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Elina [12.6K]
3 years ago
12

16) The Millers plan on retiring soon. When they were first married they purchased 1,000 shares in a mutual fund that was at tha

t time paying $10 a share. Now, thirty years later it is paying $78 a share. What is the percent increase for the fund? A) 6.8% B) 7.8% C) 78% D) 680%
Mathematics
1 answer:
OlgaM077 [116]3 years ago
4 0

Answer:

D

Step-by-step explanation:

1,000 × 10 = $10,000

1,000 × 78 = $78,000

increase ÷ original number x 100

68,000 ÷ 10,000 × 100 = 680

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Dvinal [7]
Rectangular pyramid and triangular pyramid.
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2 years ago
Help with number two
BabaBlast [244]

Answer:

y = -2x+1

Step-by-step explanation:

We have a point and a slope, so we can use the point slope form of a line

y-y1 = m(x-x1)  where m is the slope and (x1,y1) is the point

y-3 = -2(x--1)

y-3=-2(x+1)

Distribute the 2

y-3 = -2x-2

Add 3 to each side

y-3+3 = -2x-2+3

y = -2x+1

This is in slope intercept form

7 0
2 years ago
What is the solution of the proportion? 6/a = 18/27
Marta_Voda [28]

Answer:

a=9

Step-by-step explanation:

To solve this proportion, we have to get the variable, a, by itself.

First, cross multiply.

6/a=18/27

Multiply the denominator of the first fraction by the numerator of the second, and the numerator of the second by the denominator of the first.

a*18=6*27

18a=162

Now, 18 and a are being multiplied. In order to get a by itself, perform the opposite of what is being done. They are being multiplied, so the opposite would be division. Divide both sides by 18.

18a/18=162/18

a=162/18

a=9

So, the proportion, with 9 substituted in for a, will be:

6/9=18/27

6 0
3 years ago
Read 2 more answers
Find an n^th degree polynomial with real coefficients satisfying the given conditions. n = 3; -2 and 2 i are zeros; f(-1) = 15.
Ira Lisetskai [31]
So, n = 3, is a 3rd degree polynomial, roots are -2 and 2i

well 2i is a complex root, or imaginary, and complex root never come all by their lonesome, their sister is always with them, the conjugate, so if 0+2i is there, 0-2i is there too

so, the roots are -2, 2i, -2i

now... \bf \begin{cases}
x=-2\implies x+2=0\implies &(x+2)=0\\
x=2i\implies x-2i=0\implies &(x-2i)=0\\
x=-2i\implies x+2i=0\implies &(x+2i)=0
\end{cases}
\\\\\\
(x+2)\underline{(x-2i)(x+2i)}=0\\\\
-----------------------------\\\\
\textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-----------------------------\\\\
(x+2)[x^2-(2i)^2]=0\implies (x+2)[x^2-(2^2i^2)]=0
\\\\\\
(x+2)[x^2-(4\cdot -1)]=0\implies (x+2)(x^2+4)=0
\\\\\\
x^3+2x^2+4x+8=0

now, if we check f(-1), we end up with 5, not 15
hmmm

so, how to turn our 5 to 15? well, 3*5, thus

\bf 3(x^3+2x^2+4x+8)=f(x)\implies 3(5)=f(-1)\implies 15=f(-1)

usually, when we get the roots, or zeros, if any common factor that is a constant is about, they get in a division with 0 and get tossed, and aren't part of the roots, thus, we can simply add one, in this case, the common factor of 3, to make the 5 turn to 15
6 0
3 years ago
Write an equation in slope-intercept form of the line that passes through the points (-2.3) and (2.7)
liraira [26]

I'm pretty sure the answer is y=x+5

8 0
2 years ago
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