What is the distance between -2 and 3?

Write a number in which the value of the 3 is ten times greater than the value of the 3 in 135, 864

Aleonysh [2.5K]

Answer:

345,456

Step-by-step explanation:

The value of 3 in 135,864 is 30,000.

We want to write a number that is ten times 30,000.

We could write any other number in which the value of 3 is 300,000.

There are infinitely many numbers that we can write.

Some examples are:

345,456

1,315,445

5,354,456

and so on and so forth.

8 0

4 years ago

What is 6 +0.2 + 0.09 written in standard form? A 0.629 B 6.029 C 6.209 D 6.29

Free_Kalibri [48]

Answer:

D

Step-by-step explanation:

The answer is D. 6 is the whole number with no decimal after it so it goes before the decimal. Then we add the numbers after the decimal. 0.2 is the same as 0.20 so we add 0.20 plus 0.09 which gets us 0.29. Lastly, we add this all together and get a total of 6.29. Hope this helps!

5 0

3 years ago

Plssss help quick i need it for sum important

Vanyuwa [196]

(a) The product of $(2x-4)\ and\ (3x^2-x+4)$ is calculated to be $=6x^3-14x^2+12x-16$

(b)The product of $(2x-4) \ and\ (3x^2-x+4)\$ is not equal to the product of $(4x-2)\ and\ (3x^2-x+4)$

Step-by-step explanation:

(a)

$(2x-4)(3x^2-x+4)\\\\=(2x.3x^2)+(2x.-x)+(2x.4)+(-4.3x^2)+(-4.-x)+(-4.4)\\\\=6x^3-2x^2+8x-12x^2+4x-16\\\\=6x^3-14x^2+12x-16$

b)

$(4x-2)(3x^2-x+4)\\\\=(4x.3x^2)+(4x.-x)+(4x.4)+(-2.3x^2)+(-2.-x)+(-2.4)\\\\=12x^3-4x^2+16x-6x^2+2x-8\\\\=12x^3-10x^2+18x-8$

The product of $(2x-4) \ and\ (3x^2-x+4)\$ is not equal to the product of $(4x-2)\ and\ (3x^2-x+4)$

7 0

3 years ago

Verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers c that s

WINSTONCH [101]

Answer: $\bold{c=\dfrac{1+\sqrt{19}}{3}\approx1.8}$

Step-by-step explanation:

There are 3 conditions that must be satisfied:

f(x) is continuous on the given interval
f(x) is differentiable
f(a) = f(b)

If ALL of those conditions are satisfied, then there exists a value "c" such that c lies between a and b and f'(c) = 0.

f(x) = x³ - x² - 6x + 2 [0, 3]

1. There are no restrictions on x so the function is continuous $\checkmark$

2. f'(x) = 3x² - 2x - 6 so the function is differentiable $\checkmark$

3. f(0) = 0³ - 0² - 6(0) + 2 = 2

f(3) = 3³ - 3² - 6(3) + 2 = 2

f(0) = f(3) $\checkmark$

f'(x) = 3x² - 2x - 6 = 0

This is not factorable so you need to use the quadratic formula:

$x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4(3)(-6)}}{2(3)}\\\\\\.\quad =\dfrac{2\pm \sqrt{4+72}}{2(3)}\\\\\\.\quad =\dfrac{2\pm \sqrt{76}}{2(3)}\\\\\\.\quad =\dfrac{2\pm 2\sqrt{19}}{2(3)}\\\\\\.\quad =\dfrac{1\pm \sqrt{19}}{3}\\\\\\.\quad \approx\dfrac{1+4.4}{3}\quad and\quad \dfrac{1-4.4}{3}\\\\\\.\quad \approx1.8\qquad and\quad -1.1$

Only one of these values (1.8) is between 0 and 3.

5 0

3 years ago

Set C is the set of two-digit even numbers greater than 34 that are divisible by 5 C=

amid [387]

In order for it to be divisible by 5 AND be an even number, it must also be divisible by 2. So you are looking for two-digit numbers greater than 34 and divisible by 10:
C = {40, 50, 60, 70, 80, 90}

8 0

4 years ago