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3 years ago

6

in a total of $580,800 last year. If they served an average of 200 customers each day over the 363 days that the deli was open,

what was the average amount that each customer spent?

2 answers:

brilliants [131]3 years ago

7 0

I think it would be $8.

Finger [1]3 years ago

7 0

Answer:

8 dollars just did unit test

Step-by-step explanation:

remember to share the love and stay safe wear a mask remember you are loved

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What is the surface area of a cube that’s measured 8 inches

MrRa [10]

Answer:

2

Step-by-step explanation:

Haven't done this in awhile but I think it's 2 because if you cube 2 it comes to 8. L x W x H = 8 | 2 x 2 x 2 = 8

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2 years ago

Please help 30 points Really hard mulitiquestions

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3 years ago

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

3 years ago

The admission fee for a charity event is $7 for children and $10 for adults. The event was attended by 700 people and the total

Hitman42 [59]

X = 7
y = 10
x + y = 700
7x + 10y = 6400

10x + 10y = 7000
(-) 7x + 10y = 6400
______________
3x = 600
x = 200

200 + y = 700
y = 500

3 0

3 years ago

The product of 43x6 can be broken down into expand form

o-na [289]

Answer:yes

Step-by-step explanation:

any number can be expanded. 200+50+08

7 0

3 years ago

Read 2 more answers