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3 years ago

Help please

Mathematics

1 answer:

nika2105 [10]3 years ago

3 0

Answer:

The correct option is Option D) A=-0.7,B=-9/16,C=-0.4,D=-5/16

Step-by-step explanation:

Consider the provided number line.

From the number line we can observed that:

1) A, B, C and D lies between -1 and 0.

2) Point A and B lies between -1 and -0.5

3) Point C and D lies between -0.5 and 0

Therefore, the value of A and B must have a negative value lying between -1 and -0.5

Also, the value of C and D must have a negative value lying between -0.5 and 0.

Now consider the provided options.

Option A) A=-5/16,B=-9/16,C=-0.7,D=-0.4

This is Incorrect representation because the value of C and D should lying between -0.5 and 0.

Option B) A=-0.4,B=-0.7,C=-5/16,D=9/16

This is Incorrect representation because the value of A and B should lying between -1 and -0.5.

Option C) A=9/16,B=-5/16,C=-0.4,D=-0.7

This is Incorrect representation because the value of C and D should lying between -0.5 and 0.

Option D) A=-0.7,B=-9/16,C=-0.4,D=-5/16

$A = -0.7\\B = -\displaystyle\frac{9}{16} = -0.5625\\C = -0.4\\D = -\frac{5}{16} = -0.3125$

Here the value of A and B lying between -1 and -0.5 also the value of C and D lying between -0.5 and 0.

Also they are in ascending order from -1 to 0.

Hence, the correct option is Option D) A=-0.7,B=-9/16,C=-0.4,D=-5/16

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3 years ago

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What is the solution to -10 x w = -50?<br> A. w = 10<br> B. w = 5<br> C. w = -5<br> D. w = -10

ZanzabumX [31]

The solution to the given equation is w = -5

<h3>Solving of expression</h3>

What is an Algebraic Expression? An algebraic expression in mathematics is an expression which is made up of variables and constants, along with algebraic operations

Given the expression below as shown;

-10w = 50

Divide both sides by -10

-10w/-10 = 50/-10

w = -5

Hence the solution to the given equation is w = -5

Learn more on equation and expression here: brainly.com/question/22048677
#SPJ1

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2 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 $\left(-6, 5\right)$

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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

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

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3 years ago

Sylvie is 3 years older than twice edders age if sylvie is 27 years old how old is edder

Mama L [17]

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3 years ago

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aleksandr82 [10.1K]

Answer:

Point of intersections are (0, -7) and (5, -2).

Step-by-step explanation:

From the graph attached,

A straight line is intersecting the circle at the two points (0, 7) and (5, -2).

Now solve algebraically,

Equation of the line → y = x - 7 -------(1)

Equation of the circle → (x - 5)² + (y + 7)² = 25 -------(2)

By substituting the value of y from equation (1) to equation (2)

(x - 5)² + (x - 7 + 7)² = 25

(x - 5)² + x² = 25

x² - 10x + 25 + x² = 25

2x² - 10x = 0

x² - 5x = 0

x(x - 5) = 0

x = 0, 5

From equation (1),

y = 0 - 7 = -7

y = 5 - 7 = -2

Therefore, point of intersections are (0, -7) and (5, -2).

6 0

3 years ago