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3 years ago

What is 68.500.000 written in scientific notation?

Mathematics

1 answer:

Levart [38]3 years ago

8 0

Answer:

7×10to the power7

Step-by-step explanation:

65500000 and put it in the calculator and press shift then mode and press 7 and 1

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Show that: (3x-1)(x+5)(4x-3) = 12x^3 + 47x^2 - 62x +15 for all values of x. (3x-1)(x+5)(4x-3) = (?????)(4x-3) Expand into 6 term

lara31 [8.8K]

(3x - 1)( x + 5)(4x - 3) = 12x^3 + 47x^2 - 62x +15 
(3x^2 – x + 15x - 5)( 4x – 3 ) = 12x^3 + 47x^2 - 62x +15 
(3x^2 + 14x - 5)( 4x – 3 ) = 12x^3 + 47x^2 - 62x +15 
12x^3 – 9x^2 + 56x^2 – 42x – 20x + 15 = 12x^3 + 47x^2 - 62x +15 
12x^3 – 47x^2 - 62x + 15 = 12x^3 + 47x^2 - 62x +15

8 0

3 years ago

Read 2 more answers

9 squared divided by 3 cubed. Give expression in symbols and evaluate it? What is a exponential mot action with positive exponen

S_A_V [24]

There are numerous ways you can do this, but take a look at the attachment to see how I did it. I hope it answers your question! :)

7 0

3 years ago

Find the distance between the point - 3, 2 and 1 - 1

dusya [7]

Answer:

5 units

Step-by-step explanation:

Find the distance between the point - 3, 2 and 1 - 1

Given data

x1= -3

y1= 2

x2= 1

y2= -1

The expression for the distance between two points is

d=√((x_2-x_1)²+(y_2-y_1)²)

subtitute

d=√((1-(-3))²+(-1-(2))²)

d=√((1+3))²+(-1-2))²)

d=√((4)²+(-3))²)

d=√16+9

d=√25

d=5

Hence the distance between the points is 5 units

7 0

2 years ago

|a|-3/4=-5/8 Solve and show your work?

Marta_Voda [28]

Answer:

{-1/8, 1/8}

Step-by-step explanation:

Solve |a|-3/4=-5/8 for |a|: Add 3/4 to both sides of this equation, obtaining:

|a| - 6/8 + 6/8 = -5/8 + 6/8, or |a| = 1/8. This absolute value function has two components:

a = 1/8 and a = -1/8.

7 0

3 years ago

A 500500-gallon tank initially contains 200200 gallons of brine containing 100100 pounds of dissolved salt. brine containing 11

mixer [17]

Let $A(t)$ denote the amount of salt in the tank at time $t$ . We're given that the tank initially holds $A(0)=100$ lbs of salt.

The rate at which salt flows in and out of the tank is given by the relation

$\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{11\text{ lb}}{1\text{ gal}}\times\dfrac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{\dfrac{A(t)}{200+(44-11)t}\times\dfrac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}$
$\implies A'(t)+\dfrac{11}{200+33t}A(t)=484$

Find the integrating factor:

$\mu(t)=\exp\left(\displaystyle\int\frac{11}{200+33t}\,\mathrm dt\right)=(200+33t)^{1/3}$

Distribute $\mu(t)$ along both sides of the ODE:

$(200+33t)^{1/3}A'(t)+11(200+33t)^{-2/3}A(t)=484(200+33t)^{-1/3}$
$\bigg((200+33t)^{1/3}A(t)\bigg)'=484(200+33t)^{-1/3}$
$A(t)=484\displaystyle\int(200+33t)^{-1/3}\,\mathrm dt$
$A(t)=22(200+33t)^{2/3}+C$

Since $A(0)=100$ , we get

$100=22(200)^{2/3}+C\implies C\approx-652.39$

so that the particular solution for $A(t)$ is

$A(t)=22(200+33t)^{2/3}-652.39$

The tank becomes full when the volume of solution in the tank at time $t$ is the same as the total volume of the tank:

$200+(44-11)t=500\implies 33t=300\implies t\approx9.09$

at which point the amount of salt in the solution would be

$A(9.09)\approx733.47\text{ lb}$

4 0

3 years ago