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serg [7]
2 years ago
6

(7y is supposed to be 7x)

Mathematics
1 answer:
Andrei [34K]2 years ago
7 0

Answer:

Step-by-step explanation:

Angle E is (7x+27) because it is corresponding.

So, 7x+27+15x-5=180 (angle on a straight line is 180 degrees

22x+22=180

22x=158

x= 158/22 = 7.18'

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The sum of two numbers is 33. Twice the first number minus the second number equals 18. Find both numbers.
SOVA2 [1]
Y+x=33
2x-y=18
3x=51
x=17
y=33-x
y=16
3 0
3 years ago
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PLS HELP! DUE TODAY AND WILL MARK BRAINLIEST
Sauron [17]

Answer:

I believe the answer is D: 10

Step-by-step explanation:

If the other line is 2 segments of 5 and from the center to both lines is the same then x and the other line should be equal

4 0
2 years ago
Eva bought a book that was on sale for 20% off the retail price. The retail price of the book was $12. What was the price Eva pa
sp2606 [1]

Answer:

$2.4

Step-by-step explanation:

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2.4

7 0
2 years ago
How do you write a polynomial function for this graph?
expeople1 [14]

first off, let's notice the graph touches the x-axis at -1 and 3, namely, those are the zeros/solutions/roots of the polynomial and therefore, the factors come from those points.

now, at -1, the graph doesn't cross the x-axis, instead it <u>simply bounces off</u> of it, that means the zero of x = -1, has an even multiplicity, could be 4 or 2 or 6, but let's go with 2.

at x = 3, the graph does cross the x-axis, meaning it has an odd multiplicity, could be 3 or 1, or 7 or 9, but let's use 1.


\bf \begin{cases} x=-1\implies &x+1=0\\ x=3\implies &x-3=0 \end{cases}~\hspace{5em}\stackrel{\textit{even multiplicity}}{(x+1)^2}\qquad \stackrel{\textit{odd multiplicity}}{(x-3)^1}=\stackrel{y}{0} \\\\\\ (x^2+2x+1)(x-3)=y\implies x^3+2x^2+x-3x^2-6x-3=y \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill x^3-x^2-5x-3=y~\hfill

5 0
3 years ago
Read 2 more answers
A rectangular box is designed to have a square base and an open top. The volume is to be 500in.3 What is the minimum surface are
mr_godi [17]

The minimum surface area that such a box can have is 380 square

<h3>How to determine the minimum surface area such a box can have?</h3>

Represent the base length with x and the bwith h.

So, the volume is

V = x^2h

This gives

x^2h = 500

Make h the subject

h = 500/x^2

The surface area is

S = 2(x^2 + 2xh)

Expand

S = 2x^2 + 4xh

Substitute h = 500/x^2

S = 2x^2 + 4x * 500/x^2

Evaluate

S = 2x^2 + 2000/x

Differentiate

S' = 4x - 2000/x^2

Set the equation to 0

4x - 2000/x^2 = 0

Multiply through by x^2

4x^3 - 2000 = 0

This gives

4x^3= 2000

Divide by 4

x^3 = 500

Take the cube root

x = 7.94

Substitute x = 7.94 in S = 2x^2 + 2000/x

S = 2 * 7.94^2 + 2000/7.94

Evaluate

S = 380

Hence, the minimum surface area that such a box can have is 380 square

Read more about surface area at

brainly.com/question/76387

#SPJ1

5 0
2 years ago
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