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3 years ago

A right triangle has legs that are 10.1 inches and 12.2 inches long. what is the approximate length of the hypotenuse?

Mathematics

2 answers:

Nikitich [7]3 years ago

8 0

A^2+b^2=c^2

(10.1)^2+(12.2)^2=c^2
102.01+148.84=c^2
250.85=c^2
15.8=c

Hypotenuse is approx 15.8 inches

dezoksy [38]3 years ago

6 0

Answer:

The approximate length of the hypotenuse is $15.8\ in$

Step-by-step explanation:

we know that

In a right triangle

The Pythagoras Theorem states that

$c^{2} =a^{2}+b^{2}$

where

c is the hypotenuse (greater side)

a,b are the legs

In this problem we have

$a=10.1\ in$

$b=12.2\ in$

substitute

$c^{2} =10.1^{2}+12.2^{2}$

$c^{2} =250.85$

$c=15.8\ in$

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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

3 years ago

Solving systems of equations using substitution p=q+2<br> 4p+3q= -27

GREYUIT [131]

Ok, so you are given the value of P=q+2

The substitution method tells us that we must insert the value we know, into the second equation, 4P+3q= -27

Doing so will give us 4(q+2)+3q= -27

For right now, lets just focus on the first part, 4(q+2)

We can simplify this by distributing(multiplying) the 4 to whats inside the variables.

This will give us 4q+8

now lets add this back to the rest of the equation >>> 4q+8+3q = -27

We can further simplify by adding like terms >>> 7q+8 = -27

subtract the 8 from both sides >>> 7q = -35

now divide both sides by 7 >>> q = -35/7

Therefor q = -5

EDIT*

now that we know q = -5 we can put q into the equation for P !

we know that p=q+2

so lets put q in now >>> p=(-5)+2

and simplify>>> p = -3

I hope this helps:)

6 0

4 years ago

2. (01.07 MC)

Elena-2011 [213]

Part A: subtract 6 from both sides
Divide by -3 on both sides
X=-3
Part B: add like terms (-2K-3k)
Add 12 to both sides
Add 5k to both sides
Divide by 5 on both sides
K=3
Part C: distribute 6 into the parentheses
Add like terms together (36-5)
Subtract 1 from both sides
Subtract 36v from both sides
Divide by -30 on both sides
-1=v