Ronnie wants to buy a computer table that measures 10 feet by 17 feet. The cost of the table is 16.00 per square feet. How much will it cost Ronnie to buy the computer table for his study room?

Answers

Answer 1
Answer: Measurement of table:
Length=17 feet
Width=10 feet
As we know that the table will be in rectangular shape.
So, Area of table= 10 feet × 17 feet
170 squared feet.
A/Q,
Cost of 1 square feet table =16($ or any unit of currency)
Cost of 170 square feet= 16×170($)
2720($)
Answer 2
Answer: The given dimensions are 10 ft and 17 ft
Now looking at the dimensions, we can assume it is rectangular in shape and therefore the formula of the area of a rectangle is:
Area = length × width
(length = 17 ft (longest side) and width = 10 ft)
1st step: Substitute the dimensions in the formula:
A = L × W
   = 17 × 10      (order doesnt matter!)
   = 170 ft²
Coming to the cost, 1 ft² costs 16 (lets keep it £, or €, or $, since the                                                              unit is not provided)
2nd step: We multiply the area by the cost to find the total amount!
(Because every squared feet is 16 and therefore we conclude with this formula)
Cost = 170 ft² × 16£
        = £2,720
Ronnie will have to pay 
£2,720 (or any other currency units)

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What element is located at b24?B=[-4 10 -3 9
[ 6 0 2 -8
[7 -1 -11 4
[-9 -5 5 -10

Answers

b24 means the element of set B in the second row and 4th column.

(-8) is at that position.

What is set theory?

The mathematical concept of well-defined groupings of objects, or sets, that are referred to as members or elements of the set. The only sets that are taken into consideration are those whose members are also sets because pure set theory only deals with sets.

Given that the set is,

B=[-4 10 -3 9

[ 6 0 2 -8

[7 -1 -11 4

[-9 -5 5 -10

B24 stands for the B element in the second row and fourth column.

There is a (-8) there.

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b24 means Element of the set B in the second row and 4th column .
(-8) is at that position.

Which expression is equivalent to x^4-y^2 ?

Answers

x^(4) - y²
x² · x² - y ·y
(x² - y)(x + y) = x^(4) - y²

For the function y=-3+4cos (5pi/6 (x+4)), what is the minimum value

Answers

interval <-1;1>, which means that this function can take values from interval <-7;-1>. The minimum value is -7.

the min of cos (5pi/6(x+4) = -1

so the min of 4cos 5pi/6(x+4 = -4

so the min of -3 + 4 cos 5pi/6(x+4 = -7

How do you find the minimum value?

If the quadratic equation has a positive term, it also has a minimum value. This minimum can be found by graphing the function or by using one of the two equations. If you have an equation of the form y = ax ^ 2 + bx + c, you can use the equation min = c --b ^ 2 / 4a to find the minimum value.

Rearrange the function using basic algebraic rules and solve the value of x when the derivative is zero. This solution provides the x-coordinates of the vertices of the function where the maximum or minimum values ​​occur. Convert to the original function and solve to find the minimum or maximum value.

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Answer:

-7

Step-by-step explanation:

A P E X

Pls help i need asap for homework
-5/6m = -60
m = ?

Answers

Step-by-step explanation:

Hey there!

Given;

-5/6m = -60

~ Multiply "6m" and "-60".

-5 = 6m*-60

-5 = -360m

~ Divide "-5" by "-360"

m = -5/-360

m = 1/72

Therefore,the value of "m"is 1/72.

Hope it helps....

An airplane pilot fell 370 m after jumping without his parachute opening. He landed in a snowbank, creating a crater 1.5 m deep, but survived with only minor injuries. Assume that the pilot's mass was 84 kg and his terminal velocity was 50 m/s.estimate

Answers

Answer:

he ded

Step-by-step explanation:

\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \tohe no alive  because ⇆ω⇆π⊂∴∨α∈\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq  \lim_(n \to \infty) a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\n4&5&6\n7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\n4&5&6\n7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_(n \to \infty) a_n \left[\begin{array}{ccc}1&2&3\n4&5&6\n7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\n4&5&6\n7&8&9\end{array}\right] \beta x_{12 \lim_{n \to

A rectangle measures 14cm by 4cm what is the area

Answers

Answer:

56cm^(2)

Step-by-step explanation:

A rectangle measures 14 cm by 4 cm

Length of Rectangle= 14 cm

Breadth  of Rectangle= 4 cm

Area of Rectangle= Length × Breadth

Area of rectangle= 14 cm× 4 cm

Hence,Area of rectangle = 56 cm^(2)

Hence, the correct answer is 56  cm^(2)

The answer is 56 so I am getting how many points?