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riadik2000 [5.3K]
4 years ago
6

The cost of a set of four tires is $192 what is the unit rate for the cost of one tire

Mathematics
1 answer:
Dmitry_Shevchenko [17]4 years ago
6 0

Answer:

I believe the unit rate is 48 dollars per tire.

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Write the expression as a single logarithm
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3\log_7y + 4(\log_7x - 5\log_7z)\\\log_7y^{3} + 4(\log_7x - 5\log_7z)\\\log_7y^{3} + 4(\log_7x - \log_7z^{5})\\\log_7y^{3} + 4(\log_7(\frac{x}{z^{5}}))\\\log_7y^{3} + \log_7(\frac{x}{z^{5}})^{4}\\\log_7(y^{3} \times (\frac{x}{z^5{5}})^{4})

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3 years ago
The quotient of 3,200 and 4 will have -blank- zeros.
Nostrana [21]
B. It will have two zeros. Hope this helps!
3 0
3 years ago
Read 2 more answers
I’m confused and don’t have an angle tool to even try to do the question
yanalaym [24]
<h3>Answer:   40</h3>

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

Explanation:

JQ is longer than QN. We can see this visually, but the rule for something like this is the segment from the vertex to the centroid is longer compared to the segment that spans from the centroid to the midpoint.

See the diagram below.

The ratio of these two lengths is 2:1, meaning that JQ is twice as long compared to QN. This is one property of the segments that form when we construct the centroid (recall that the centroid is the intersection of the medians)

We know that JN = 60

Let x = JQ and y = QN

The ratio of x to y is x/y and this is 2/1

x/y = 2/1

1*x = y*2

x = 2y

Now use the segment addition postulate

JQ + QN = JN

x + y = 60

2y + y = 60

3y = 60

y = 60/3

y = 20

QN = 20

JQ = 2*y = 2*QN = 2*20 = 40

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

We have

JQ = 40 and QN = 20

We see that JQ is twice as larger as QN and that JQ + QN is equal to 60.

7 0
3 years ago
PLEASE ANSWERRRRRR
Ierofanga [76]

Answer: 4.8

Step-by-step explanation:

4 0
3 years ago
sam is eating a hamburger. He at 20% of the hamburger with his first bite, 20% of what is left in his second bite, and so on, ea
ryzh [129]

The hamburger will weigh 250 g, if there was originally 160g of it remaining after the second bite given that 20% of the hamburger with his first bite, 20% of what is left in his second bite, and so on, eating 20% of what remains with each bite. This can be obtained by assuming the total weight as x and multiplying the percentages.

<h3>How much did the hamburger weigh?</h3>

Let the total weight of the hamburger be x.

  • First it is given that Sam ate 20% of the hamburger.

The first bite will be,

20% of the hamburger = 20% × x

20% of the hamburger = 20/100 × x

20% of the hamburger = 20x/100

⇒ first bite = 20x/100

Therefore the remaining hamburger will be,

remaining hamburger = x - 20x/100

remaining hamburger = (100x - 20x)/100

⇒ remaining hamburger after first bite = 80x/100

  • Next it is given that Sam ate 20% of the remaining hamburger.

The second bite will be,

20% of the remaining hamburger = 20% × 80x/100

20% of the remaining hamburger = 20/100 × 80x/100

20% of the remaining hamburger = 16x/100

⇒ second bite = 16x/100

Therefore the remaining hamburger will be,

remaining hamburger = 80x/100 - 16x/100

remaining hamburger after second bite = 64x/100

  • It is given that there was originally 160g of it remaining after the second bite.

Then, 64x/100 = 160

x = 160 × 100/64

x = 250 g

Hence the hamburger will weigh 250 g, if there was originally 160g of it remaining after the second bite given that 20% of the hamburger with his first bite, 20% of what is left in his second bite, and so on, eating 20% of what remains with each bite.

Learn more about percentages here:

brainly.com/question/25395

#SPJ4

5 0
2 years ago
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