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ryzh [129]
3 years ago
14

Please help me on this

Mathematics
1 answer:
Shkiper50 [21]3 years ago
5 0

Answer: 3


Step-by-step explanation:

x+4=2x+1

  -1=     -1

x+3=2x

-x     -x

3=x

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Solve for b₂, in terms of the area of a trapezoid, A, the height, h, and b₁: A =1/2 (b₁ + b₂)h
nikklg [1K]

Answer: b₂=2A/h-b₁

Step-by-step explanation:

A =1/2 (b₁ + b₂)h

2A=(b₁ + b₂)h

2A/h=b₁ + b₂

b₂=2A/h-b₁

8 0
2 years ago
Solve the equation x = -2y + 6 for y
klio [65]

Answer:

x=2y=6

Step-by-step explanation:

x+2y=-2y+6+2y

5 0
3 years ago
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If p is the smallest of four integers, what is their sum interms of P​
xxMikexx [17]

Answer:

If  p  is the smallest of  n  consecutive integers of the same sign than we have  p ,  p+1 ,  p+2 ,  … ,  p+(n−1) ,

So the sum is

∑k=0n−1(p+k)=∑k=0n−1p+∑k=0n−1k=np+n2−n2  

Here  n=4  

So we have  4p+6  

And checking

p+(p+1)+(p+2)+(p+3)=4p+6  

Note if  p=−v  

Than you have the same thing as if  p=v−n+1  just negative for example  3  consecutive integers the smallest is  −5  so the sum is  −5+(−4)+(−3)=3×−5+32−32=−15+3=−12  

On the other hand:

−(3+4+5)=−(3×3+32−32)=−(9+3)=−12  

If  p=−v  the sum of next  v+1  integers is  −(∑k=0vk)=−(v2+v2)  

Than needs an other  v  integers to bring it up to  0  again. From there it is

∑k=0hk=h2+h2  

Where  h=n−(2v+1) .

So recap if  p  is the smallest of  n  consecutive integers their sum is

p+(p+1)+(p+2)+…+(p+(n−1))=⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪np+n2+n2−(n|p|+n2+n2)((n−|p|)2+(n−|p|)2)−(|p|p+|p|2+|p|2)((n−|p|)2+(n−|p|)2)n≥0p<0∧n<|p|+1p<0∧|p|<n<2|p|+1p<0∧n>2|p|+1

Step-by-step explanation:

6 0
3 years ago
3) -x + 5 = -11x + 25
konstantin123 [22]

Answer:

x=2

Step-by-step explanation:

-x+5=-11x+25

add +x on both sides -11x+x =-10x

-10x+25=5

subtract 25 on both sides

you get -10x=-20

x=2

6 0
3 years ago
Davie read for 30 minutes on Monday, 60 minutes on Tuesday, 15 minutes on Thursday, and 25 on Friday. what is the total time Dav
mariarad [96]

Answer: the Answer of this question is 130

Step-by-step explanation:

7 0
4 years ago
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