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3 years ago

The price of a house decreased from $500,000 to $400,000 in one year. What is the percent decrease?

Mathematics

2 answers:

-BARSIC- [3]3 years ago

6 0

Answer:

A) 20%

Step-by-step explanation:

A) ✅

500000(0.2) = 100,000

500000 - 100,000 = 400,000

B) ❌

500000(0.25) = 125,000

500000 - 125,000 = 375,000

C) ❌

500000(0.3) = 150,000

500000 - 150,000 = 350,000

D) ❌

500000(0.35) = 175,000

500000 - 175,000 = 325,000

Vika [28.1K]3 years ago

3 0

Answer:

Step-by-step explanation:

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2 years ago

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COMPLETE<br> (s – r)(x) =

iren [92.7K]

Answer:

sx-rx

Step-by-step explanation:

Since s, r, and x are unknown variables, you just multiply them. s × x = sx.

r × x = rx. then when you put the together, it will look like this:

sx-rx

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2 years ago

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$

From equation(1), substituting x with (4x), Then:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}$

The first three terms of cos (4x) is:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}$

$\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}$

$\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}$

Multiplying equation (2) with (3); we have :

$\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Finally , multiplying 5 with $\mathtt{ e^{-x^2} cos (4x) }$ ; we have:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

7 0

3 years ago

Please help!! I don’t know what i’m doing at all

Shkiper50 [21]

Answer:

53 + 3g ≤ 65, g ≤ 4

Step-by-step explanation:

To solve for g:

53 + 3g ≤ 65

First, subtract 53 from both sides.

53 - 53 + 3g ≤ 65 - 53

=> 3g ≤ 12

Divide 3 by both sides

=> 3g / 3 ≤ 12 /3

=> g ≤ 4

Therefore, g ≤ 4

Hope this helps :)

8 0

2 years ago

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