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3 years ago

Solve for x: 2 over 5 (x - 4) = 2x

Mathematics

1 answer:

xeze [42]3 years ago

5 0

$\frac{2}{5}(x-4)=2x^{(5} \\\\ 2(x-4)=10x \\\\ 2x-8=10x \\\\ 2x-10x=8 \\\\ -8x=8 \ |:8 \\\\ -x=1 \\\\ \boxed{x=-1}$

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Which three measurements could be the lengths of the sides of a triangle?

laila [671]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

I need the answer for this

Svetach [21]

Answer:

535.9 ft²

Step-by-step explanation:

Since there are 360° in a circle, and 240° is 2/3 of 360°, we can say that the area of the bolded sector is <em>2/3 the area of the whole circle</em>. The area of a circle with a radius of r is πr², or approximately 3.14r², so the area of the whole circle is ≈ 3.14(16)² = 3.14(256) = 803.84 ft². Taking 2/3 of this gets us 803.84 * (2/3) ≈ 535.9 ft²

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3 years ago

Factor completely p^2-13p+42

ira [324]

Factoring the expression would look like this :
(p - 7) (p - 6)

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3 years ago

I am having trouble with this relative minimum of this equation.<br>

Norma-Jean [14]

Answer:

So the approximate relative minimum is (0.4,-58.5).

Step-by-step explanation:

Ok this is a calculus approach. You have to let me know if you want this done another way.

Here are some rules I'm going to use:

$(f+g)'=f'+g'$ (Sum rule)

$(cf)'=c(f)'$ (Constant multiple rule)

$(x^n)'=nx^{n-1}$ (Power rule)

$(c)'=0$ (Constant rule)

$(x)'=1$ (Slope of y=x is 1)

$y=4x^3+13x^2-12x-56$

$y'=(4x^3+13x^2-12x-56)'$

$y'=(4x^3)'+(13x^2)'-(12x)'-(56)'$

$y'=4(x^3)'+13(x^2)'-12(x)'-0$

$y'=4(3x^2)+13(2x^1)-12(1)$

$y'=12x^2+26x-12$

Now we set y' equal to 0 and solve for the critical numbers.

$12x^2+26x-12=0$

Divide both sides by 2:

$6x^2+13x-6=0$

Compaer $6x^2+13x-6=0$ to $ax^2+bx+c=0$ to determine the values for $a=6,b=13,c=-6$ .

$a=6$

$b=13$

$c=-6$

We are going to use the quadratic formula to solve for our critical numbers, x.

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-13 \pm \sqrt{13^2-4(6)(-6)}}{2(6)}$

$x=\frac{-13 \pm \sqrt{169+144}}{12}$

$x=\frac{-13 \pm \sqrt{313}}{12}$

Let's separate the choices:

$x=\frac{-13+\sqrt{313}}{12} \text{ or } \frac{-13-\sqrt{313}}{12}$

Let's approximate both of these:

$x=0.3909838 \text{ or } -2.5576505$ .

This is a cubic function with leading coefficient 4 and 4 is positive so we know the left and right behavior of the function. The left hand side goes to negative infinity while the right hand side goes to positive infinity. So the maximum is going to occur at the earlier x while the minimum will occur at the later x.

The relative maximum is at approximately -2.5576505.

So the relative minimum is at approximate 0.3909838.

We could also verify this with more calculus of course.

Let's find the second derivative.

$f(x)=4x^3+13x^2-12x-56$