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Delvig [45]
3 years ago
6

A total of $6000 $6000 is invested: part at 8% 8% and the remainder at 13% 13%. how much is invested at each rate if the annual

interest is $650 $650?
Mathematics
1 answer:
liq [111]3 years ago
8 0
0.13x + 0.08(6000-x) = 650 Solve for x
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neonofarm [45]
Ok, so...what is the question?...
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Please help with this question
never [62]

Answer:

2,744 feet³

Step-by-step explanation:

Volume = Length * Width* Hight

A cube has all the sides equal in measure so Length and Width and Hight are equal and are given in the picture to be 14 feet.

V = 14 *14*14 = 2,744 cubic feet

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AnnZ [28]

Answer:

you need to simplify them like 2n+6 simplifird if 2.(n+3)

Step-by-step explanation:

8 0
3 years ago
Two thirds of a number reduced by 11 is equal to 4 more than the number. Find the number. n= Answer
Ira Lisetskai [31]

Answer:

-45

Step-by-step explanation:

2/3x - 11 = x + 4

  • I created an inequality representing the above statement first. This makes things look less complicated than what the question is asking.

2/3x = x + 15

  • solve for x. I started by adding 11 to both sides.

-1/3x = 15

  • multiply both sides by -3

x = -45

8 0
3 years ago
Read 2 more answers
Help please solve<br> <img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B6x%5E5%2B11x%5E4-11x-6%7D%7B%282x%5E2-3x%2B1
Shkiper50 [21]

Answer:

\displaystyle  -\frac{1}{2} \leq x < 1

Step-by-step explanation:

<u>Inequalities</u>

They relate one or more variables with comparison operators other than the equality.

We must find the set of values for x that make the expression stand

\displaystyle \frac{6x^5+11x^4-11x-6}{(2x^2-3x+1)^2} \leq 0

The roots of numerator can be found by trial and error. The only real roots are x=1 and x=-1/2.

The roots of the denominator are easy to find since it's a second-degree polynomial: x=1, x=1/2. Hence, the given expression can be factored as

\displaystyle \frac{(x-1)(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)^2(x-\frac{1}{2})^2} \leq 0

Simplifying by x-1 and taking x=1 out of the possible solutions:

\displaystyle \frac{(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)(x-\frac{1}{2})^2} \leq 0

We need to find the values of x that make the expression less or equal to 0, i.e. negative or zero. The expressions

(6x^3+14x^2+10x+12)

is always positive and doesn't affect the result. It can be neglected. The expression

(x-\frac{1}{2})^2

can be 0 or positive. We exclude the value x=1/2 from the solution and neglect the expression as being always positive. This leads to analyze the remaining expression

\displaystyle \frac{(x+\frac{1}{2})}{(x-1)} \leq 0

For the expression to be negative, both signs must be opposite, that is

(x+\frac{1}{2})\geq 0, (x-1)

Or

(x+\frac{1}{2})\leq 0, (x-1)>0

Note we have excluded x=1 from the solution.

The first inequality gives us the solution

\displaystyle  -\frac{1}{2} \leq x < 1

The second inequality gives no solution because it's impossible to comply with both conditions.

Thus, the solution for the given inequality is

\boxed{\displaystyle  -\frac{1}{2} \leq x < 1 }

7 0
3 years ago
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