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3 years ago

Can anyone help me with this problem?

Mathematics

1 answer:

Snowcat [4.5K]3 years ago

7 0

This uses the geometric mean for right triangles...the one involving the altitude. For us the pattern is $\frac{r}{h}=\frac{h}{s}$ . "s" is our answer.

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Solve for x in the triangle. Round your answer to the nearest tenth.

rewona [7]

Answer:

i think the answer might be 9

8 0

3 years ago

(6x² + 36x² + 12) ÷ (x+6)

astra-53 [7]

Answer:

=42x2+1

Step-by-step explanation:

6x2+36x2+12x+6

=42x2+1

8 0

3 years ago

Find the roots 5x^2+16=7x^2/3

TEA [102]

Answer: I Got No solution for this problem

Step-by-step explanation:

Graph each side of the equation. The solution is the x-value of the point of intersection.

No solution

3 0

3 years ago

Jeffrey is using a copy machine to make posters for a school play. The machine prints the posters at a constant rate. After 3 mi

jasenka [17]

Answer:

B: 11 minutes.

Step-by-step explanation:

So, we know that the machine prints at a constant rate.

After 3 minutes, the machine prints 108 posters.

So, let’s find the <em>unit rate. </em>We will divide the output over the input. So:

$108\text{ posters}/3\text{ minutes}=36\text{ posters/min}$

So, the unit rate of the machine is 36 posters per minute.

We can write an equation. If t represent the number of minutes and p represent the amount of posters, then:

$p=36t$

Gives the amount of posters p and t minutes.

We want to know how long it will take Jeffrey to make 396 posters.

So, we can substitute 396 for p and solve for t. Hence:

$396=36t$

Divide both sides by 36:

$t=11$

Therefore, it will take Jeffrey 11 minutes to make 396 posters.

7 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