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4 years ago

Exponential Growth/Decay Function The value of a car is $21,500. It loses 12% of its value every year.

Mathematics

1 answer:

Allushta [10]4 years ago

8 0

(a) 21500(0.88) ᣔx

y = A(1-r) ᣔx

= 21500 (1-0.12) ᣔx

= 21500(0.88) ᣔx

(b) $9,985

21500(0.88) ᣔ6 = 9985

· 0.88ᣔ6 = 0.46440408

· 21500 * 0.46440408 = 9984.68 -> 9985

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Find the Area: 12 ft 12 ft

BigorU [14]

a=12 ft

Po=(0.25×π×a²)

Pc=a²+0.5×Po

Pc=a²+0.5×(0.25×π×a²)

Pc=a²×(1+0.5×0.25×π)

Pc=144×(1+0.5×0.25×π)

Pc≈200.55 ft²

8 0

3 years ago

X-2y=1;3x-6y=3 substitution

emmainna [20.7K]

Answer:

All solutions

Step-by-step explanation:

When solving by substitution, we will first set one equation equal to a variable, and then plug that value into the second equation as the variable. Here is what I mean;

x - 2y = 1

3x - 6y = 3

x = 1 + 2y

3x - 6y = 3

3(1 + 2y) - 6y = 3

3 + 6y - 6y = 3

3 = 3

[] Oh no! This doesn't work well. Let's graph it and see what is happening;

-> See attached

-> I have made the lines very thick so you can see the overlap, they are the same size in reality

The answer is all solutions because the graphs are exactly the same.

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.

- Heather

8 0

3 years ago

What multiplies to -29 and adds to 1

Margarita [4]

First set up equation
x times y=xy

so
xy=-29
and
x+y=1
subtract x from both sides
y=1-x
subsitute 1-x for y in first equation
x(1-x)=-29
distribute
x-x^2=-29
add x^2 to both sides
x=-29+x^2
subtract x from both sides
0=x^2-x-29
so we can use the quadratic formula to solve for x if the equation=0 and it is in ax^2+bx+c form so

if
ax+bx+c=0 then x= $\frac{ -b+/-\sqrt{b^2-4ac} }{2a}$ that means x= $\frac{ -b-\sqrt{b^2-4ac} }{2a}$ or x= $\frac{ -b-\sqrt{b^2-4ac} }{2a}$ so

x^2-x-29
a=1
b=-1
c=-29
$\frac{ -(-1)-\sqrt{-1^2-4(1)(-29)} }{2(1)}$ = $\frac{ +1-\sqrt{1^2-(-116)} }{2(1)}=\frac{ +1-\sqrt{1^2+116} }{2}=\frac{ +1-\sqrt{117} }{2}= \frac{1-10.816653826392}{2} = [tex] \frac{-9.816653826392}{2}= -4.908326913196$

the second number is
$\frac{ -(-1)+\sqrt{-1^2-4(1)(-29)} }{2(1)}=\frac{ +1+\sqrt{1^2+(-116)} }{2(1)}= \frac{1+10.816653826392}{2} = \frac{ +1+\sqrt{117} }{2}= \frac{10.816653826392}{2}=5.908326913196$

the two numbers are
5.908326913196 and
-4.908326913196

7 0

3 years ago

For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

C. $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

From the above expression, the power of (5) is 2

Express 2 as 6 - 4

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

Check the list of options for the expression on the left hand side

The correct answer is $(5-\frac{1}{2})^6$

3 0

3 years ago

Which of the following statistics would provide a good comparison between data sets?

White raven [17]

Answer:

ans is mean

Step-by-step explanation:

pls make me brainlest pls i beg you

8 0

3 years ago