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3 years ago

Alexis washes 10 1/2 windows in 3/4 hours. At this rate, how many windows can she wash in one hour?

Mathematics

1 answer:

Zepler [3.9K]3 years ago

4 0

Alexis can wash a total of 14 windows in 1 hour

<h3>Solution:</h3>

Given that Alexis washes $10\frac{1}{2}$ windows in $\frac{3}{4}$ hours

To find: number of windows washed in one hour

To simplify the calculations we will convert the given mixed fraction to improper fraction

$\begin{array}{l}{=10 \frac{1}{2}} \\\\ {=\frac{2 \times 10+1}{2}} \\\\ {=\frac{21}{2}=10.5}\end{array}$

Hence, to wash 10.5 windows, Alexis takes 0.75 hours

Let "n" be the number of windows washed in 1 hour

10.5 windows ⇒ 0.75 hours

"n" windows ⇒ 1 hour

By cross-multiplication we get,

$n \times 0.75 = 10.5 \times 1$

$n = \frac{10.5}{0.75} \\\\n = 14$

Thus Alexis can wash 14 windows in one hour

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3 years ago

A house blueprint has a scale of 1 in. : 4 ft. The length and width of each room in the actual house are shown in the table. Com

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We know that
scale------------> 1 in/4 ft
[blueprint]=scale*[actual]

Part 1) Living room
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Part 2) Kitchen
length (16 ft) -------------> [blueprint]=(1/4)*16=4 in
width (16 ft) ----------------> [blueprint]=(1/4)*16=4 in

Part 3) Office
length (8 ft) -------------> [blueprint]=(1/4)*8=2 in
width (16 ft) ----------------> [blueprint]=(1/4)*16=4 in

Part 4) Bedroom 1
length (8 ft) -------------> [blueprint]=(1/4)*8=2 in
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Part 5) Bedroom 2
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width (20 ft) ----------------> [blueprint]=(1/4)*20=5 in

Part 6) Bathroom
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see the attached figure to view the table

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3 years ago

Read 2 more answers

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.

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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.

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