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3 years ago

Please solve this asap!

Mathematics

1 answer:

laiz [17]3 years ago

4 0

X is equal to 8. This is because you need to set up the ratio of 60=7x+4 because the angles are equal to each other. If you subtract the 4 from both sides you are left with 56=7x and 56 divided by 7 is 8

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(2x+3)(4x-1)<8x^(2)+1

Sidana [21]

Answer: $x$

Step-by-step explanation:

$(2x+3)(4x-1)$

Begin by multiplying parentheses.

$(2x*4x)+(2x*-1)+(3*4x)+(3*-1)$

$(8x^2)+(-2x)+(12x)+(-3)$

$8x^2-2x+12x-3$

Subtract $8x^2$

$8x^2-2x+12x-3+(-8x^2)$

$-2x+12x-3$

Add 3

$-2x+12x-3+3$

$-2x+12x$

Combine like terms;

$10x$

Divide by 10

$x$

Simplify;

$x$

8 0

3 years ago

Help? 10 points please ❤

hammer [34]

Question 23

Lateral Surface Area = πD x height
Lateral Surface Area = π (4) (6) = 75.40 in²

Question 24

Total Surface Area = 2(8.2 x 3.4) + 2(8.2 x 2) + 2(3.4 x 2)
Total Surface Area = 36.56 in²

Question 25

Total Surface Area = 2(10 x 12) + 2(12 x 4) + 2(10 x 4)
Total Surface Area = 416 in²

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4 years ago

Hours Worked vs Salary hours worked-left salary-right

Evgesh-ka [11]

Answer:

for every ten hours there is an extra 300 dollars.

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5 0

3 years ago

Read 2 more answers

What is the perimeter of the triangle

nevsk [136]

Answer:

There’s no image, but to find the perimeter of the triangle just add up the totals of each side.

Step-by-step explanation:

3 0

3 years ago

Consider the equation below. (If you need to use -[infinity] or [infinity], enter -INFINITY or INFINITY.)f(x) = 2x3 + 3x2 − 180x

soldier1979 [14.2K]

Answer:

(a) The function is increasing $\left(-\infty, -6\right) \cup \left(5, \infty\right)$ and decreasing $\left(-6, 5\right)$

(b) The local minimum is x = 5 and the maximum is x = -6

(c) The inflection point is $x = -\frac{1}{2}$

(d) The function is concave upward on $\left(- \frac{1}{2}, \infty\right)$ and concave downward on $\left(-\infty, - \frac{1}{2}\right)$

Step-by-step explanation:

(a) To find the intervals where $f(x) = 2x^3 + 3x^2 -180x$ is increasing or decreasing you must:

1. Differentiate the function

$\frac{d}{dx}f(x) =\frac{d}{dx}(2x^3 + 3x^2 -180x) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f'(x)=\frac{d}{dx}\left(2x^3\right)+\frac{d}{dx}\left(3x^2\right)-\frac{d}{dx}\left(180x\right)\\\\f'(x) =6x^2+6x-180$

2. Now we want to find the intervals where f'(x) is positive or negative. This is done using critical points, which are the points where f'(x) is either 0 or undefined.

$f'(x) =6x^2+6x-180 =0\\\\6x^2+6x-180 = 6\left(x-5\right)\left(x+6\right)=0\\\\x=5,\:x=-6$

These points divide the number line into three intervals:

$(-\infty,-6)$ , $(-6,5)$ , and $(5, \infty)$

Evaluate f'(x) at each interval to see if it's positive or negative on that interval.

$\left\begin{array}{cccc}Interval&x-value&f'(x)&Verdict\\(-\infty,-6)&-7&72&Increasing\\(-6,5)&0&-180&Decreasing\\(5, \infty)&6&72&Increasing\end{array}\right$

Therefore f(x) is increasing $\left(-\infty, -6\right) \cup \left(5, \infty\right)$ and decreasing $\left(-6, 5\right)$

(b) Now that we know the intervals where f(x) increases or decreases, we can find its extremum points. An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We know that:

f(x) increases before x = -6, decreases after it, and is defined at x = -6. So f(x) has a relative maximum point at x = -6.

f(x) decreases before x = 5, increases after it, and is defined at x = 5. So f(x) has a relative minimum point at x = 5.

(c)-(d) An Inflection Point is where a curve changes from Concave upward to Concave downward (or vice versa).

Concave upward is when the slope increases and concave downward is when the slope decreases.

To find the inflection points of f(x), we need to use the f''(x)

$f''(x)=\frac{d}{dx}\left(6x^2+6x-180\right)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f''(x)=\frac{d}{dx}\left(6x^2\right)+\frac{d}{dx}\left(6x\right)-\frac{d}{dx}\left(180\right)\\\\f''(x) =12x+6$

We set f''(x) = 0

$f''(x) =12x+6 =0\\\\x=-\frac{1}{2}$

Analyzing concavity, we get

$\left\begin{array}{cccc}Interval&x-value&f''(x)\\(-\infty,-1/2)&-2&-18\\(-1/2,\infty)&0&6\\\end{array}\right$

The function is concave upward on $(-1/2,\infty)$ because the f''(x) > 0 and concave downward on $(-\infty,-1/2)$ because the f''(x) < 0.

f(x) is concave down before $x = -\frac{1}{2}$ , concave up after it. So f(x) has an inflection point at $x = -\frac{1}{2}$ .

7 0

3 years ago