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3 years ago

What is... 55.

Mathematics

1 answer:

sukhopar [10]3 years ago

3 0

Answer:

one more would be 56

one less would be 54

ten more would 65

ten less would be 45

Step-by-step explanation:

add one more to 55 to get 56

subtract one from 55 to get 54

add ten more to 55 to get 65

subtract ten from 55 to get 45

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Helppp last question only (c). Maths

MAVERICK [17]

I cannot just answer part (c) because it is dependent on previous answers. So, let's go through parts (a) and (b) so we can then answer (c).

For part a, understand that critical points are points in a function f(x) where the derivative of the function f'(x) is either equal to 0 or it does not exist. Let's take the derivative of this function, using quotient rule which states that:

$z(x) = (\frac{f(x)}{g(x)})' = \frac{f'(x)g(x) - g'(x)f(x)}{(g(x))^2}$

In our case, f(x) = x + 7 and g(x) = x + 1. Following the rule, our derivative will be:

$z(x) = \frac{(1(x+1)) - (1(x+7))}{(x+1)^2}$

Let's simplify that:

$z(x) = \frac{x+1 - x-7}{(x-1)^2} = \frac{-6}{(x+1)^2}$

Notice that the above function is never equal to 0 (numerator never equals 0, so the function follows). It is, however, undefined at x = -1 as the denominator is 0:

$z(1) = \frac{-6}{(-1+1)^2} = \frac{-6}{0^2} = \frac{-6}{0}$

This gives us a divide by 0 error, therefore it is undefined. This is a critical point, so your answer to part (a) should be changed to "The critical number(s) is/are -1.

For part (b) it asks if the function is ever increasing. For a function to be increasing over some interval, the derivative of that function must be positive. Notice that even if we plug in a negative number for the denominator, the derivative will never be positive because the denominator is squared, meaning we will always have a positive denominator. Our numerator will always be negative no matter what, so a negative number over a positive number will always equal a negative number suggesting that our function will never be increasing. Let's try a negative number: x = -2:

$z(-1) = \frac{-6}{(-2-1)^2} = \frac{-6}{(-3)^2} = -\frac{6}{9} =-\frac{2}{3}$

Therefore, your answer to part (b) is correct! Good job!

For part (c) we must identify any interval for which the function is decreasing. We just figured out that our function is always decreasing. Even if we plug in a negative number. But, we have a little "plot-twist." The function is undefined at x = -1, therefore, it is not decreasing at that single point. We write that as:

$-\infty < x < \infty, x \neq -1$

or, we can also write it as:

$-\infty < x < -1$ and $-1 < x < \infty$

We write it in interval notation as: (-∞, -1) and (-1, ∞).

So, your answer to part (c) must be one of the above. We have answered all of the questions. Hope I could help! :)

4 0

3 years ago

IMA GIVE 1,000 COOKIES TO THE PERSON WHO GETS THIS QUESTION RIGHT

Ahat [919]

Answer:

I answered this wrong and don't have time to fix it now

Step-by-step explanation:

I will come back to it later. I can't delete this answer i'm sorry

4 0

4 years ago

Mr Smith class 1/3 were absent on Monday and Mrs. Brown class 2/5 were absent on Monday. If 4 students from each class were abse

anygoal [31]

Answer:

Mr. Smiths class has 12 students and Mrs. Brown's class has 10 students.

Step-by-step explanation:

if 1/3 of the class is absent and that amount is four you would multiply 4 by 3 and get 12. If 2/5 of the class are missing and there 4 students first you would divide 4 by 2 and get 2, then you would take the 2 and multiply it by 5 and get 10.

3 0

4 years ago

The graph of the continuous function g, the derivative of the function f, is shown above. The function g is piecewise linear for

4vir4ik [10]

A) $g=f'$ is continuous, so $f$ is also continuous. This means if we were to integrate $g$ , the same constant of integration would apply across its entire domain. Over $0$ , we have $g(x)=2x$ . This means that

$f_{0$

For $f$ to be continuous, we need the limit as $x\to1^-$ to match $f(1)=3$ . This means we must have

$\displaystyle\lim_{x\to1}x^2+C=1+C=3\implies C=2$

Now, over $x$ , we have $g(x)=-3$ , so $f_{x$ , which means $f(-5)=17$ .

b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,

$\displaystyle\int_1^6g(x)=4+\int_3^62(x-4)^2\,\mathrm dx=4+6=10$

c) $f$ is increasing when $f'=g>0$ , and concave upward when $f''=g'>0$ , i.e. when $g$ is also increasing.

We have $g>0$ over the intervals $0$ and $x>4$ . We can additionally see that $g'>0$ only on $0$ and $x>4$ .

d) Inflection points occur when $f''=g'=0$ , and at such a point, to either side the sign of the second derivative $f''=g'$ changes. We see this happening at $x=4$ , for which $g'=0$ , and to the left of $x=4$ we have $g$ decreasing, then increasing along the other side.

We also have $g'=0$ along the interval $-1$ , but even if we were to allow an entire interval as a "site of inflection", we can see that $g'>0$ to either side, so concavity would not change.

5 0

3 years ago

A circular pool has a radius of 20 feet. What is the area of the pool? Use 3.14 for π.

GrogVix [38]

The answer i think is 50

3 0

4 years ago