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dexar [7]
2 years ago
15

LeAnn is purchasing giftwrap for a box that measures 8 inches long, 6 inches wide, and 6 inches tall. Calculate the total area o

f the box that will be covered in gift wrap. Show all of your work for full credit.
Mathematics
2 answers:
Pavel [41]2 years ago
8 0

Let

L---------> the length side of the box

W--------> the width side of the box

H-------> the height of the box

we know that

L=8\ in\\W=6\ in\\H=6\ in

the surface area of the box is equal to

S=2*area\ of\ the\ base+perimeter\ of\ the\ base*H

<u>Find the area of the base</u>

A=L*W=8*6=48\ in^{2}

<u>Find the perimeter of the base</u>

P=2*L+2*W=2*8+2*6=28\ in

<u>Find the surface area</u>

S=2*48+28*6

S=264\ in^{2}

therefore

<u>the answer is</u>

the total area of the box that will be covered in gift wrap is 264\ in^{2}

sertanlavr [38]2 years ago
5 0
A box shape in three dimensional space. Formally, a polyhedron for which all faces are rectangles. 
Rectangular Parallelepiped
Volume = lwh 8*6*6=288Lateral Surface Area = 2lh + 2wh=94Surface Area = 2lw + 2lh + 2wh=288
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Answer:

Probability of a flowering plant bearing flowers in January if it is a peony= 70 %

Step-by-step explanation:

We know that:

Probability =Total favourable outcomes/ Total possible outcomes

Probability of flowering plants bearing flowers in January is given in the table. It means that out of 100 peony  70 peony flowers in January.

There are five kinds of flowers.

We have to select peony and find it's Probability.

As all flowers are independent, their flowering phenomenon does not depend on each other.

So,Probability of a flowering plant bearing flowers in January if it is a peony= 70 %

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50 % of them participated

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Gavin has 460 baseball players in his collection of baseball cards,and %15 of the players are pitchers.How many pitchers are in
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460 times 0.15
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An observation of the Moon was conducted from Friday, November 8, 2013 to Thursday, November 14, 2013. The study of the Moon during this period occurred consistently between the hours of 8 and 9 p.m. EST within the Northern Hemisphere at 37.3346° N, 79.5228° W (Bedford, V.A.). The Moon was noted to be illuminated on the right side and had a dark shadow on the left side indicating a waxing phase. The light region grew over the surface of the Moon with each subsequent night. The first night’s phase was waxing crescent with over 25 percent of the Moon lit up. The next night, the light had grown to cover more of the Moon as it continued through its waxing crescent phase. On November 10th, the Moon exhibited traits of being at first-quarter or half-moon status because at least 50 percent of its surface was illuminated. In the following nights, the Moon displayed characteristics of waxing gibbous as the light continued to grow across the moon’s surface from right to left. The Moon was nearing closer to the full moon phase on November 14th as only a very small dark shadow was visible on the left side. 
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4 0
2 years ago
Find the values of k for which the line y=1-2kx does not meet the curve y=9x²-(3k+1)x+5
Lilit [14]

Let's equate the two given functions and attempt to solve for x:

y = 1 -2kx = y = 9x^2 -(3k+1)x + 5

Eliminating y, 1 -2kx = 9x^2 -(3k+1)x + 5

Rearranging terms in descending order by powers of x:

0 = 9x^2 - (3k+1)x + 2kx + 5 - 1 , or

0 = 9x^2 - kx - x + 4

This is a quadratic equation with coefficients a = 9, b = -(k+1) and c = 4.

For certain k, not yet known, solutions exist. Solutions here implies points at which the two curves intersect.

k+1 plus or minus sqrt( [-(k+1)]^2 - 4(9)(4) )

x = -----------------------------------------------------------------

2(9)

The discriminant is k^2 + 2k + 1 - 144, or k^2 + 2k - 143.

If the discriminant is > 0, there are two real, unequal roots. We don't want this, since we're interested in finding k value(s) for which there's no solution.

If the discr. is = 0, there are two real, equal roots. Again, we don't want this.

If the discr. is < 0, there are no real roots. This is the case that interests us.

So our final task is to determine the k values for which the discr. is < 0:

Determine the k value(s) for which the discriminant, k^2 + 2k - 143, is 0.

This k^2 + 2k - 143 factors as follows: (k-11)(k+13), and when set = to 0, results in k: {-13,11}.

Set up intervals on the number line: (-infinity, - 13), (-13, 11) and (11, infinity).

Choosing a test number from each interval, determine the interval or intervals on which the discriminant is negative:

Case 1: k = -15; the discriminant (k^2 + 2k - 143) is (-15)^2 + 2(-15) - 143 = +52. Reject this interval

Case 2: k = 0; the discriminant is then 0 + 0 - 143 (negative); thus, the discriminant is negative on the interval (-13,11).

Case 3: k = 20; the discriminant is positive. Reject this interval.

Summary: The curves do not intersect on the interval (-13,11).

4 0
2 years ago
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