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stealth61 [152]
3 years ago
11

Please help top and bottom.

Mathematics
1 answer:
Helga [31]3 years ago
3 0
1. Perimeter
22a + 2(6a + 8) -4 = p

[I got 22a from 12a + 10a]

2. If p = 250, what is a?

22a + 2(6a + 8) - 4 = 250
Distributive property
22a + 12a + 16 - 4 = 250
Add like terms
34a + 12 = 250
Subtract 12 by both sides
34a = 238
Divide 34 by both sides
a = 7
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Find the slope of the given line, if it is defined. 2x + 1 = 0
svlad2 [7]

Given :

A line 2x + 1 =0 .

To Find :

The slope of the given line .

Solution :

We know , slope of line is given by the tangent of the angle between the x-axis and the line .

Now, for line 2x + 1 =0 i.e x=\dfrac{-1}{2} .

The line is perpendicular to x-axis and cuts the x-axis at \dfrac{-1}{2} .

Therefore , the angle between the line and x-axis is 90^o .

So , slope m=\tan\ 90^o i.e undefined .

Therefore , the slope of given line is not defined .

Hence , this is the required solution .

5 0
3 years ago
I need help. this is the problem x^2/3+10=7x^1/3
Arlecino [84]

\it x^{\frac{2}{3}}+10=7x^{\frac{1}{3}} \Leftrightarrow  (x^{\frac{1}{3}})^2-7x^{\frac{1}
  {3}} +10=0 
\\\;\\
We\ note\ x^{\frac{1}{3}} =t \ \ and \ the \ equation \ will \ be:
\\\;\\
t^2-7t+10 = 0 \Leftrightarrow t^2 -2t-5t+10=0 \Leftrightarrow t(t-2) -5(t-2)=0

\it \Leftrightarrow (t-2)(t-5)=0 
\\\;\\
t-2=0 \Rightarrow t=2 \Rightarrow x^{\frac{1}{3}}=2 \Rightarrow (x^{\frac{1}{3}})^3 =2^3 \Rightarrow x = 8
\\\;\\
t-5=0 \Rightarrow t=5 \Rightarrow x^{\frac{1}{3}}= 5  \Rightarrow (x^{\frac{1}{3}})^3 = 5^3 \Rightarrow x = 125

S = {8; 125}


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Zarrin [17]
To add a negative and a positive I usually just subtract them so ex: -5+4=-1
and to subtract them I add the numbers together so -5-4=-9 so its basically the opposite   
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3 years ago
2x+y=10<br> Y=5x-11<br> Solve the system of equations by substitution
antoniya [11.8K]
The answer would be (3,4)
8 0
3 years ago
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