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asambeis [7]
4 years ago
12

How far are 27 and -54 apart on the number line?

Mathematics
2 answers:
goldenfox [79]4 years ago
5 0
The answer would have to be B negative 27 (-27) because 27 minus 54 is -27 so it is B
olasank [31]4 years ago
3 0
The answer is D) 81 because you have to add up the absolute values together to get the answer
Hope it helped!!!!
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1.678 rounded to the nearest hundredth​
Talja [164]

Good morning,

Answer:

1.678 rounded to the nearest hundredth is 1.68

Step-by-step explanation:

By looking at the number 1.678 it’s clear that the hundredth digit is 7

since 8∈{5,6,7,8,9} then we delete the 8 and add one to 7

we get 1.68 .

:)

7 0
3 years ago
Read 2 more answers
The graph represents two complex numbers, z1 and z2, as solid line vectors. Which points represent their complex conjugates?
REY [17]

Point <em>A</em> represents the complex conjugate z₁ and point L represents the complex conjugate of z₂ respectively

The complex conjugate of a complex number is a complex number that having equal magnitude in the real and imaginary part as the complex number to which it is a conjugate, but the imaginary part of the complex conjugate has an opposite sign to the original complex number

Therefore, graphically, the complex conjugate is a reflection of the original complex number across the x-axis because the transformation for a reflection of the point (x, y) across the x-axis is given as follows;

Preimage (x, y) reflected across the <em>x</em> axis give the image (x, -y)

Where in a complex number, we have;

x = The real part

y = The imaginary part

The reflection of z₁ across the x-axis gives the point <em>A</em>, while the reflection of z₂ across the x-axis gives the point <em>L</em>

Therefore;

Point <em>A</em> represents the complex conjugate z₁ and point L represents the complex conjugate of z₂

Learn more about complex numbers here;

brainly.com/question/20365080

4 0
3 years ago
What is the area of the rectangle? Round to the nearest square inch.
Mademuasel [1]
Asking the Math Gods...

A=W*L

9*3=27in^2
9*5=45in^2

3 0
3 years ago
Replace ? with =, &gt;, or &lt; to make the statement true.
alekssr [168]
The answer is 12 ÷ 4 + 13 > 2 + 22 ÷ 2
8 0
3 years ago
<img src="https://tex.z-dn.net/?f=%24a%2Ba%20r%2Ba%20r%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D15%24%24a%5E%7B2%7D%2B%28a%20r%29%5E%7B
riadik2000 [5.3K]

Let

S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n

where we assume |r| < 1. Multiplying on both sides by r gives

r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}

and subtracting this from S_n gives

(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}

As n → ∞, the exponential term will converge to 0, and the partial sums S_n will converge to

\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}

Now, we're given

a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a

a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}

We must have |r| < 1 since both sums converge, so

\dfrac{15}a = \dfrac1{1-r}

\dfrac{150}{a^2} = \dfrac1{1-r^2}

Solving for r by substitution, we have

\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)

\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}

Recalling the difference of squares identity, we have

\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}

We've already confirmed r ≠ 1, so we can simplify this to

\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15

It follows that

\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12

and so the sum we want is

ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

7 0
2 years ago
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