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3 years ago

Write 47.5 in decimal words

Mathematics

2 answers:

ad-work [718]3 years ago

6 0

Forty-seven and five tenths

rjkz [21]3 years ago

4 0

Forty-seven and five tenths.

47 = forty-seven
. = and
.5 = five tenths

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Related Questions (More Answers Below)

5 0

3 years ago

In ΔSTU, the measure of ∠U=90°, ST = 33 feet, and US = 20 feet. Find the measure of ∠T to the nearest tenth of a degree.

lubasha [3.4K]

Answer:

Step-by-step explanation:

sin β = $\frac{20}{33}$ ⇒ β = arcsin $\frac{20}{33}$ = <em>37.3°</em>

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2 years ago

The lines shown below are perpendicular. If the green line has a slop of 2/5, what is the slope of the red line?

aivan3 [116]

The slope of the red line that is perpendicular to the green line is: -5/2.

<h3>What are the Slope Values of Perpendicular Lines?</h3>

When one line lies perpendicular to another line, the slope of one must be the negative reciprocal of the other line.

<h3>What is the Negative Reciprocal of a Number?</h3>

If given a number, i.e. a/b, the negative reciprocal of a/b would the opposite value of the reciprocal of a/b.

Reciprocal of a/b is b/a. Negative reciprocal of a/b would therefore be: -b/a.

Given that the slope of the green line is: 2/5. And it is perpendicular to the red line. The slope of the red line would be the negative reciprocal of 2/5.

Negative reciprocal of 2/5 is -5/2.

Therefore, the slope of the red line is: -5/2.

Learn more about slope of perpendicular lines on:

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1 year ago

About Exercise 2.3.1: Proving conditional statements by contrapositive Prove each statement by contrapositive

Sladkaya [172]

Answer:

See proofs below

Step-by-step explanation:

A proof by counterpositive consists on assuming the negation of the conclusion and proving the negation of the hypothesis.

a) Assume that n is not odd. Then n is even, that is, n=2k for some integer k. Hence n²=4k²=2(2k²)=2t for some integer t=2k². Then n² is even, therefore n² is not odd. We have proved the counterpositive of this statement.

b) Assume that n is not even, then n is odd. Thus, n=2k+1 for some integer k. Now, n³=(2k+1)³=8k³+6k²+6k+1=2(4k³+3k²+3k)+1=2t+1 for the integer t=4k³+3k²+3k. Thus n³ is odd, that is, n³ is not even.

c) Suppose that n is not odd, that is, n is even. Now, n=2k for some integer k. Then 5n+3=10k+3=2(5k+1)+1, thus 5n+3 is odd, then 5n+3 is not even.

d) Suppose that n is not odd, then n is even. Now, n=2k for some integer k. Then n²-2n+7=4k²-4k+7=2(2k²-2k+3)+1. Hence n²-2n+7 is odd, that is, n²-2n+7 is not even.

e) Assume that -r is not irrational, then -r is rational. Since -1 is rational, then (-1)(-r)=r is rational. Thus r is not irrational.

f) Assume that 1/z is not irrational. Then 1/z is rational. Multiplucative inverses of rational numbers are rational, hence z is rational, that is, z is not irrational.

g) Suppose that z>y. We will prove that z³+zy²≤z²y+y³ is false, that is, we will prove that z³+zy²>z²y+y³. Multiply by the nonnegative number z² in the inequality z>y to get z³>z²y (here we assume z and y nonzero, in this case either z³>0=y³ is true or z³=0>y³ is true). On the other hand, multiply by z² (positive number) to get zy²>y³. Add both inequalities to obtain z³+zy²>z²y+y³ as required.

h) Suppose than n is even. Then n=2k, and n²=4k² is divisible by 4.

i) Assume that "z is irrational or y is irrational" is false. Then z is rational and y is rational. Rational numbers are closed under sum, then z+y is rational, that is, z+y is not irrational.

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3 years ago

If A(2,4) is reflected over the x-axis, what are it’s new coordinates ?

OverLord2011 [107]

(2,-4) is new reflected coordinates

6 0

3 years ago