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3 years ago

What is the value of the dependent variable if the value of the independent variable is –4?

1 answer:

8 0

If you would like to know the value of the dependent variable if the value of the independent variable is - 4, you can calculate this using the following steps:

independent variable ... - 4 ... x
dependent variable ... f(x)

f(x)= -3x^2 + 5x
f(-4) = -3 * (-4)^2 + 5 * (-4) = -3 * 16 - 20 = -48 - 20 = - 68

The correct result would be C. –68.

You might be interested in

Find the greatest solution for x+y when x^2+y^2 = 7, x^3+y^3=10

damaskus [11]

Answer:

Step-by-step explanation:

set

$f(x,y)=x+y\\$

constrain:

$g(x,y)=x^2+y^2 = 7\\h(x,y)=x^3+y^3=10$

Partial derivatives:

$f_{x}=1\\f_{y} =1 \\g_{x}=2x \\g_{y}=2y\\h_{x}=3x^2 \\h_{y}=3y^2$

Lagrange multiplier:

$grad(f)=a*grad(g)+b*grad(h)\\$

$\left[\begin{array}{ccc}1\\1\end{array}\right]=a\left[\begin{array}{ccc}2x\\2y\end{array}\right]+b\left[\begin{array}{ccc}3x^2\\3y^2\end{array}\right]$

4 equations:

$1=2ax+3bx^2\\1=2ay+3by^2\\x^2+y^2=7\\x^3+y^3=10$

By solving:

$a=4/9\\b=-2/27\\x+y=4$

Second mathod:

Solve for x^2+y^2 = 7, x^3+y^3=10 first:

$x=\frac{1}{2} -\frac{\sqrt{13}}{2} \ or \ y=\frac{1}{2} +\frac{\sqrt{13}}{2} \\x=\frac{1}{2} +\frac{\sqrt{13}}{2} \ or \ y=\frac{1}{2} -\frac{\sqrt{13}}{2} \\x+y=-5\ or\ 1 \or\ 4$

The maximum is 4

6 0

3 years ago

An owner of a key rings manufacturing company found that the profit earned (in thousands of dollars) per day by selling n number

vovikov84 [41]

Answer:

The number of key rings sold on a particular day when the total profit is $5,000 is 4,000 rings.

Step-by-step explanation:

The question is incomplete.

An owner of a key rings manufacturing company found that the profit earned (in thousands of dollars) per day by selling n number of key rings is given by

$P=n^2-2n-3$

where n is the number of key rings in thousands.

Find the number of key rings sold on a particular day when the total profit is $5,000.

We have the profit defined by a quadratic function.

We have to calculate n, for which the profit is $5,000.

$P=n^2-2n-3=5\\\\n^2-2n-8=0$

We have to calculate the roots of the polynomial we use the quadratic equation:

$n=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\n= \frac{-2\pm\sqrt{4-4*1*(-8)}}{2}= \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{36}}{2} =\frac{-2\pm6}{2} \\\\n_1=(-2-6)/2=-8/2=-4\\\\n_2=(-2+6)/2=4/2=2$

n1 is not valid, as the amount of rings sold can not be negative.

Then, the solution is n=4 or 4,000 rings sold.

7 0

3 years ago

The school you want to go to has an acceptance rate of 15%. That means that 15% of the students who apply to the school get in.

lisov135 [29]

12,000 Students Applied...

To find the answer, consider that for every 1000 students that apply, 150 students are accepted.

Divide 1800 by 150 and you get 12....which can be translated to 12,000 students

4 0

3 years ago

I NEED HELP WITH THIS ASAP

grin007 [14]

hopefully someone helps u with this tho sorry

3 0

3 years ago

Please help me with this math question!

barxatty [35]

1) Change radical forms to fractional exponents using the rule:
The nth root of "a number" = "that number" raised to the reciprocal of n.
For example $\sqrt[n]{3} = 3^{ \frac{1}{n} }$ .

The square root of 3 ( $\sqrt{3}$ ) = 3 to the one-half power ( $3^{ \frac{1}{2} }$ ).
The 5th root of 3 ( $\sqrt[5]{3}$ ) = 3 to the one-fifth power ( $3^{ \frac{1}{5} }$ ).

2) Now use the product of powers exponent rule to simplify:
This rule says $a^{m} a^{n} = a^{m+n}
$ . When two expressions with the same base (a, in this example) are multiplied, you can add their exponents while keeping the same base.

You now have $(3^{ \frac{1}{2} })*(3^{ \frac{1}{5} })$ . These two expressions have the same base, 3. That means you can add their exponents:
$(3^{ \frac{1}{2} })(3^{ \frac{1}{5} })\\
= 3^{(\frac{1}{2} + \frac{1}{5}) }\\
= 3^{\frac{7}{10}}$

3) You can leave it in the form $3^{\frac{7}{10}}$ or change it back into a radical $\sqrt[10]{3^7}$

------

Answer: $3^{\frac{7}{10}}$ or $\sqrt[10]{3^7}$

6 0

3 years ago