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3 years ago

What is the coeffiect of 5/3 xy

1 answer:

8 0

The coefficient is the constant part of a monomial, basically the number in front of the variable, in this expression it is 5/3

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The answer to the problem

larisa86 [58]

She puts 8% of her salary in, so multiply her pay by 8% to get her yearly amount she saves:

45000 x 0.08 = $3,600 per year.

The company puts 6% of that in the account:

3600 x 0.06 = $216

So per year 3600 + 216 = $3,816 is saved.

3,816 x 2 = $7,632 is the 2 year total.

The answer is A.

6 0

3 years ago

Jasmine took a cab home from her office. The cab charged a flat fee of $4, plus $2 per mile. Jasmine paid $32 for the trip. How

vaieri [72.5K]

To solve, make an equation. The equation would be
$$32 = 4 + 2x$
x represents the miles driven
to solve, subtract 4 from each side of the equation.
$32-4 = 4-4 + 2x
The equation is now
$28 = 2x
Now you divide by 2 on each side.
$28 ÷ 2 = 2x ÷ 2
simplify
14 = x
The answer is 14 miles

3 0

3 years ago

Find the indicated arc and angle measures:

nevsk [136]

Answer:

arc AB = 70°

arc BC = 110°

arc ABC = 180°

arc CDB = 250°

Step-by-step explanation:

Since these angles are not inscribed, but are at the center point of the circle, the angles and arcs will be the same measure.

Solve for arc AB:

Angle AFB = 70°, so arc AB = 70°

Solve for arc BC:

Angle AFC is a straight angle so it is 180°. To find angle BFC, subtract angle AFB from 180°.

180 - 70 = 110°

Since angle BFC = 110°, arc BC = 110°

Solve for arc ABC:

Add arc AB and arc BC together.

70° + 110° = 180°

arc ABC = 180°

Solve for arc CDB:

There are 360° in a cirlce. To find arc CDB, subtract arc BC from 360°.

360° - 110° = 250°

arc CDB = 250°

7 0

2 years ago

The endpoints of (MP)are M(2,1) and P(12,6). If point K partitions (MP) in a ratio of MK:KP = 3:2, what are the coordinates of K

GREYUIT [131]

Answer:

K(8, 4)

Step-by-step explanation:

Given:

M(2, 1), P(12, 6)

MK:KP = 3:2

Required:

Coordinates of K

SOLUTION:

Coordinates of K can be determined using the formula below:

$x = \frac{mx_2 + nx_1}{m + n}$

$y = \frac{my_2 + ny_1}{m + n}$

Where,

$M(2, 1) = (x_1, y_1)$

$P(12, 6) = (x_2, y_2)$

$m = 3, n = 2$

Plug in the necessary values to find the coordinates of K:

$x = \frac{mx_2 + nx_1}{m + n}$

$x = \frac{3(12) + 2(2)}{3 + 2}$

$x = \frac{36 + 4}{5}$

$x = \frac{40}{5}$

$x = 8$

$y = \frac{my_2 + ny_1}{m + n}$

$y = \frac{3(6) + 2(1)}{3 + 2}$

$y = \frac{18 + 2}{5}$

$y = \frac{20}{5}$

$y = 4$

The coordinates of K = (8, 4)

3 0

3 years ago

(c). Under a set of controlled laboratory conditions, the size of the population P of a certain bacteria culture at time t (in s

Bezzdna [24]

(i) Since P(t) gives the population of the culture after t seconds, the population after 1 second is

P(1) = 3•1² + 3e¹ + 10 = 13 + 3e ≈ 21.155

In Mathematica, it's convenient to define a function:

P[t_] := 3t^2 + 3E^t + 10

(E is case-sensitive and must be capitalized. Alternatively, you could use Exp[t]. You can also specify that the argument t must be non-negative by entering a condition via P[t_ ;/ t >= 0], but that's not necessary.)

Then just evaluate P[1], or N[P[1]] or N <at symbol> P[1] or P[1] // N to get a numerical result.

(ii) The average rate of change of P(t) over an interval [a, b} is

(P(b) - P(a))/(b - a)

Then the ARoC between t = 2 and t = 6 is

(P(6) - P(2))/(6 - 2) ≈ 321.030

In M,

(P[6] - P[2])/(6 - 2)

and you can also include N just as before.

(iii) You want the instantaneous rate of change of P when t = 60 (since 1 minute = 60 seconds). Differentiate P :

P'(t) = 6t + 3e^t

Evaluate the derivative at t = 60 :

P'(60) = 6•60 + 3e⁶⁰ = 360 + 3e⁶⁰

The approximate value is quite large, so I'll just leave its exact value.

In M, the quickest way would be P'[60], or you can differentiate and replace (via ReplaceAll or /.) t with 60 as in D[P[t], t] /. t -> 60.

5 0

3 years ago