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4 years ago

8. A number whose fifth part is increased by 5 is equal to its fourth part diminished by 5 . find the number

Mathematics

1 answer:

Ilya [14]4 years ago

3 0

8. A number whose fifth part is increased by 5 is equal to its fourth part diminished by 5 . find the number

the number = X

fifth part = X/5

fifth part is increased by 5 = X/5 + 5

fourth part = X/4

fourth part diminished by 5 = X/4 - 5

fifth part is increased by 5 is equal to its fourth part diminished by 5

X/5 + 5 = X/4 - 5

X/5 - X/4 = - 5 - 5

(4X - 5X)/20 = - 10

- X/20 = - 10

X = 200

Check:

X/5 + 5 = X/4 - 5

200/5 + 5 = 200/4 - 5

40 + 5 = 50 - 5

45 = 45

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What are the domain and range of this function y=(x+3)^2 -5

n200080 [17]

Answer: Choice A

Note: the range should be $[-5, \infty)$ . See explanation below.

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We can plug in any real number for x to get some output for y. The domain is the set of all real numbers in which we say $(-\infty, \infty)$ which is interval notation. It represents the interval from negative infinity to positive infinity.

The range is the set of possible outputs. The smallest output possible is y = -5 which occurs at the vertex (3,-5). We can get this y value or larger. So we can describe the range as the set of y values such that $y \ge -5$ and that translates to the interval notation $[-5, \infty)$ .

The square bracket says "include this endpoint" while the curved parenthesis says to exclude the endpoint. Your teacher mistakenly wrote $(-5, \infty)$ for choice A, when they should have written $[-5, \infty)$

I think either your teacher made a typo or somehow the formatting messed up. Either way, choice A is the closest to the answer.

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3 years ago

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The ratio of boys to girls in the ninth grade is 6:5. If there are 185 girls in the class, how many boys are there?

xeze [42]

Answer:

$222$ boys

Step-by-step explanation:

Let $x$ = total number of ninth graders. By knowing how many students there are in the grade, we can find for the number of boys.

$\frac{5}{6+5}x = 185$ (number of girls in ninth grade)

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To find for the number of boys, we can use substitution:

$\frac{6}{6+5}x = \frac{6}{11}(407)$

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∴ There are $222$ boys in the class.

Hope this helps :)

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Aditya's dog routinely eats Aditya's leftovers, which vary seasonally. As a result, his weight fluctuates throughout

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Based on the weight and the model that is given, it should be noted that W(t) in radians will be W(t) = 0.9cos(2πt/366) + 8.2.

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From the information, W(t) = a cos(bt) + d. Firstly, calculate the phase shift, b. At t= 0, the dog is at maximum weight, so the cosine function is also at a maximum. The cosine function is not shifted, so b = 1.

Then calculate d. The dog's average weight is 8.2 kg, so the mid-line d = 8.2. W(t) = a cos t + 8.2. Then calculate a, the dog's maximum weight is 9.1 kg. The deviation from the average is 9.1 kg - 8.2 kg = 0.9 kg. W(t) = 0.9cost + 8.2

Lastly, calculate t. The period p = 2π/b = 2π/1 = 2π. The conversion factor is 1 da =2π/365 rad. Therefore, the function with t in radians is W(t) = 0.9cos(2πt/365) + 8.2.

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