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ser-zykov [4K]
3 years ago
11

Write a compound interest function to model each situation. Then find the balance after the given number of years.

Mathematics
1 answer:
Reika [66]3 years ago
7 0

Answer:

hi

Step-by-step explanation:

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Triangle PQR where P(1,4), Q(-3,-4),R(7,k) is right-angled at Q. find the value of k
NikAS [45]

Answer:

k = -9.

Step-by-step explanation:

As the triangle is right-angled at Q, by Pythagoras:

PR^2 = PQ^2 + RQ^2

So, substituting the given data and using the distance formula between 2 points:

(7 - 1)^2 + (k - 4)^2 =  (-4-4)^2 + (-3-1)^2 + (7 - (-3))^2 + (k - (-4))^2

36 + (k - 4)^2 = 64 + 16 + 100 + ( k + 4)^2

(k - 4)^2 - (k + 4)^2 = 180 - 36

k^2 - 8k + 16 - (k^2 + 8k + 16) = 144

-16k = 144

k = -9.

8 0
3 years ago
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What is c? Please answer this
Black_prince [1.1K]

Answer:

C is a letter on the alphabet

Step-by-step explanation:

you havent included any pictures so i dont know what you mean

6 0
3 years ago
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PR = 9x - 31 and QR = 43; find x
lesya692 [45]
PR=9(43)-31

PR=387-31

PR=356

I think that's the answer :)
3 0
3 years ago
CALC- limits<br> please show your method
gladu [14]
A. Factor the numerator as a difference of squares:

\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6

c. As x\to\infty, the contribution of the terms of degree less than 2 becomes negligible, which means we can write

\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4

e. Let's first rewrite the root terms with rational exponents:

\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}

Next we rationalize the numerator and denominator. We do so by recalling

(a-b)(a+b)=a^2-b^2
(a-b)(a^2+ab+b^2)=a^3-b^3

In particular,

(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3
(x^{1/2}-x)(x^{1/2}+x)=x-x^2

so we have

\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}

For x\neq0 and x\neq1, we can simplify the first term:

\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x

So our limit becomes

\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43
3 0
3 years ago
Please help with this math problem!! :)
Elza [17]

Answer:

x + 150 deg = 180deg (being co-interior angles)

:. x = 30 deg

2. y^2 + 7= 32 ( opposite sides of parallelogram are equal)

or, y^2 = 25

or y^2 = 5^2

: . y = 5

3. k= 2y^2 ( opposite sides of parallelogram are equal)

or, k = 2× 5^2

: . k = 50

5 0
3 years ago
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