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2 years ago

15

The formula for converting temperature from Fahrenheit (f) to celsius (c) is c= 5/4(f-32). if the temperature is 77 Fahrenheit,

what is it in celsius?
a) 60 5/9 degrees c
B) 45 degrees c
c) 41 2/3 degrees c
D) 25 degrees c

2 answers:

spin [16.1K]2 years ago

6 0

Answer:

the answer is D) 25 degrees c

stepan [7]2 years ago

6 0

Answer: D) 25 degrees c

Step-by-step explanation:

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Plot two points that are 3units from Point C and also share the same y-coordinate as Point C.

egoroff_w [7]

The two points at a distance of 3 units from C with the same y coordinate are (-3, -5) and (-9, -5).

Step-by-step explanation:

Step 1:

First, we must plot the point C on the graph. Point C is on the third quadrant so the x value and the y value are negative.

The coordinate of C is (-6, -5).

We need to plot two points that are 3 units away from point C but have the same y coordinate i.e. y = -5.

Step 2:

So assume the two other points are A and B.

The y coordinates of A and B are -5. Only the x coordinate varies. So in order to get the coordinates of point A and B, we add and subtract 3 from the x coordinate of point C.

Point A = (-6+3, -5) = (-3,-5).

Point B = (-6-3, -5) = (-9, -5).

So two points at a distance of 3 units from C with the same y coordinate are (-3, -5) and (-9, -5).

6 0

3 years ago

Davin ran the marathon in 4 3/5 hours. Melba ran the marathon in 226 minutes. After she crossed the finish line, Melba waited fo

emmasim [6.3K]

(by far the most painful question all day) 226 mins = 3.77 hours 4 3/5 hours = 4.6 hours 4.6- 3.77 = 0.83 hours 0.83 hours = 49.8 mins = 2988 seconds.

7 0

3 years ago

Find two unit vectos that are orthogonal to both [0,1,2] and [1,-2,3]

alekssr [168]

Answer:

Let the vectors be

a = [0, 1, 2] and

b = [1, -2, 3]

( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.

Let the cross product be another vector c.

To find the cross product (c) of a and b, we have

$\left[\begin{array}{ccc}i&j&k\\0&1&2\\1&-2&3\end{array}\right]$

c = i(3 + 4) - j(0 - 2) + k(0 - 1)

c = 7i + 2j - k

c = [7, 2, -1]

( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:

c / | c |

Where | c | = √ (7)² + (2)² + (-1)² = 3√6

Therefore, the unit vector is

$\frac{[7,2,-1]}{3\sqrt{6} }$

or

[ $\frac{7}{3\sqrt{6} }$ , $\frac{2}{3\sqrt{6} }$ , $\frac{-1}{3\sqrt{6} }$ ]

The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:

[ $\frac{-7}{3\sqrt{6} }$ , $\frac{-2}{3\sqrt{6} }$ , $\frac{1}{3\sqrt{6} }$ ]

In conclusion, the two unit vectors are;

[ $\frac{7}{3\sqrt{6} }$ , $\frac{2}{3\sqrt{6} }$ , $\frac{-1}{3\sqrt{6} }$ ]

and

[ $\frac{-7}{3\sqrt{6} }$ , $\frac{-2}{3\sqrt{6} }$ , $\frac{1}{3\sqrt{6} }$ ]

<em>Hope this helps!</em>

7 0

2 years ago

How is this expression written with exponents? 3 . 3 . x . x . x . x

marin [14]

Solve:-

Write it down again:-

3 · 3 · x · x · x · x

We will write it like this:-

3 · 3 · x³

7 0

3 years ago

Suppose that a company needs 1, 200,000 items during a year and that preparation for each production run costs $500. Suppose als

MAVERICK [17]

Answer:

The number of items in each production run so that the total costs of production and storage are minimized is 8165 items/run

Step-by-step explanation:

We will use the following variables:

Q = Quantity being ordered

Q* = the optimal order Quantity: the result being sought

D = annual Demand for the item, over the year

P = unit Production cost

S = cost of setting up a production run, regardless of the number of units in the production run (fixed cost per production run)

H = annual cost to Hold one unit

It is important to note which variables are annualized, which are per-order and which are per-unit.

Using the variables, here are the components of the first equation

Total Cost, TC = PC + SC + HC

PC = P x D : Production Cost = unit Production cost times the annual Demand

SC = (D x S)/Q : Setting up Cost = annual Demand times cost per production setup, divided by the order Quantity (number of units)

HC = (H x Q)/2: Holding Cost = annual unit Holding cost times order Quantity (number of units), divided by 2 (because throughout the year, on average the warehouse is half full).

So TC = PC + SC + HC = (P x D) + ((D x S)/Q) + ((H x Q)/2) = PD + (DS/Q) + HQ/2

To obtain the optimal order quantity, Q* that minimizes TC, at the minimum TC, dTC/dQ = 0

dTC/dQ = (H/2) – (D x S)/(Q²) = 0

(H/2) – (D x S)/(Q²) = 0

Solving for Q, which is Q* at this point.

(Q*)² = 2DS/H

Q* = √(2DS/H)

D = annual demand for the item = 200000

S = cost of setting up a production run, regardless of the number of units in the production run (fixed cost per production run) = $500

H = annual cost to Hold one unit = $3

Q* = √(2×200000×500/3) = 8164.97 = 8165 items.

3 0

3 years ago