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kherson [118]
3 years ago
6

HELPPP!!! ASAPPP!! PLEASE!! WILL GIVE EXTRA POINTS!!

Mathematics
1 answer:
Zina [86]3 years ago
5 0

Answer:

both of those lines intersect each other and are equal lines

Step-by-step explanation:

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I don’t know if this is correct !!!!!!! Please answer correctly !!!!!!! Will mark Brianliest !!!!!!!!!!!!!!!!!!
Molodets [167]

Answer:

yes i think

Step-by-step explanation:

4 0
2 years ago
13=5+n please help meeeeee
Nat2105 [25]
You subtract 5 from both sides to get n=8
4 0
3 years ago
Read 2 more answers
I am having trouble with this problem. I need to find the lengths of this right triangle. How would you do this?
podryga [215]
You labeled the triangle wrong sides 'a' and 'b' are supposed to be the sides that make the right angle.  the other side is called the hypotenuse which is the longest side which you should have labeled 'c'

so Pythagorean theorem says
a^2+b^2=c^2
so
(2x+1)^2+(11x+5)^2=(12x+1)^2
distribute
(4x^2+4x+1)+(121x^2+110x+25)=(144x^2+24x+1)
add like terms
125x^2+114x+26=144x^2+24x+1
subtract 125x^2 from both sides
114x+26=19x^2+24x+1
subtract 114x from both sides
26=19x^2-90x+1
subtract 26 from both sides
0=19x^2-90-25
factor
(x-5)(19x+5)=0
therefor x-5=0 and/or 19x+5=0
so
x-5=0 add 5 to both sides
x=5
19x+5=0
subtract 5 from both sides
19x=-5
divide both sides by 19
x=-5/19
since side legnths can't be negative, we can cross this solution out

so x=5
subtitute
1+2x
1+2(5)
1+10=11
side a=11

11x+5
11(5)+5
55+5=60
side b=60

12x+1
12(5)+1
60+1=60
side c=61
add them all up
side a+b+c=11+60+61=132=total legnth
5 0
3 years ago
What the answer to this "difficult" math equation?
Ksju [112]

Answer:

1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 + 5 + 5 + 6 + 6 + 7 + 7 + 8 + 8 + 9 + 9 + 10 + 10 - 90 = 20

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Let E = {(x, y) e R2|xy > 0}. Determine whether E is a subspace of R2 . X3
gayaneshka [121]

Answer:

E is not a subspace of \mathbb{R}^2

Step-by-step explanation:

E is not a subspace of  \mathbb{R}^2

In order to see this, we must find two points (a,b), (c,d) in  E such that (a,b) + (c,d) is not in E.

Consider

(a,b) = (1,1)

(c,d) = (-1,-1)

It is easy to see that both (a,b) and (c,d) are in E since 1*1>0 and (1-)*(-1)>0.  

But (a,b) + (c,d) = (1-1, 1-1) = (0,0)

and (0,0) is not in E.

By the way, it can be proved that in any vector space all sub spaces must have the vector zero.

5 0
3 years ago
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