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adell [148]
3 years ago
13

Dedra said that 3x + 5 and 18 + x are equivalent expressions because you can replace the x in both equations with a 4. Tomas sai

d they are not equivalent because when you replace x with 5 in each expression, they don’t have the same value. Who is correct?
A. Dedra is correct
B. Tommy is correct
C. Neither person is correct
D. Both people are correct

Mathematics
2 answers:
pogonyaev3 years ago
6 0
If x = 4, then

3x + 5 = 3*4+5 = 17
and,
18 + x = 18 + 4 = 22

Dedra is not correct then.

If x = 5

3x + 5 = 20
and,
18 + x = 23

Therefore, Tommy is correct.
Oksi-84 [34.3K]3 years ago
3 0

Answer:

B

Step-by-step explanation:

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20 points best answer gets brainliest
lana66690 [7]

<em>Note: The second image should be titled "Data Set #2"</em>

It seems like you've already answered a part of this question yourself, but let's get into the details.

<h3>Parts A and B: 5 Number Summary</h3>

The five values the questions ask you to find the minimum, maximum, median, and the first and third quartiles for both of the data sets. These data are frequently called the <em>five number summary</em> of a data set, and we can use them to create a <em>box plot</em> of our data. The meanings of maximum and minimum are pretty obvious - they're just the biggest and smallest values in the set - but the median and the first and third quartiles all refer to different "middles" in the set.

The <em>median</em> is the "middle value" of a set of ordered data. When we have an odd number of data points, the median is simply the middle number of the set, but when we have an even number, as is the case with these two data sets, we have to find the number halfway between the two middle values. In data set 1, that number is

\dfrac{16+25}{2}=\dfrac{41}{2}=20.5

In data set 2, it's the number halfway between 8 and 10, which is 9.

The median splits any set of data into two parts: all the data points <em>smaller </em>than the median, and all of those <em>larger</em> than the median. In data set one, it's the two subsets {1, 4, 9, 16} and {25, 36, 49, 64}. The median of the smaller set gives us the <em>first quartile</em>, and the median of the larger one gives us the <em>third quartile.</em>

Why "first" and "third" quartile? Where are the second and fourth ones? While they don't go be the titles officially, those values are already part of our five number summary:

  • First quartile
  • Second quartile (the median)
  • Third quartile
  • Fourth quartile (the maximum)
<h3>Part C: Range, spread, and box plots</h3>

To get a visual for how our data is spread out, we can visualize our five-number-summary with a <em>box plot</em>. I've created a box plot for each of the data sets in the first two image uploads. The little nubs on the far ends, sometimes called the "whiskers" of the plot, are the minimum and maximum of the data set; the "box" represents the <em>interquartile range</em> of the data: all the values between the first and third quartile of the data; and the notch going down through the box is the median of the data.

We can see at a glance that data set 1 spans a far greater range of values that data set 2, and that its data points tend to be more concentrated in the lower values. Data set 2, by contrast, is much more uniform; its median lies right in the center of its range, and the "box" is centered similarly along it.

Comparing the medians of two data sets, especially those with the same number of values, can give us valuable information as to how much "larger" or "smaller" one set is than the other, but we need to bring in the other numbers in the five-number summary for a better picture about how that data is spread out.

<h3>Part D: Histograms vs. Box Plots: Which one is better?</h3>

There's no correct answer to this, because each type of graph gives us insight into different aspect of a data set.

A box and whisker plot is great for understanding:

  • The range of a set of data
  • Its spread
  • Its center

While a histogram can reveal:

  • <em>How </em>and <em>where</em> values are concentrated
  • Gaps and outliers

The histogram of data set 1, set to constant intervals of 7 units, shows us that many of the values at the lower end, and get more spread out as we go further - the empty patches become more frequent as we continue to the right, suggesting that our values will become more sparse as they get larger.

Contrast that with data set 2, which has a totally flat, uniform distribution when viewed at a constant interval of 4 units. The box plot and histogram work in tandem to give us a visual, quantitative picture of our data which we can use to make informed conclusions about it.

6 0
3 years ago
Solve the equation for all real solutions in simplest form.<br> 3x2 - 12x + 11 = x
Ivanshal [37]

Answer:

(13 + or - (sign) square root(37)) / 6

Step-by-step explanation:

First search up the quadratic formula which is

x = (- b + or - square root(b^2 - 4ac)) / 2a

and to find a, b, and c you need this formula

ax^2 + bx + c = 0 (btw this is the safest solution to finding your answer with a number in-front of a quadratic equation.)

Math:

a = 3, b = -13, c = 11

b = -13, because of the x on the other end of the equation as the equation above "ax^2 + bx + c = 0" you need the zero to proceed to this next step.

Side Note: You only take the number and not the variable, variables are x, y, w, z, or etc.

(- (- 13) + or - square root((-13)^2 - 4(3)(11) ) ) / 2 * 3

13 + or - square root(37) / 6

5 0
3 years ago
Graph the equation (x - 3)^2 + (y - 2)^2 = 25 and y = 6.
Dmitriy789 [7]

Answer:

x=6, x=0

Step-by-step explanation:

(x - 3) ^{2}+ (y - 2) ^{2}= 25

(x - 3) ^{2}+ (6 - 2) ^{2}= 25

x=6, x=0

3 0
3 years ago
On a morning of a day when the sun will pass directly​ overhead, the shadow of a 60​-ft building on level ground is 25 ft long.
AysviL [449]

Answer:

The shadow is decreasing at the rate of 3.55 inch/min

Step-by-step explanation:

The height of the building = 60ft

The shadow of the building on the level ground is 25ft long

Ѳ is increasing at the rate of 0.24°/min

Using SOHCAHTOA,

Tan Ѳ = opposite/ adjacent

= height of the building / length of the shadow

Tan Ѳ = h/x

X= h/tan Ѳ

Recall that tan Ѳ = sin Ѳ/cos Ѳ

X= h/x (sin Ѳ/cos Ѳ)

Differentiate with respect to t

dx/dt = (-h/sin²Ѳ)dѲ/dt

When x= 25ft

tanѲ = h/x

= 60/25

Ѳ= tan^-1(60/25)

= 67.38°

dѲ/dt= 0.24°/min

Convert the height in ft to inches

1 ft = 12 inches

Therefore, 60ft = 60*12

= 720 inches

Convert degree/min to radian/min

1°= 0.0175radian

Therefore, 0.24° = 0.24 * 0.0175

= 0.0042 radian/min

Recall that

dx/dt = (-h/sin²Ѳ)dѲ/dt

= (-720/sin²(67.38))*0.0042

= (-720/0.8521)*0.0042

-3.55 inch/min

Therefore, the rate at which the length of the shadow of the building decreases is 3.55 inches/min

8 0
4 years ago
Hey i am not smart but what is <br> <img src="https://tex.z-dn.net/?f=83%5Cfrac%7B1%7D%7B3%7D%20minus%20%5Cfrac%7B2%7D%7B3%7D" i
Ray Of Light [21]
It would be 82 2/3 since 1/3 minis 2/3 would equal -1/3 you subtract that from 83 and get 82 2/3
6 0
3 years ago
Read 2 more answers
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