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Aleksandr-060686 [28]

2 years ago

13

The Washingtons but a studio apartment for $240,000. They pay a down payment of $60,000. What percent of the purchase price woul

d a $12,000 down payment be?

1 answer:

Ludmilka [50]2 years ago

5 0

It would be 5%. I know the answer is right I just don’t know how to explain it. Sry. But I hope this helps.

You might be interested in

Alejandro was asked to solve a system of linear equations by using the substitution method. What is the error in his work below?

JulsSmile [24]

Step 1:
4y + x = 15
−9x + 5y = 20

Step 2: correct
4y + x = 15
x = 15 − 4y

Step 3: wrong
4y + x = 15
4y + (15 − 4y) = 15 wrong step
4y + 15 − 4y = 15
15 = 15
***once you have solved for a variable in one equation, you substitute it in the other equation

Corrected Step 3:
−9x + 5y = 20
-9(15 - 4y) + 5y= 20
(-9*15) + (-9*-4y) + 5y= 20
-135 + 36y + 5y= 20
-135 + 41y= 20
add 135 to both sides
41y= 155
divide both sides by 41
y= 155/41 or 3 32/41 or 3.78

ANSWER: The x value in step 2 should have been substituted in the opposite equation in step 3. The y solution would then be 155/41 or 3 32/41 or 3.78

Hope this helps! :) Would love to be marked brainliest.

3 0

2 years ago

Read 2 more answers

Find the limit (enter 'DNE' if the limit does not exist)

vaieri [72.5K]

Answer:

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\sqrt{-2x^2-6y^2+1}+1\right)=2$

Step-by-step explanation:

We need to first simplify the expression using rationalization(i.e. if a square root term exists in the denominator, then multiply and divide the whole expression by the denominator(but the change the sign of its middle term))

here, we need to find:

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\dfrac{-2x^2-6y^2}{\sqrt{-2x^2-6y^2+1}-1}\right)$

first we'll rationalize our expression:

$\dfrac{-2x^2-6y^2}{\sqrt{-2x^2-6y^2+1}-1}\left(\dfrac{\sqrt{-2x^2-6y^2+1}+1}{\sqrt{-2x^2-6y^2+1}+1}\right)$

$\dfrac{-(2x^2+6y^2)(\sqrt{-2x^2-6y^2+1}+1)}{(\sqrt{-2x^2-6y^2+1}+1)^2-(1)^2}$

$\dfrac{-(2x^2+6y^2)(\sqrt{-2x^2-6y^2+1}+1)}{-2x^2-6y^2+1-1}$

$\dfrac{-(2x^2+6y^2)(\sqrt{-2x^2-6y^2+1}+1)}{-(2x^2+6y^2)}$

$\sqrt{-2x^2-6y^2+1}+1$

this is our simplified expression, now we can apply our limits:

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\sqrt{-2x^2-6y^2+1}+1\right)$

$\sqrt{-2(0)^2-6(0)^2+1}+1$

$1+1$

$2$

the limit does exists and it is 2.

5 0

2 years ago

Please help ASAP 25 pts + brainliest to right/best answer

BigorU [14]

Answer:

d

Step-by-step explanation:

write in exponential form

log2(2^-1)

simplify

-1

4 0

2 years ago

Anthony buys one pound of oranges for $12 and sells it for $14. How can he write the profit in his account book?

Aleonysh [2.5K]

The answer would be, A) +$2 because he spend $12 but earned $14 so when you subtract 14 and 12 you get $2, which would be his profit

5 0

2 years ago

Solve: 14 = k(0.4) k = _____

KatRina [158]

<h2><u>k = 35</u></h2>

Let's solve your equation step-by-step.

14=k(0.4)

Step 1: Simplify both sides of the equation.

14=0.4k

Step 2: Flip the equation.

0.4k=14

Step 3: Divide both sides by 0.4.

$\frac{0.4k}{0.4}$ = $\frac{14}{0.4}$

<h3><u>k=35</u></h3>

4 0

2 years ago

Read 2 more answers