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myrzilka [38]
3 years ago
9

Which expressions are equivalent to 5+(−3)(6x−5)

Mathematics
1 answer:
boyakko [2]3 years ago
8 0

Answer:

<h2>C. None of the above</h2>

Step-by-step explanation:

5+(-3)(6x-5)\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\\\=5+(-3)(6x)+(-3)(-5)\\\\=5-18x+15\qquad\text{combine like terms}\\\\=-18x+(5+15)\\\\=-18x+20

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What is this one??<br> -5-(15y-1)=2(7y-16)-y
sesenic [268]

Answer:

y=1

Step-by-step explanation:

Let's solve by opening the bracket first

-5-(15y-1)=2(7y-16)-y

Open the bracket

-5-15y+1=14y-32-y

Collect like terms

-5+1+32=14y-y+15y

-5+33=14y+14y

28=28y

Divide both sides by 28

y=1

Therefore the final answer is 1

3 0
3 years ago
The number of classified documents has increased approximately linearly from 8.2 million documents in 2001 to 17.4 million docum
Andrei [34K]

Answer:

  n -8.2 = 2.3(t -1)

Step-by-step explanation:

You are given definitions for the variables t and n, along with two data points:

  (t, n) = (1, 8.2) and (5, 17.4)

The slope of the line between these points is ...

  m = (y2 -y1)/(x2 -x1)

  m = (17.4 -8.2)/(5 -1) = 9.2/4 = 2.3

The equation can be written in point-slope form as ...

  y -k = m(x -h) . . . . . . line with slope m through point (h, k)

  n -8.2 = 2.3(t -1) . . . . a linear model describing the data

______

<em>Additional comment</em>

If we eliminate the parentheses and add 8.2, we get the point-slope equation ...

  n = 2.3t +5.9

4 0
3 years ago
what is (1/2 times 14 times 11) times 9.5 I'm doing volume of triangular prisms in math and i need help doing the muliplcaion pr
Andre45 [30]
\frac{1}{2} X 14=  \frac{1}{28}
\frac{1}{28} X 11=  \frac{1}{308}
\frac{1}{308} X 9.5 =  \frac{1}{2926} or 2926.
Hope this helped!
8 0
3 years ago
Question 10 of 10<br> Which of the following is most likely the next step in the series?
RUDIKE [14]
The answer is most likely b
4 0
3 years ago
Read 2 more answers
In a quadrilateral, a student can draw two diagonals. In a pentagon, a student can draw five diagonals. In a hexagon, a student
ohaa [14]
Hello,

Let's assume a polygonal of n vertices.

Starting from of vertice we can reach n-1 vertices by 2 must be excluded.

So there are n*(n-1-2) possibilities but each  possibility is counted twice

Number of diagonals = n(n-3)/2

Examples:
n=4==>#=4*(4-3)/2=4*1/2=2
n=5==>#=5(5-3)/2=5*2/2=5
n=6==>#=6*(6-3)/2=6*3/2=9

n=17==>#=17*(17-3)/2=119.
6 0
3 years ago
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