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4 years ago

Find a, (x-3) + (x-a) = 15

Mathematics

1 answer:

slega [8]4 years ago

8 0

a = 2x - 18

remove the parenthesis and simplify

x - 3 + x - a = 15

2x - 3 - a = 18 ( add 3 to both sides

2x - a = 18 ( subtract 2x from both sides )

- a = 18 - 2x ( multiply through by - 1 )

a = - 18 + 2x

a = 2x - 18

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Create an exponential to describe $100 at 2% interest, compounded annually, for x years. y=100(.98)^x y=100(.8)^x y=100(1.2)^x y

zysi [14]

The exponential to describe $100 at 2% interest, compounded annually, for x years is $y=100(1.02)^{x}$

<h3><u>Solution:</u></h3>

Given that $ 100 at 2 % interest , compounded annually for "x" years

<em><u>The formula for compound interest, including principal sum, is:</u></em>

$A=P\left(1+\frac{r}{n}\right)^{n t}$

A = the future value of the investment/loan, including interest

P = the principal investment amount (the initial deposit or loan amount)

r = the annual interest rate (decimal)

n = the number of times that interest is compounded per unit t

t = the time the money is invested or borrowed for

Here in this sum,

P = $ 100

$r = 2 \% = \frac{2}{100} = 0.02$

number of years = x

Here given that compounded annually , so n = 1

Let "y" be the amount after "x" years

Substituting the values in formula we get,

$\begin{aligned}&y=100\left(1+\frac{0.02}{1}\right)^{1 \times x}\\\\&y=100(1+0.02)^{x}\\\\&y=100(1.02)^{x}\end{aligned}$

Thus the exponential to describe is $y=100(1.02)^{x}$

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4 years ago

Root 54 to the nearest tenth

kobusy [5.1K]

The square root of 54 to the nearest tenth is : 7.3

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3 years ago

Read 2 more answers

Find the solution set of the following quadratic equations using the quadratic formula.

oksian1 [2.3K]

Answer:

<h3> 1) $x_1=\dfrac{7+\sqrt{13}}2\,,\quad x_2=\dfrac{7-\sqrt{13}}2$ </h3><h3> 2) $x_1=-\dfrac13\,,\quad x_2=-3$ </h3>

Step-by-step explanation:

$x^2 - 7x + 9 = 0\quad\implies\quad a=1\,,\ b = -7\,,\ c=9\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-7)\pm\sqrt{(-7)^2-4\cdot1\cdot9}}{2\cdot1}=\dfrac{7\pm\sqrt{49-36}}2\\\\x_1=\dfrac{7+\sqrt{13}}2\,,\quad x_2=\dfrac{7-\sqrt{13}}2$

<h3>2)</h3><h3> $3x^2 + 10x=-3\\\\3x^2+10x+3=0\quad\implies\quad a=3\,,\ b =10\,,\ c=3\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-10\pm\sqrt{10^2-4\cdot3\cdot3}}{2\cdot3}= \dfrac{-10\pm\sqrt{100-36}}6\\\\x_1=\dfrac{-10+\sqrt{64}}6=\dfrac{-10+8}6=-\dfrac13\,,\qquad x_2=\dfrac{-10-8}6=-3$ </h3>