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borishaifa [10]
3 years ago
5

Again, easy but desperate

Mathematics
2 answers:
Ymorist [56]3 years ago
8 0
Yes it’s proportional

2 pounds = 9.98
9.98 divided by 2 = 4.99
5 x 4.99 = 24.95
hammer [34]3 years ago
6 0

Answers:

  • A) Yes, the total cost is proportional to the number of pounds of chocolate.
  • B) Explanation is below

=================================================

Let,

x = number of pounds of chocolate

y = cost of buying x pounds of chocolate

If you buy x = 2 pounds, then you spend y = 9.98 dollars.

So k = y/x = 9.98/2 = 4.99 is the cost per pound. This is known as the unit price or unit cost. It's the cost of one pound of chocolate.

Now if you buy x = 5 pounds, then you'll be charged a total of y = 24.95

So, k = y/x = 24.95/5 = 4.99

We get the same constant of proportionality here, and therefore, this is a direct proportional relationship. Each time x goes up by 1, y goes up by 4.99; in other words, each time you buy 1 extra pound of chocolate, you pay $4.99 more. This is to be expected whenever you're at the grocery store as long as you ignore things like coupons or bulk discounts or things of that nature.

The equation of this line is y = 4.99x

This shows the slope is 4.99 and the y intercept is 0.

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

An alternative approach:

Again we define x and y as stated above. We see that the two points are (2,9.98) and (5,24.95)

Computing the slope gets us

m = (y2-y1)/(x2-x1)

m = (24.95-9.98)/(5-2)

m = 14.97/3

m = 4.99

So this matches what we got earlier.

Now turn to y = mx+b. Plug in that value of m along with (x,y) = (2,9.98)

Doing so gets us 9.98 = 4.99*2+b

Let's solve for b

9.98 = 4.99*2+b

9.98 = 9.98+b

0 = b

b = 0

I subtracted 9.98 from both sides to isolate b. The fact we get the y intercept is 0 helps show that we have a direct proportion. All direct proportion linear equations go through the origin (0,0). This is to reflect that buying 0 pounds of chocolate means you pay 0 dollars.

We get the same result as the previous section.

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Analyze the diagram below and complete the instructions that follow.<br><br> Find m
Stells [14]

Answer:

D. 70.53

Step-by-step explanation:

To solve this problem, one must use the right angle trigonometric rations. These ratios can be used to describe the ratio between the sides and an angle in a right triangle. Please note, each side is named relative to the angle that one is looking at, thus the same side can acquire different names based on the angle one uses to describe it. The trigonometric ratios are the following:

sin(\theta)=\frac{opposite}{hypotenuse}\\\\cos(\theta)=\frac{adjacent}{hypotenuse}\\\\tan(\theta)=\frac{opposite}{adjacent}

In this case, one is asked to find the measure of (<CBA), one is given the hypotenuse (the side opposite the right angle), and the side adjacent to this angle. Therefore, one should use the cosine (cos) function to find this angle.

cos(\theta)=\frac{adjacent}{hypotenuse}

Substitute,

cos(CBA) = \frac{3}{9}

Simplify

cos(CBA) = \frac{1}{3}

Inverse operations,

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7 0
3 years ago
Please help me, I’ll give brainliest
kolbaska11 [484]

Answer:

<em>x</em><em> </em><em>=</em><em> </em><em>4</em><em>5</em>

<em>y</em><em> </em><em>=</em><em> </em><em>5</em>

Step-by-step explanation:

<em>it</em><em> </em><em>is</em><em> </em><em>b</em><em>ecause</em>

<em>9</em><em>0</em><em>/</em><em>2</em><em> </em><em>=</em><em> </em><em>4</em><em>5</em>

<em>(</em><em>9</em><em>0</em><em>+</em><em>1</em><em>0</em><em>)</em><em>/</em><em>2</em><em>0</em><em> </em>

<em>(</em><em>1</em><em>0</em><em>0</em><em>)</em><em>/</em><em>2</em><em>0</em><em> </em><em>=</em><em> </em><em>,</em><em>5</em>

7 0
4 years ago
Can someone give me the answer to this
vredina [299]
You solve this problem via PEMDAS (Parentheses Exponents Multiplication Division Addition Subtraction.) 

To start you simplify everything inside parentheses.

9+3*5+5^2

Then you simplify the exponents.

9+3*5+25

Next, you simplify the multiplication signs.

9+15+25

Then you simplify all the addition signs from left to right.

24+25

49
6 0
4 years ago
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