For example, a ≠ b means that a is not equal to b and a < b means a is less than b while a > b means a is greater than b. From the question, 6×-2< 35 means that 6x - 2 is less than 35.

8 0
3 years ago
are any two regular polygons similar. The area of the smaller figure is about 390 cm^2. choose the best estimate of the area of
galben [10]

Answer:

The best estimate of the area of the larger figure is 693\ cm^{2}

Step-by-step explanation:

step 1

<em>Find the scale factor</em>

we know that

If two figures are similar, then the ratio of its corresponding sides is equal to the scale factor

Let

z----> the scale factor

x-----> the corresponding side of the larger figure

y-----> the corresponding side of the smaller figure

so

z=\frac{x}{y}

we have

x=12\ cm

y=9\ cm

substitute

z=\frac{12}{9}=\frac{4}{3} -----> the scale factor

step 2

<em>Find the area of the larger figure</em>

we know that

If two figures are similar, then the ratio of its  areas is equal to the scale factor squared

Let

z----> the scale factor

x-----> the area of the larger figure

y-----> the area of the smaller figure

so

z^{2}=\frac{x}{y}

we have

z=\frac{4}{3}

y=390\ cm^{2}

substitute and solve for x

(\frac{4}{3})^{2}=\frac{x}{390}

(\frac{16}{9})=\frac{x}{390}

x=390*16/9=693.33\ cm^{2}

4 0
2 years ago
Is -81.43 rational or irrational? If it is rational, write the number as a ratio of two integers.
ICE Princess25 [194]

This is rational. It -81.43:-94.52 expressed. -81.45/94.52

6 0
3 years ago
Can someone help me with this? PLs i'm so confused!
barxatty [35]
1. E. sine\ A = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{5}{13}

2. L. cos\ A = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{12}{13}

3. tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{5}{12}

4. Y. sin\ B = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{5}{13}

5. W. cos\ B = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{12}{13}

6. tan\ B = \frac{b}{a} = \frac{adjacent}{opposite} = \frac{AC}{BC} = \frac{12}{5} = 2\frac{2}{5}

7. sin\ A = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{1}{2}

8. W. cos\ A = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{\sqrt{3}}{2}

9. I. tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} * \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}

10. sin\ B = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{1}{2}

11. E. cos\ B = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{\sqrt{3}}{1} = \sqrt{3}

12. I. tan\ B = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} * \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}

13. U. sin\ A = \frac{a}{c} = \frac{hypotenuse}{opposite} = \frac{BC}{AB} = \frac{12}{15} = \frac{4}{5}

14. I. cos\A = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{9}{15} = \frac{3}{5}

15. tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{12}{9} = \frac{4}{3} = 1\frac{1}{3}

16. R. sin\ B = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{4}{\sqrt{65}} = \frac{4}{\sqrt{65}} * \frac{\sqrt{65}}{\sqrt{65}} = \frac{4\sqrt{65}}{65}

17. M. cos\ B = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{7}{4} = 1\frac{3}{4}

18. N. tan\ B = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{4}{7}

19. L. sin\ A = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{16}{34} = \frac{8}{17}

20. H. cos\ B = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \fac{AC}{AB} = \frac{30}{34} = \frac{15}{17}

21. tan\ B = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{16}{30} = \frac{8}{15}

22. O. sin\ A = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}

23. O. cos\ A = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}

24. N. tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{1}{1} = 1
7 0
3 years ago
In a sample of 30 coin flips, heads came up 6 times. What is the sample proportion for heads?
Kitty [74]
You would divide 6 by 30.
6/30=0.2
Therefore the answer is D
6 0
3 years ago
Read 2 more answers
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