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aleksklad [387]

3 years ago

10

LeAnn is purchasing.giftwrap for a box that measures 10 inches long, 6 inches wide, and 6 inches tall. Calculate the total area

of the box that will be covered in.giftwrap. Show all of your work for full credit.

2 answers:

tigry1 [53]3 years ago

6 0

Answer:

the answers is 312 in²

Step-by-step explanation:

hope this help

choli [55]3 years ago

3 0

Answer:

$\boxed{312~in^2}$

Step-by-step explanation:

(P) Long = 10 in

(L) Wide = 6 in

(T) Tall = 6 in

.

So, Total area is

$= 2(P\times L + P \times T + L\times T)$

$= 2(10\times 6 + 10 \times 6 + 6\times 6)$

$= 2(60 + 60 + 36)$

$= 2(156)$

$= 312~in^2$

.

Happy to help :)

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alukav5142 [94]

6+2=8
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8 0

3 years ago

Read 2 more answers

Nora's paper airplane flew 9 1/6 feet. That was 2 3/4 feet farther than Max's plane flew. Write and solve an equation to show ho

stepan [7]

Answer:

Max's paper airplane flew $\frac{77}{12}$ feet or $6\frac{5}{12}$ feet.

Step-by-step explanation:

As Nora's paper airplane flew 9 1/6 feet

i.e.

$9\frac{1}{6}=\frac{55}{6}$

Nora's paper was farther than Max's plane flew by = 2 3/4 feet

i.e.

$2\frac{3}{4}=\frac{11}{4}$

Thus the equation to show how far Max's paper airplane flew is:

$\:\:9\frac{1}{6}-2\frac{3}{4}$

$=\frac{55}{6}-\frac{11}{4}$

$\mathrm{Least\:Common\:Multiplier\:of\:}6,\:4:\quad 12$

$\mathrm{Adjust\:Fractions\:based\:on\:the\:LCM}$

$=\frac{110}{12}-\frac{33}{12}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$=\frac{110-33}{12}$

$\mathrm{Subtract\:the\:numbers:}\:110-33=77$

$=\frac{77}{12}$

$=6\frac{5}{12}$

Therefore, Max's paper airplane flew $\frac{77}{12}$ feet or $6\frac{5}{12}$ feet.

6 0

3 years ago

I need help for this, thanks

gulaghasi [49]

(a) Yes all six trig functions exist for this point in quadrant III. The only time you'll run into problems is when either x = 0 or y = 0, due to division by zero errors. For instance, if x = 0, then tan(t) = sin(t)/cos(t) will have cos(t) = 0, as x = cos(t). you cannot have zero in the denominator. Since neither coordinate is zero, we don't have such problems.

---------------------------------------------------------------------------------------

(b) The following functions are positive in quadrant III:

tangent, cotangent

The following functions are negative in quadrant III

cosine, sine, secant, cosecant

A short explanation is that x = cos(t) and y = sin(t). The x and y coordinates are negative in quadrant III, so both sine and cosine are negative. Their reciprocal functions secant and cosecant are negative here as well. Combining sine and cosine to get tan = sin/cos, we see that the negatives cancel which is why tangent is positive here. Cotangent is also positive for similar reasons.

5 0

3 years ago

The probability that a lab specimen contains high levels of contamination is 0.12. A group of 4 independent samples are checked.

sergiy2304 [10]

Answer:

0.5996 is the probability that none contain high levels of contamination.

Step-by-step explanation:

We are given the following information:

We treat lab specimen containing high levels of contamination as a success.

P( lab specimen containing high levels of contamination) = 0.12

Then the number of lab specimens follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 4

We have to find the probability that none of the lab specimen consist of high level of contamination.

We have to evaluate:

$P(x = 0)\\= \binom{4}{0}(0.12)^0(1-0.12)^{(4-0)}\\= 0.5996$

0.5996 is the probability that none contain high levels of contamination.

5 0

4 years ago

Rectangle perimeter please

-Dominant- [34]

Answer:

624

Step-by-step explanation:

Width: 104

Length: 208

104+104+208+208=624

3 0

3 years ago

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