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3 years ago

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Three friends decide to rent an apartment and split the cost evenly. They each paid $710 towards the total move in cost of first

and last month's rent and a security deposit.
If rent is $690 per month, how much was the security deposit?

1 answer:

lina2011 [118]3 years ago

7 0

Answer:

Help me with this please

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BabaBlast [244]

Pythagorean's theorem says the hypotenuse of a triangle is A^2+B^2=C^2 in this case I'll put the L,W, and D so it's easy to understand. L^2+W^2=D^2 is the exact same formula as above and is how we will solve this. (30^2)+(10^2)=1000 then we take the square root since 1000=D^2, and we get 36.62 as D.

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3 years ago

austin ran "a" miles. joshua ran at least one more than 3 times the number of miles that austin ran. write an inequality that co

Mnenie [13.5K]

3a > a is the inequality

7 0

3 years ago

Use the intersect method to solve the equation. 14x^3-53x^2+41x-4=-4x^3-x^2+1x+4

UNO [17]

Answer:

x = (68 2^(1/3) + (27 i sqrt(591) + 445)^(2/3))/(27 (1/2 (27 i sqrt(591) + 445))^(1/3)) + 26/27 or x = (68 (-2)^(2/3) - (-2)^(1/3) (27 i sqrt(591) + 445)^(2/3))/(27 (27 i sqrt(591) + 445)^(1/3)) + 26/27 or x = 1/27 ((-2)/(27 i sqrt(591) + 445))^(1/3) ((-1)^(1/3) (27 i sqrt(591) + 445)^(2/3) - 68 2^(1/3)) + 26/27

Step-by-step explanation:

Solve for x over the real numbers:

14 x^3 - 53 x^2 + 41 x - 4 = -4 x^3 - x^2 + x + 4

Subtract -4 x^3 - x^2 + x + 4 from both sides:

18 x^3 - 52 x^2 + 40 x - 8 = 0

Factor constant terms from the left hand side:

2 (9 x^3 - 26 x^2 + 20 x - 4) = 0

Divide both sides by 2:

9 x^3 - 26 x^2 + 20 x - 4 = 0

Eliminate the quadratic term by substituting y = x - 26/27:

-4 + 20 (y + 26/27) - 26 (y + 26/27)^2 + 9 (y + 26/27)^3 = 0

Expand out terms of the left hand side:

9 y^3 - (136 y)/27 - 1780/2187 = 0

Divide both sides by 9:

y^3 - (136 y)/243 - 1780/19683 = 0

Change coordinates by substituting y = z + λ/z, where λ is a constant value that will be determined later:

-1780/19683 - 136/243 (z + λ/z) + (z + λ/z)^3 = 0

Multiply both sides by z^3 and collect in terms of z:

z^6 + z^4 (3 λ - 136/243) - (1780 z^3)/19683 + z^2 (3 λ^2 - (136 λ)/243) + λ^3 = 0

Substitute λ = 136/729 and then u = z^3, yielding a quadratic equation in the variable u:

u^2 - (1780 u)/19683 + 2515456/387420489 = 0

Find the positive solution to the quadratic equation:

u = (2 (445 + 27 i sqrt(591)))/19683

Substitute back for u = z^3:

z^3 = (2 (445 + 27 i sqrt(591)))/19683

Taking cube roots gives 1/27 2^(1/3) (445 + 27 i sqrt(591))^(1/3) times the third roots of unity:

z = 1/27 2^(1/3) (445 + 27 i sqrt(591))^(1/3) or z = -1/27 (-2)^(1/3) (445 + 27 i sqrt(591))^(1/3) or z = 1/27 (-1)^(2/3) 2^(1/3) (445 + 27 i sqrt(591))^(1/3)

Substitute each value of z into y = z + 136/(729 z):

y = (68 2^(2/3))/(27 (27 i sqrt(591) + 445)^(1/3)) + 1/27 (2 (27 i sqrt(591) + 445))^(1/3) or y = (68 (-2)^(2/3))/(27 (27 i sqrt(591) + 445)^(1/3)) - 1/27 (-2)^(1/3) (27 i sqrt(591) + 445)^(1/3) or y = 1/27 (-1)^(2/3) (2 (27 i sqrt(591) + 445))^(1/3) - (68 (-1)^(1/3) 2^(2/3))/(27 (27 i sqrt(591) + 445)^(1/3))

Bring each solution to a common denominator and simplify:

y = (2^(1/3) ((27 i sqrt(591) + 445)^(2/3) + 68 2^(1/3)))/(27 (445 + 27 i sqrt(591))^(1/3)) or y = (68 (-2)^(2/3) - (-2)^(1/3) (27 i sqrt(591) + 445)^(2/3))/(27 (445 + 27 i sqrt(591))^(1/3)) or y = 1/27 2^(1/3) (-1/(445 + 27 i sqrt(591)))^(1/3) ((-1)^(1/3) (27 i sqrt(591) + 445)^(2/3) - 68 2^(1/3))

Substitute back for x = y + 26/27:

Answer: x = (68 2^(1/3) + (27 i sqrt(591) + 445)^(2/3))/(27 (1/2 (27 i sqrt(591) + 445))^(1/3)) + 26/27 or x = (68 (-2)^(2/3) - (-2)^(1/3) (27 i sqrt(591) + 445)^(2/3))/(27 (27 i sqrt(591) + 445)^(1/3)) + 26/27 or x = 1/27 ((-2)/(27 i sqrt(591) + 445))^(1/3) ((-1)^(1/3) (27 i sqrt(591) + 445)^(2/3) - 68 2^(1/3)) + 26/27

5 0

3 years ago

What is one fourth of 52

AlekseyPX

Your answer will be 13

7 0

3 years ago

Read 2 more answers

a-plus auto body is painting desings on a customer's car. they had 18 pints of blue paint on hand. they used 1/2 of it for the f

netineya [11]

4³/₄ pints

<h3>Further explanation</h3>

<u>Given:</u>

A-Plus auto body is painting desigs on a customer's car.

They had 18 pints of blue paint on hand.
They used $\frac{1}{2}$ of it for the flames, and $\frac{1}{3}$ of it for the sparks.
They needed $7\frac{3}{4}$ pints of blue paint to paint the next design.

<u>Question:</u>

How many more pints of blue paint will they need?

<u>The Process:</u>

From the denominators 2 and 3 above, we get LCM = 6.

Let us draw a diagram that contains all the pints.

$18 \ pints \div 6 \ units = 3 \rightarrow \boxed{3}\boxed{3}\boxed{3}\boxed{3}\boxed{3}\boxed{3} = 18 \ pints$

The flames: $\frac{1}{2}$ or half of 6 unit above, that is $\boxed{3}\boxed{3}\boxed{3} = 9 \ pints$

The sparks: $\frac{1}{3} = \frac{2}{6}$ or 2 of 6 unit above, that is $\boxed{3}\boxed{3} = 6 \ pints$

So, the remainder is 18 - 9 - 6 = 3 pints.

They needed $7\frac{3}{4}$ pints of blue paint to paint the next design.

Let us count how many more pints of blue paint will they need.

$\boxed{7\frac{3}{4} - 3 = ?}$

$\boxed{= \bigg( 7 + \frac{3}{4} \bigg) - 3}$

$\boxed{= 7 - 3 + \frac{3}{4}}$

$\boxed{\boxed{ \ 4\frac{3}{4}}}$

Thus, they $4\frac{3}{4}$ pints more of blue paint.

<h3>Learn more</h3>

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Keywords: A-Plus, auto body, painting, designs, on a customer's car, they had 18 pints, blue paint, 1/2, for the flames, 1/3, the sparks, they needed, 7³/₄, the next design, how many more, will they need

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3 years ago