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marishachu [46]
1 year ago
12

What is the product?(4y − 3)(2y2 + 3y − 5)8y3 + 3y + 158y3 − 23y + 158y3 − 6y2 − 17y + 158y3 + 6y2 − 29y + 15

Mathematics
1 answer:
motikmotik1 year ago
5 0

We need to find the product of :

\mleft(4y-3\mright)\mleft(2y2+3y-5\mright)

So, the result as following:

\begin{gathered} \mleft(4y-3\mright)\mleft(2y^2+3y-5\mright) \\ =4y\cdot(2y^2+3y-5)-3\cdot(2y^2+3y-5) \\ =8y^3+12y^2-20y-(6y^2+9y-15) \\ =8y^3+12y^2-20y-6y^2-9y+15 \\  \\ =8y^3+6y^2-29y+15 \end{gathered}

So, the answer is the option 4. 8y3 + 6y2 − 29y + 15​

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Answer: 145.8

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1. P(x)=x^2+2<br> 2. P(x)= 2x +10
ruslelena [56]

Answer:

see explanation

Step-by-step explanation:

Since p(x) = x² + 2 , if x ≤ 4

For x = 4 then this is included in the inequality x ≤ 4

whereas x > 4 does not include x = 4 but values greater than 4

Thus to evaluate x = 4 use p(x) = x² + 2

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3 years ago
Shawna won 40 super bouncy balls playing
patriot [66]

Answer:

13 not including Shawna

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5 0
2 years ago
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Angi has 32 nickels and dimes. Their total value is $1.90. How many of each coin does she have?
Radda [10]

Answer:

6 dimes, 26 nickels

Step-by-step explanation:

The number of nickels is N, and the number of dimes is D.

N + D = 32

5N + 10D = 190

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5 0
3 years ago
Identify the lower class​ limits, upper class​ limits, class​ width, class​ midpoints, and class boundaries for the given freque
maw [93]

Answer:

The number of individuals included in the summary is 146.

Step-by-step explanation:

The frequency distribution table provided is as follows:

Class Intervals    Frequency

    100 - 199               24

   200 - 299              90

   300 - 399              27

   400 - 499                1

   500 - 599               4

The lower class limit it the smallest value of each class interval.

Lower class limit = {100, 200, 300, 400, 500}

The upper class limit it the highest value of each class interval.

Upper class limit = {199, 299, 399, 499, 599}

The lower class boundaries are the lower class limits decreased by 0.5 and the upper class boundaries are the upper class limits increased by 0.5.

Class boundaries:

   99.5 - 199.5  

  199.5 - 299.5

  299.5 - 399.5

  399.5 - 499.5

  499.5 - 599.5

The class width is the difference between the class boundaries of each class.

Class width = 199.5 - 99.5 = 100

So, the class width is 100.

The midpoints of a class is the average value of the boundaries of a class.

\text{Midpoint}_{100-199}=\frac{\text{Lower class boundary} + \text{Upper class boundary}}{2}\\

                         =\frac{99.5+199.5}{2}\\\\=149.5

\text{Midpoint}_{200-299}=\frac{\text{Lower class boundary} + \text{Upper class boundary}}{2}\\

                         =\frac{199.5+299.5}{2}\\\\=249.5

\text{Midpoint}_{300-399}=\frac{\text{Lower class boundary} + \text{Upper class boundary}}{2}\\

                         =\frac{299.5+399.5}{2}\\\\=349.5

\text{Midpoint}_{400-499}=\frac{\text{Lower class boundary} + \text{Upper class boundary}}{2}\\

                         =\frac{399.5+499.5}{2}\\\\=449.5

\text{Midpoint}_{500-599}=\frac{\text{Lower class boundary} + \text{Upper class boundary}}{2}\\

                         =\frac{499.5+599.5}{2}\\\\=549.5

The number of individuals included in the summary is the sum of all frequencies.

\text{Number of Individuals}=24 + 90 + 27 + 1 + 4=146

Thus, the number of individuals included in the summary is 146.

3 0
3 years ago
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