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3 years ago

Explain how multiplying fractions is similar to mulitiplying whole numbers and how it is different

Mathematics

2 answers:

tigry1 [53]3 years ago

3 0

You could find the difference on the internet if not found tell me and I would explain for you

Allisa [31]3 years ago

3 0

Multiplying fractions and whole are the same cause it’s both multiplying

You might be interested in

which is the rationalized form of the expression the square root of x divided by the square root of x +the square root of 7

Ilya [14]

Answer:

I assume your goal is to rationalize the denominator -

So you need to multiply the denominator by its conjugate.

So we have:

$\frac{\sqrt{x} }{\sqrt{x}-\sqrt{7} } *\frac{\sqrt{x} +\sqrt{7} }{\sqrt{x} +\sqrt{7}}$

$\frac{x-\sqrt{7}\sqrt{x} }{x-7}$

8 0

3 years ago

PLEASE HELP, GOOD ANSWERS GET BRAINLIEST. +40 POINTS WRONG ANSWERS GET REPORTED

MA_775_DIABLO [31]

1. Ans:(A) 123

Given function: $f(x) = 8x^2 + 11x$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(8x^2 + 11x)$
=> $\frac{d}{dx} f(x) = \frac{d}{dx}(8x^2) + \frac{d}{dx}(11x)$
=> $\frac{d}{dx} f(x) = 2*8(x^{2-1}) + 11$
=> $\frac{d}{dx} f(x) = 16x + 11$

Now at x = 7:
$\frac{d}{dx} f(7) = 16(7) + 11$

=> $\frac{d}{dx} f(7) = 123$

2. Ans:(B) 3

Given function: $f(x) =3x + 8$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(3x + 8)$
=> $\frac{d}{dx} f(x) = \frac{d}{dx}(3x) + \frac{d}{dx}(8)$
=> $\frac{d}{dx} f(x) = 3*1 + 0$
=> $\frac{d}{dx} f(x) = 3$

Now at x = 4:
$\frac{d}{dx} f(4) = 3$ (as constant)

=>Ans: $\frac{d}{dx} f(4) =$ 3

3. Ans:(D) -5

Given function: $f(x) = \frac{5}{x}$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{5}{x})$
or
$\frac{d}{dx} f(x) = \frac{d}{dx}(5x^{-1})$
=> $\frac{d}{dx} f(x) = 5*(-1)*(x^{-1-1})$
=> $\frac{d}{dx} f(x) = -5x^{-2}$

Now at x = -1:
$\frac{d}{dx} f(-1) = -5(-1)^{-2}$

=> $\frac{d}{dx} f(-1) = -5 *\frac{1}{(-1)^{2}}$
=> Ans: $\frac{d}{dx} f(-1) = -5$

4. Ans:(C) 7 divided by 9

Given function: $f(x) = \frac{-7}{x}$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{-7}{x})$
or
$\frac{d}{dx} f(x) = \frac{d}{dx}(-7x^{-1})$
=> $\frac{d}{dx} f(x) = -7*(-1)*(x^{-1-1})$
=> $\frac{d}{dx} f(x) = 7x^{-2}$

Now at x = -3:
$\frac{d}{dx} f(-3) = 7(-3)^{-2}$

=> $\frac{d}{dx} f(-3) = 7 *\frac{1}{(-3)^{2}}$
=> Ans: $\frac{d}{dx} f(-3) = \frac{7}{9}$

5. Ans:(C) -8

Given function:
$f(x) = x^2 - 8$

Now if we apply limit:
$\lim_{x \to 0} f(x) = \lim_{x \to 0} (x^2 - 8)$

=> $\lim_{x \to 0} f(x) = (0)^2 - 8$
=> Ans: $\lim_{x \to 0} f(x) = - 8$

6. Ans:(C) 9

Given function:
$f(x) = x^2 + 3x - 1$

Now if we apply limit:
$\lim_{x \to 2} f(x) = \lim_{x \to 2} (x^2 + 3x - 1)$

=> $\lim_{x \to 2} f(x) = (2)^2 + 3(2) - 1$
=> Ans: $\lim_{x \to 2} f(x) = 4 + 6 - 1 = 9$

7. Ans:(D) doesn't exist.

Given function: $f(x) = -6 + \frac{x}{x^4}$
In this case, even if we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

Check:
$f(x) = -6 + \frac{x}{x^4} \\ f(x) = -6 + \frac{1}{x^3} \\ f(x) = \frac{-6x^3 + 1}{x^3} \\ Rationalize: \\ f(x) = \frac{-6x^3 + 1}{x^3} * \frac{x^{-3}}{x^{-3}} \\ f(x) = \frac{-6x^{3-3} + x^{-3}}{x^0} \\ f(x) = -6 + \frac{1}{x^3} \\ Same$

If you apply the limit, answer would be infinity.

8. Ans:(A) Doesn't Exist.

Given function: $f(x) = 9 + \frac{x}{x^3}$
Same as Question 7
If we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

9, 10.
Please attach the graphs. I shall amend the answer. :)

11. Ans:(A) Doesn't exist.

First We need to find out: $\lim_{x \to 9} f(x)$ where,
$f(x) = \left \{ {{x+9, ~~~~~x \textless 9} \atop {9- x,~~~~~x \geq 9}} \right.$

If both sides are equal on applying limit then limit does exist.

Let check:
If x $\textless$ 9: answer would be 9+9 = 18
If x $\geq$ 9: answer would be 9-9 = 0

Since both are not equal, as $18 \neq 0$ , hence limit doesn't exist.

12. Ans:(B) Limit doesn't exist.

Find out: $\lim_{x \to 1} f(x)$ where,

$f(x) = \left \{ {{1-x, ~~~~~x \textless 1} \atop {x+7,~~~~~x \textgreater 1} } \right. \\ and \\ f(x) = 8, ~~~~~ x=1$

If all of above three are equal upon applying limit, then limit exists.

When x < 1 -> 1-1 = 0
When x = 1 -> 8
When x > 1 -> 7 + 1 = 8

ALL of the THREE must be equal. As they are not equal. 0 $\neq$ 8; hence, limit doesn't exist.

13. Ans:(D) -∞; x = 9

f(x) = 1/(x-9).

Table:

x f(x)=1/(x-9)

----------------------------------------

8.9 -10

8.99 -100

8.999 -1000

8.9999 -10000

9.0 -∞

Below the graph is attached! As you can see in the graph that at x=9, the curve approaches but NEVER exactly touches the x=9 line. Also the curve is in downward direction when you approach from the left. Hence, -∞, x =9 (correct)

14. Ans: -6

s(t) = -2 - 6t

Inst. velocity = $\frac{ds(t)}{dt}$

Therefore,

$\frac{ds(t)}{dt} = \frac{ds(t)}{dt}(-2-6t) \\ \frac{ds(t)}{dt} = 0 - 6 = -6$

At t=2,

Inst. velocity = -6

15. Ans: +∞, x =7

f(x) = 1/(x-7)^2.

Table:

x f(x)= 1/(x-7)^2

--------------------------

6.9 +100

6.99 +10000

6.999 +1000000

6.9999 +100000000

7.0 +∞

Below the graph is attached! As you can see in the graph that at x=7, the curve approaches but NEVER exactly touches the x=7 line. The curve is in upward direction if approached from left or right. Hence, +∞, x =7 (correct)

-i

7 0

3 years ago

Read 2 more answers

How many groups of ten are there in 230?

Artyom0805 [142]

Answer:

I am pretty sure the answer is 23

Step-by-step explanation:

The answer is 23 because if you multiply 23 by 10 you get 230 which is the same if you divided 230 by 10 and you get 23.

4 0

3 years ago

What is the area of a triangle with vertices (3,0) (9,0) (7,6)

Nezavi [6.7K]

Answer:

The area of the triangle is of 21 units of area.

Step-by-step explanation:

The area of a triangle with three vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by the determinant of the following matrix:

$A = \pm 0.5 \left|\begin{array}{ccc}x_1&y_1&1\\x_2&y_2&1\\1_3&y_3&1\end{array}\right|$

In this question:

Vertices (3,0) (9,0) (7,6). So

$A = \pm 0.5 \left|\begin{array}{ccc}3&0&1\\9&0&1\\7&7&1\end{array}\right|$

$A = \pm 0.5(3*0*1+0*1*7+1*9*7-0*7*1-0*9*1-3*1*7)$

$A = \pm 0.5*(63-21)$

$A = \pm 0.5*42 = 21$

The area of the triangle is of 21 units of area.

5 0

2 years ago

A null hypothesis is that the mean nose lengths of men and women are the same. The alternative hypothesis is that men have a lon

algol [13]

Answer:

b. There is not enough evidence to say that the populations of men and women have different mean nose lengths.

See explanation below.

Step-by-step explanation:

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the means for the two groups are different (men have longer mean nose length than women), the system of hypothesis would be:

Null hypothesis: $\mu_{men} \leq \mu_{women}$

Alternative hypothesis: $\mu_{men} > \mu_{women}$

Assuming that we know the population deviations for each group, for this case is better apply a z test to compare means, and the statistic is given by:

$z=\frac{\bar X_{men}-\bar X_{women}}{\sqrt{\frac{\sigma^2_{men}}{n_{men}}+\frac{\sigma^2_{women}}{n_{women}}}}$ (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Let's assume that the calculated statistic is $z_{calc}$

Since is a right tailed test test the p value would be:

$p_v =P(Z>z_{calc})=0.225$

And we know that the p value is 0.225. If we select a significance level for example 0.05 or 0.1 we see that $p_v >\alpha$

And on this case we have enough evidence to FAIl to reject the null hypothesis that the means are equal. So then the best conclusion would be:

b. There is not enough evidence to say that the populations of men and women have different mean nose lengths.

8 0

3 years ago