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Alexandra [31]
2 years ago
11

The model represents the equation 3x - 2 < 4 what is the solution for this inequality

Mathematics
1 answer:
musickatia [10]2 years ago
8 0

Answer:

<em>x < 2</em>

<em />

Step-by-step explanation:

3x - 2 < 4

 <em>  + 2 | + 2</em>

------------------

    3x < 6

         ↓

<em>(3x) / 3  <  6 / 3</em>

         ↓

       x < 2

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A cylinder has a height of 4 inches greater than the radius of its base. Find the radius and the height to the nearest inch if t
anastassius [24]

Answer:

h=5in, r=1in

Step-by-step explanation:

The volume of a cylinder is:

V=\pi*r^2*h

where r is the radius and h is the height.

The height is 4 inches greater that the radius, so:

h=r+4in

Substituting this in the formula for volume:

V=\pi*r^2*(r+4)

the volume is V=5\pi in^3

thus:

5\pi in^3=\pi*r^2*(r+4in)

Dividing everything between pi:

5in^3=r^2*(r+4in)

5in^3=r^3*+4r^2

We can see that the solution for this is r=1in

since (1)^3*+4(1)^2=1+4=5

We have the radius, now we find the height:

h=r+4in=1in + 4in = 5in

h=5in, r=1in

7 0
3 years ago
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Write as a mathematical expression he product of 5 and x
Vaselesa [24]
The answer to the question

8 0
2 years ago
Help me with this math problem please, only this part because I can't seem to solve the last one, I'd like an explanation but it
jonny [76]

Why....

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4 0
3 years ago
Prove each of the following statements below using one of the proof techniques and state the proof strategy you use.
pochemuha

Answer:

See below

Step-by-step explanation:

a) Direct proof: Let m be an odd integer and n be an even integer. Then, there exist integers k,j such that m=2k+1 and n=2j. Then mn=(2k+1)(2j)=2r, where r=j(2k+1) is an integer. Thus, mn is even.

b) Proof by counterpositive: Suppose that m is not even and n is not even. Then m is odd and n is odd, that is, m=2k+1 and n=2j+1 for some integers k,j. Thus, mn=4kj+2k+2j+1=2(kj+k+j)+1=2r+1, where r=kj+k+j is an integer. Hence mn is odd, i.e, mn is not even. We have proven the counterpositive.

c) Proof by contradiction: suppose that rp is NOT irrational, then rp=m/n for some integers m,n, n≠. Since r is a non zero rational number, r=a/b for some non-zero integers a,b. Then p=rp/r=rp(b/a)=(m/n)(b/a)=mb/na. Now n,a are non zero integers, thus na is a non zero integer. Additionally, mb is an integer. Therefore p is rational which is contradicts that p is irrational. Hence np is irrational.

d) Proof by cases: We can verify this directly with all the possible orderings for a,b,c. There are six cases:

a≥b≥c, a≥c≥b, b≥a≥c, b≥c≥a, c≥b≥a, c≥a≥b

Writing the details for each one is a bit long. I will give you an example for one case: suppose that c≥b≥a then max(a, max(b,c))=max(a,c)=c. On the other hand, max(max(a, b),c)=max(b,c)=c, hence the statement is true in this case.

e) Direct proof: write a=m/n and b=p/q, with m,q integers and n,q nonnegative integers. Then ab=mp/nq. mp is an integer, and nq is a non negative integer. Hence ab is rational.

f) Direct proof. By part c), √2/n is irrational for all natural numbers n. Furthermore, a is rational, then a+√2/n is irrational. Take n large enough in such a way that b-a>√2/n (b-a>0 so it is possible). Then a+√2/n is between a and b.

g) Direct proof: write m+n=2k and n+p=2j for some integers k,j. Add these equations to get m+2n+p=2k+2j. Then m+p=2k+2j-2n=2(k+j-n)=2s for some integer s=k+j-n. Thus m+p is even.

7 0
2 years ago
Please help I am very confused:)
Fynjy0 [20]

Answer:

the one solution is -20

Step-by-step explanation:

-15x - 27 + 7 = -25 - 15x +5

-27 + 7 = -25 +5

-20 = -20

3 0
2 years ago
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