What is the distance between the points (2,-3) and (-6,4) on the coordinate plane

Write an expression to represent: The sum of four and the product of three and a number x xx.

Ira Lisetskai [31]

Answer:

4+3x

It's pretty straight forward, you have the number 4 that is added to the rest of the expression because sum is addition, and the product of 2 numbers is multiplication so it would make 3(x) or 3x, which results in 4+3x.

8 0

2 years ago

Emily purchased lemons for $600. She sold 3/4 of these at a loss of 20% and the remaining at a gain of 20%. How much percent doe

MAVERICK [17]

Answer:

10%

Step-by-step explanation:

Given: CP of lemon is $600.

3/4 of lemon sold at 20% loss

Remaining lemon at 20% gain.

Considering the quantity of lemon remain constant.

Cost price of 3/4 of lemon= $\frac{3}{4} \times 600= \$450$

As given, 3/4 of lemon sold at 20% loss.

∴ Selling price of $\frac{3}{4}\ of\ lemon= 450- \frac{20}{100}\times 450$

Selling price= $450- 90= \$ 360$

Hence, selling price of 3/4 lemon is $360.

Now, the cost price of remaining lemon $(1-\frac{3}{4} )= (\$ 600-\$ 450)$

∴ The cost price of $\frac{1}{4}\ lemon = \$ 150$

As given, remaining $\frac{1}{4} lemon\ sold\ at\ gain\ of\ 20\%$

∴ Selling price of $\frac{1}{4} \ lemon= (150\times \frac{20}{100}+150)$

Selling price of $\frac{1}{4} \ of\ lemon= (30+150)$

Hence, selling price of 1/4 lemon is $180

Loss\profit percent= $\frac{(SP-CP)}{CP} \times 100$

∴ Loss\profit percent= $\frac{60}{600} \times 100= 10\%$

Hence, the loss percentage is 10%

8 0

3 years ago

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'

vodka [1.7K]

Answer:

a) $v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

b) $0$

c) $a(t) = x''(t) = v'(t) =6t-12$

When the acceleration is 0 we have:

$6t-12=0, t =2$

And if we replace t=2 in the velocity function we got:

$v(t) = 3(2)^2 -12(2) +9=-3$

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

$x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10$

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

$v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

$v(t) = 3t^2 -12t +9$

We can factorize this function as $v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)$

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is $0\leq t\leq 10$ we need to analyze the following intervals: