All categories

3 years ago

Without multiplying, determine the sign of the product (356,864)(−194,758).

Mathematics

2 answers:

STatiana [176]3 years ago

8 0

I think it would be A.) but at the same time i think it's wrong

nydimaria [60]3 years ago

7 0

B) the sign of the product is negative because a positive multiplied by a negative is a negative.

You might be interested in

Solve for X 2/3x+4=1/2x+5

mote1985 [20]

For this case we must find the value of the variable of the following equation:

$\frac {2} {3} x + 4 = \frac {1} {2} x + 5$

Subtracting $\frac {1} {2} x$ from both sides:

$\frac {2} {3} x- \frac {1} {2} x + 4 = 5\\\frac {4-3} {6} x + 4 = 5\\\frac {1} {6} x + 4 = 5$

Subtracting 4 from both sides:

$\frac {1} {6} {x} = 5-4\\\frac {1} {6} x = 1$

Multiplying by 6 on both sides:

$x = 6 * 1\\x = 6$

Answer:

$x = 6$

5 0

3 years ago

Read 2 more answers

Find the area under the standard normal curve between z = 1 and z = 2

emmasim [6.3K]

Answer:

Step-by-step explanation:

Find the area under the standard normal curve between z = 1 and z = 2.

13.59% or 0.1359
THANXXX
HEYAAA

3 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:

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