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3 years ago

12

Which expression is equivalent to -6(3x-2/3)?

2 answers:

nlexa [21]3 years ago

6 0

D

i got it right on the unit test review

kupik [55]3 years ago

3 0

The answer is D -18x+4

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__________ multiplied 20 = 360<br> full explanation answer required

Setler79 [48]

18 x 20 = 360
360 / 20 = 18
So 18 is the answer

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3 years ago

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What is this? Someone please help me

Reil [10]

Answer and Step-by-step explanation:

Haven't done this in a while, so here we go:

The rule (I'm not really sure) would have to be negative number and negative number.

This means that:

(4, 0) would become (-4, 0) A'

(6, -2) becomes (-6, 2) B'

(8, -7) becomes (-8, 7) C'

and (0, -5) becomes (0, 5) D'

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3 years ago

Simplify the expression below.<br><br> 13y+x-7y

Zigmanuir [339]

13y+x-7y. The answer to this expression is 6y+x

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3 years ago

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Nidhi is thinking of a number n, and she wants her sister to guess the number. Her first clue is that eight less than three time

Dmitriy789 [7]

Answer:

The range of numbers that Nidhi might be thinking of is between 3 and 9.

Step-by-step explanation:

We are given that Nidhi is thinking of a number n, and she wants her sister to guess the number.

Her first clue is that eight less than three times her number is between one and nineteen.

Let the number Nidhi is thinking be 'n'.

So, eight less than three times her number means that;

$(3\times \text{n}) - 8 = 3\text{n}-8$

Now, according to the question; the above equation is between one and nineteen, so;

$1< 3\text{n}-8 < 19$

Adding 8 to the whole expression;

$1+8< 3\text{n}-8 +8< 19+8$

$9< 3\text{n}< 27$

Dividing the whole expression by 3;

$\frac{9}{3} < \frac{3\text{n}}{3} < \frac{27}{3}$

$3< \text{n}< 9$

Hence, the range of numbers that Nidhi might be thinking of is between 3 and 9.

3 0

3 years ago

How do you determine the area under a curve in calculus using integrals or the limit definition of integrals?

RSB [31]

Answer:

Please check the explanation.

Step-by-step explanation:

Let us consider

$y = f(x)$

To find the area under the curve $y = f(x)$ between $x = a$ and $x = b$ , all we need is to integrate $y = f(x)$ between the limits of $a$ and $b$ .

For example, the area between the curve y = x² - 4 and the x-axis on an interval [2, -2] can be calculated as:

$A=\int _a^b|f\left(x\right)|dx$

= $\int _{-2}^2\left|x^2-4\right|dx$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=\int _{-2}^2x^2dx-\int _{-2}^24dx$

solving

$\int _{-2}^2x^2dx$

$\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1$

$=\left[\frac{x^{2+1}}{2+1}\right]^2_{-2}$

$=\left[\frac{x^3}{3}\right]^2_{-2}$

computing the boundaries

$=\frac{16}{3}$

Thus,

$\int _{-2}^2x^2dx=\frac{16}{3}$

similarly solving

$\int _{-2}^24dx$

$\mathrm{Integral\:of\:a\:constant}:\quad \int adx=ax$

$=\left[4x\right]^2_{-2}$

computing the boundaries

$=16$

Thus,

$\int _{-2}^24dx=16$

Therefore, the expression becomes

$A=\int _a^b|f\left(x\right)|dx=\int _{-2}^2x^2dx-\int _{-2}^24dx$

$=\frac{16}{3}-16$

$=-\frac{32}{3}$

$=-10.67$ square units

Thus, the area under a curve is -10.67 square units

The area is negative because it is below the x-axis. Please check the attached figure.

6 0

3 years ago