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Maslowich
3 years ago
9

Help pls D: i'll worship you if you do <33

Mathematics
1 answer:
Stells [14]3 years ago
7 0

Answer:

b

Step-by-step explanation:

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A vendor sells pizza and soft drinks.A customer buys 8 slices of pizza and 6 drinks for$ 39.50.Another customer buys 5 slices of
otez555 [7]

Answer:

$3.61

Step-by-step explanation:

let p represent pizza and d represent drink, then

8p + 6d = 39.5 → (1)

5p + 7d = 30.44 → (2)

Multiplying (1) by 7 and (2) by - 6 and adding will eliminate d

56p + 42d = 276.5 → (3)

- 30p - 42d = - 182.64 → (4)

Add (3) and (4) term by term

(56p - 30p) + (42d- 42d) = (276.5 - 182.64), that is

26p = 93.86 ( divide both sides by 26 )

p = 3.61

That is a slice of pizza costs $3.61

7 0
4 years ago
What is 6/8 divided by 78/9
aliya0001 [1]
So, when you divide by a fraction, you are basically multiplying by the reciprocal of said fraction. Ergo:
6/8 * 9/78 = 54 / 624
4 0
4 years ago
Use the squared identities to simplify 2sin2x cos2x. Check below.
PilotLPTM [1.2K]

To simplify 2sin2x cos2, you need to use (1 - cos(4x))/4. The correct answer between all the choices given is the first choice or letter A. I am hoping that this answer has satisfied your query about and it will be able to help you.

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I need help on a equation −4a5⋅6a5​
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7 0
3 years ago
Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by
a_sh-v [17]

Answer:

Dimension of the box is 16.1\times 7.1\times 2.45

The volume of the box is 280.05 in³.

Step-by-step explanation:          

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the box?

Solution :

Let h be the height of the box which is the side length of a corner square.

According to question,

A sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

The length of the box is L=21-2h

The width of the box is W=12-2h

The volume of the box is V=L\times W\times H

V=(21-2h)\times (12-2h)\times h

V=(21-2h)\times (12h-2h^2)

V=252h-42h^2-24h^2+4h^3

V=4h^3-66h^2+252h

To maximize the volume we find derivative of volume and put it to zero.

V'=12h^2-132h+252

0=12h^2-132h+252

Solving by quadratic formula,

h=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

h=\frac{-(-132)\pm\sqrt{132^2-4(12)(252)}}{2(12)}

h=\frac{132\pm72.99}{24}

h=2.45,8.54

Now, substitute the value of h in the volume,

V=4h^3-66h^2+252h

When, h=2.45

V=4(2.45)^3-66(2.45)^2+252(2.45)

V\approx 280.05

When, h=8.54

V=4(8.54)^3-66(8.54)^2+252(8.54)

V\approx -170.06

Rejecting the negative volume as it is not possible.

Therefore, The volume of the box is 280.05 in³.

The dimension of the box is

The height of the box is h=2.45

The length of the box is L=21-2(2.45)=16.1

The width of the box is W=12-2(2.45)=7.1

So, Dimension of the box is 16.1\times 7.1\times 2.45

6 0
4 years ago
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