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Ivanshal [37]
3 years ago
14

The cost for fudge at Candy Lane is represented by the table. The graph shows the cost of a similar fudge that Best Fudge Shop s

ells.
Which is the difference in cost when purchasing 5 pounds of fudge at each store?
Mathematics
2 answers:
Vladimir [108]3 years ago
5 0

Can you add an attachment to the table, so we can see the prices.

Brilliant_brown [7]3 years ago
5 0

Answer:The answer is B) i got it right o USATESTPREP

Step-by-step explanation:

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Find the volume. SHOW WORK
Anettt [7]

Answer:

827.39

Step-by-step explanation:

Volume = L X W X H

So 6.2 X 8.5 X 15.7

6.2 X 8.5 = 52.7

52.7 X 15.7 = 827.39

So the volume is <u>827.39</u>

3 0
2 years ago
Read 2 more answers
Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______
Ludmilka [50]

Answer:

(a)\ \sec^2(\theta) = 82

(b)\ \cot(\theta) = \frac{1}{9}

(c)\ \cot(\frac{\pi}{2} - \theta) = 9

(d)\ \csc^2(\theta) = \frac{82}{81}

Step-by-step explanation:

Given

\tan(\theta) = 9

Required

Solve (a) to (d)

Using tan formula, we have:

\tan(\theta) = \frac{Opposite}{Adjacent}

This gives:

\frac{Opposite}{Adjacent} = 9

Rewrite as:

\frac{Opposite}{Adjacent} = \frac{9}{1}

Using a unit ratio;

Opposite = 9; Adjacent = 1

Using Pythagoras theorem, we have:

Hypotenuse^2 = Opposite^2 + Adjacent^2

Hypotenuse^2 = 9^2 + 1^2

Hypotenuse^2 = 81 + 1

Hypotenuse^2 = 82

Take square roots of both sides

Hypotenuse =\sqrt{82}

So, we have:

Opposite = 9; Adjacent = 1

Hypotenuse =\sqrt{82}

Solving (a):

\sec^2(\theta)

This is calculated as:

\sec^2(\theta) = (\sec(\theta))^2

\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2

Where:

\cos(\theta) = \frac{Adjacent}{Hypotenuse}

\cos(\theta) = \frac{1}{\sqrt{82}}

So:

\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2

\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2

\sec^2(\theta) = (\sqrt{82})^2

\sec^2(\theta) = 82

Solving (b):

\cot(\theta)

This is calculated as:

\cot(\theta) = \frac{1}{\tan(\theta)}

Where:

\tan(\theta) = 9 ---- given

So:

\cot(\theta) = \frac{1}{\tan(\theta)}

\cot(\theta) = \frac{1}{9}

Solving (c):

\cot(\frac{\pi}{2} - \theta)

In trigonometry:

\cot(\frac{\pi}{2} - \theta) = \tan(\theta)

Hence:

\cot(\frac{\pi}{2} - \theta) = 9

Solving (d):

\csc^2(\theta)

This is calculated as:

\csc^2(\theta) = (\csc(\theta))^2

\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2

Where:

\sin(\theta) = \frac{Opposite}{Hypotenuse}

\sin(\theta) = \frac{9}{\sqrt{82}}

So:

\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2

\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2

\csc^2(\theta) = \frac{82}{81}

4 0
3 years ago
Please help right away
andrew-mc [135]

Answer:

100

Step-by-step explanation:

Mixed candy question... Skittles jar... to be filled with Jelly beans.

Let's first calculate the volume of the jar.  We'll assume it's a regular cylindrical prism jar, unlike the one on the photo which is narrower on top.

V = π * r² * h = π * (3.5)² * 11.5 = 140.875 π = 442.6 cubic cm

Now, we don't have the precise measurement of a jelly bean, but we know it's roughly 2-3 cubic cm.  The precision isn't needed to answer this question, just to have a rough idea... it's no 300 cu cm per jelly bean.

So, let's assume a 3 cu cm per jelly bean (2 cu cm wouldn't the final answer)....

442.6 / 3 = 147.5 jelly beans, approximately.

So, can they fit 100,000?  No

Can we fit 10,000 in there?  No

Can we fit 100?  Yes.

Can we fit 1?  Certainly

The most reasonable lower-limit would then be 100.

4 0
3 years ago
Find three consecutive even integers such that the sum of the least integer and the middle integer is 2424 more than the greates
SIZIF [17.4K]
Let the smallest integer be x
The second integer = x + 2
The third integer = x + 4

x + x + 2 = x + 4 + 24
2x + 2 = x + 28
2x - x = 28 - 2
x = 26

So the numbers are 26, 28, 30
5 0
3 years ago
What is the solution for x in the equation? 16x − 4 + 5x = -67 A.x=-3 B.x=3 C.x=1/3 D. x=-1/3
lozanna [386]

Answer:

A. X=-3

Step-by-step explanation:

16x-4+5x=-67

21x=-67+4

21x=-63

Divide through by 21

X=-3

8 0
3 years ago
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