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Zinaida [17]
3 years ago
6

Which set of rational numbers are orders from least to greatest? A) -1, -1 1/2, -1 1/4, -1 7/8 B)-1 7/8, -1 1/2, -1 1/4, -1

Mathematics
2 answers:
Semmy [17]3 years ago
6 0

Answer:

-1 7/8, -1 1/2, -1 1/4, -1

Step-by-step explanation:

First note that the negative value closer to zero on the number is greater. an those far from zero will be the least

According to the given option, we can first express the fractions as percentage as shown

-1 7/8 = -15/8 * 100 = -187.5%

-1 1/2 = -3/2 * 100 = -150%

-1 1/4 = -5/4 * 100 = -125%

-1 = -1 * 100 = -100%

Based on the explanation above, the least value is  -187.5%  and the greatest value is -100%

Rearranging from least to greatest is expressed as  -187.5%,-150%, -125%, -100%

According to their respective fractions;

-1 7/8, -1 1/2, -1 1/4, -1

miss Akunina [59]3 years ago
6 0

Answer:

b

Step-by-step explanation:

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Y=x+5<br> x-5y=-9<br> Can someone please help ASAP
Dmitriy789 [7]
X = -4
Y = 1

1 = -4 + 5
-4 - 5(1) = -9
7 0
3 years ago
Suppose a and b are both non zero real numbers. Find real numbers c and d such that 1/a+ib= c+id
Thepotemich [5.8K]

\begin{gathered} c=\frac{a}{a^2+b^2} \\ d=\frac{-b}{a^2+b^2} \end{gathered}

Explanation

\frac{1}{a+bi}=c+di

Step 1

multiplicate by the conjugate

\begin{gathered} \frac{1}{a+bi}\cdot\frac{a-bi}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2-(bi)^2} \\ \frac{1}{a+bi}\cdot\frac{a-bi}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2-(-b^2)}=\frac{a-bi}{a^2+b^2} \end{gathered}

notice that

\begin{gathered} \frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i \\ \frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i=c+di \\ so \\  \end{gathered}\begin{gathered} c=\frac{a}{a^2+b^2} \\ d=\frac{-b}{a^2+b^2} \end{gathered}

I hope this helsp you

6 0
1 year ago
How do you solve #21?
Inessa05 [86]
5x^3+8x^2/3x^4-16x^2=x^2(5x+8)/x^2(3x^2-16)=5x+8/3x^2-16
lim x--->0 5x+8/3x^2-16=8/-16=-1/2
6 0
3 years ago
Which of the following is equal to the expression (8x)^-2/3 (27x)¯1/3
LiRa [457]

The equivalent expression of (8x)^-2/3 * (27x)^-1/3 is 1/12x

<h3>How to evaluate the expression?</h3>

The expression is given as:

(8x)^-2/3 * (27x)^-1/3

Evaluate the exponent 8^-2/3

(8x)^-2/3 * (27x)^-1/3 = 1/4(x)^-2/3 * (27x)^-1/3

Evaluate the exponent (27x)^-1/3

(8x)^-2/3 * (27x)^-1/3 = 1/4(x)^-2/3 * 1/3(x)^-1/3

Multiply 1/4 and 1/3

(8x)^-2/3 * (27x)^-1/3 = 1/12(x)^-2/3 * (x)^-1/3

Evaluate the exponent

(8x)^-2/3 * (27x)^-1/3 = 1/12(x)^(-2/3 -1/3)

This gives

(8x)^-2/3 * (27x)^-1/3 = 1/12(x)^(-1)

So, we have

(8x)^-2/3 * (27x)^-1/3 = 1/12x

Hence, the equivalent expression of (8x)^-2/3 * (27x)^-1/3 is 1/12x

Read more about equivalent expression at

brainly.com/question/2972832

#SPJ1

3 0
1 year ago
Find the solutions to the equation below.
Volgvan

Answer:

A and C

Step-by-step explanation:

Move all terms to the left side and set equal to zero. Then set each factor equal to zero

4 0
3 years ago
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