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4 years ago

15

One of the solutions of the system of equations shown in the graph has an x-value of -4. What is its corresponding integer y-val

ue?
-1

-3

0

3

2 answers:

MrRissso [65]4 years ago

8 0

Answer:

-3

Step-by-step explanation:

There are 2 point of intersection:

(-1,0)

(-4,-3)

Alenkasestr [34]4 years ago

5 0

Answer:

The 1st intersection P(x, y) has x = -4 and y = -3 (as shown in graph).

Option B = -3 is correct.

Hope this helps!

:)

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An art history professor assigns letter grades on a test according to the following scheme. A: Top 13%13% of scores B: Scores be

amm1812

Answer:

The numerical limits for a B grade is between 81 and 89.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 79.7, \sigma = 8.4$

B: Scores below the top 13% and above the bottom 56%

Below the top 13%:

Below the 100-13 = 87th percentile. So below the value of X when Z has a pvalue of 0.87. So below X when Z = 1.127. So

$Z = \frac{X - \mu}{\sigma}$

$1.127 = \frac{X - 79.7}{8.4}$

$X - 79.7 = 8.4*1.127$

$X = 89$

Above the bottom 56:

Above the 56th percentile, so above the value of X when Z has a pvalue of 0.56. So above X when Z = 0.15. So

$Z = \frac{X - \mu}{\sigma}$

$0.15 = \frac{X - 79.7}{8.4}$

$X - 79.7 = 8.4*0.15$

$X = 81$

The numerical limits for a B grade is between 81 and 89.

3 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

Mr. Brown purchased 15 concert tickets for a total of $200. The concert tickets cost $15 for adults and $10 for

jok3333 [9.3K]

Answer:

St2ep-by-step explanation:

5 0

3 years ago

6m - 3p + 10 when m = 6, p = 4

asambeis [7]

Answer:

The answer is 58.

Step-by-step explanation:

Plug in and Solve.

$6m - 3p + 10 \\ 6(6) + 3(4) + 10 \\ 36 + 12 + 10 \\ 48 + 10 \\ 58$

6 0

3 years ago

Read 2 more answers

A can in the shape of a right cylinder is filled with 10W-40 oil. The weight of 10W-40 oil is 0.857 gram per cubic centimeter. I

disa [49]

First find the area of the cylinder using the following formula
$A = 2\pi rh + 2\pi r^2$
$A = (2 \times \pi \times 6 \times 10) + (2 \times \pi \times 6^2)$

After finding the area, <span>calculate the weight of the oil (in grams) in the can
</span>
weight in grams = A times 0.857

4 0

3 years ago