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adelina 88 [10]
3 years ago
13

A box has a length of 15 in., a width of 7 in., and a height of 9 in. If Tara plans to cover the box with wrapping paper, how mu

ch wrapping paper will she need?
Mathematics
1 answer:
maxonik [38]3 years ago
4 0
We have to calculate the area of the box.

area (box)=2(lenght x width) + 2(lenght x height) +2(width x height)

area (box)=2(15 in x 7 in) + 2(15 in x 9 in) + 2(7 in x 9 in)=
=2(105 in²)+2(135 in²)+2(63 in³)=210 in²+270 in²+126 in²=606 in².

Answer: she need 606 in² of wrapping paper.

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Solve for theta <br> Please answer both parts
Vika [28.1K]

Answer:

Step-by-step explanation:

a)  tanθ = x/3

θ = arctan(x/3)

:::::

b)  cosθ = 9/x

θ = arccos(9/x)

4 0
3 years ago
Describe at least two differences between constructing an inscribed regular hexagon and constructing an inscribed square.
liberstina [14]

Correct answer is A.

The diameter of the circle is used for the square construction, but the radius of the circle is used for the regular hexagon construction.

<h3>How to calculate the diameter?</h3>

If you know the radius of the circle, multiply by 2 to get the diameter. The radius is the distance from the center of the circle to the edge. For example, if the radius of a circle is 4 cm, then the diameter is 4 cm x 2 = 8 cm. If you know the circumference of the circle, divide by π to get the diameter.

<h3>Briefing:</h3>

The diameter of the circle is used for the square construction, but the radius of the circle is used for the regular hexagon construction which is A.

The diameter of the circle is used for the square as l=2r therefore corresponds to the diameter, while for a hexagon r=r being the radius of the circle.

To know more about Diameter visit:

brainly.com/question/5501950

#SPJ4

I understand that the question you are looking for is:

Describe at least two differences between constructing an inscribed regular hexagon and constructing an inscribed square.

A) The diameter of the circle is used for the square construction, but the radius of the circle is used for the regular hexagon construction.

B) The radius of the circle is used for the square construction, but the diameter of the circle is used for the regular hexagon construction.

C) The square will need two arcs along the circle, but the hexagon will need two arcs above and two arcs below the diameter of the circle.

D) The square will need six arcs along the circle, but the hexagon will need two arcs above and two arcs below the diameter of the circle.

7 0
2 years ago
I think of a number multiply it by 2, subtract 5 and multiply the result by 2
zalisa [80]
Call our original number n. Then here's what we did in an equation:
(n*2-5)*2=2

We can solve this for n:
n*2-5=1
n*2=6
n=3
7 0
3 years ago
Find the limit, if it exists. (If an answer does not exist, enter DNE.) lim (x, y) → (0, 0) x4 − 34y2 x2 + 17y2
HACTEHA [7]

Answer:

<h2>DNE</h2>

Step-by-step explanation:

Given the limit of the function \lim_{(x,y) \to (0,0)} \frac{x^4-34y^2}{x^2+17y^2}, to find the limit, the following steps must be taken.

Step 1: Substitute the limit at x = 0 and y = 0 into the function

= \lim_{(x,y) \to (0,0)} \frac{x^4-34y^2}{x^2+17y^2}\\=  \frac{0^4-34(0)^2}{0^2+17(0)^2}\\= \frac{0}{0} (indeterminate)

Step 2: Substitute y = mx int o the function and simplify

= \lim_{(x,mx) \to (0,0)} \frac{x^4-34(mx)^2}{x^2+17(mx)^2}\\\\= \lim_{(x,mx) \to (0,0)} \frac{x^4-34m^2x^2}{x^2+17m^2x^2}\\\\\\= \lim_{(x,mx) \to (0,0)} \frac{x^2(x^2-34m^2)}{x^2(1+17m^2)}\\\\\\= \lim_{(x,mx) \to (0,0)} \frac{x^2-34m^2}{1+17m^2}\\

= \frac{0^2-34m^2}{1+17m^2}\\\\=  \frac{34m^2}{1+17m^2}\\\\

<em>Since there are still variable 'm' in the resulting function, this shows that the limit of the function does not exist, Hence, the function DNE</em>

4 0
3 years ago
Factor.
Veseljchak [2.6K]
The answer is B. (6x + 4)(x - 1). I hope this helps!
5 0
3 years ago
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