Locating Zeros of Polynomial Functions:
Determine the zeros to the nearest tenth

The roots of equation are: x= -3+ √29/2 and x= -3-√29/2.

Quadratic equation's roots The roots of a quadratic equation are the values of the variables that fulfil the equation. In other words, if f(a) = 0, then x = a is a root of the quadratic equation f(x). The x-coordinates of the sites where the curve y = f(x) intersects the x-axis are the real roots of an equation f(x) = 0.

given:

x² + 3x - 5 = 0

So, solving for x

x² + 3x - 5 = 0

D= (3)² - 4(1)(-5)= 9 + 20= 29

Now, using the quadratic equation

x= -b ± √ b² -4ac/ 2a

x= -3 ±√(3)² -4(1)(-5) / 2

x= -3±√29/2

x= -3+ √29/2 and x= -3-√29/2

Hence, the roots of equation are: x= -3+ √29/2 and x= -3-√29/2.

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If s = 1/4 unit and A = 80s^2

A = 80 (1/4)^2

A = 80 (1/16)

A = 5

Answer

5 square units

5 square units

substitute s = [tex]\frac{1}{4}[/tex] into the equation

A= 80 × ([tex]\frac{1}{4}[/tex])² = 80 × [tex]\frac{1}{16}[/tex] = [tex]\frac{80}{16}[/tex]= 5 square units

Yes ,

side ×side×side= side^3 =volume of cube  

=>side^3=30

=>side = 30^1/3-answer

Hi!

[tex]3x=18[/tex]

[tex]\frac{3x}{3}=\frac{18}{3}[/tex]

[tex]x=6[/tex]

Explanation: This question is super easy on this question. First you had to divide by 3 from both sides. And simplify, it gave us the answer is x=6 is the right answer. Hope this helps! And thank you for posting your question at here on Brainly. And have a great day. -Charlie

Answer:

  y -8 = -5(x -6)

Step-by-step explanation:

The point-slope form of the equation for a line is generally written ...

  y -k = m(x -h)

for slope m and point (h, k).

The slope of your parallel line is the same as the slope of the reference line, -5. So your equation is ...

  y -8 = -5(x -6)

Answer:

Points (0,-3) and (-3,-1) work, I got a 100% on the test.

Note that a:b = c:d is the same as a/b = c/d

y:1.5=0.2:0.75 same as y/1.5 = 0.2/0.75

y = 0.4

x: 1 2/3=y:3 1/3 same as x/1 2/3 = y / 3 1/3

x = y/2 = 0.2

y:1.5=0.2:0.75

so y=1.5*0.2/0.75=0.4

x:1 2 /3=y:3 1/3

so x=1 2/3*0.4/3 1/3=0.2

When you replace the comparison symbol (<) with an equal sign (=), you get the equation of a line in slope-intercept form:

... y = mx + b

where m is the slope, and b is the y-intercept.

Your equation has m = -1/4 and b = -1. To graph this line, find the point (0, -1) on the y-axis. To find another point on the line, you can use the slope value (rise/run = -1/4), which tells you the line "rises" -1 for each "run" of +4. That is, another point on the line will be 4 units to the right and 1 unit down, at (4, -2). Working in the other direction (to the left, instead of to the right), the -1/4 slope tells you the point 4 units left and 1 unit up (-4, 0) will also be on the line. Draw a dashed line through these points,

The dashed line you just drew is the boundary of the solution region. It is dashed because the line itself is not part of the solution. (Those points do not meet the requirement for "less than.")

Appropriate values of y are ones that are less than those on the line, so the solution region is indicated as being the half-plane below the line. You indicate this by shading the solution region. (See the attachment for an example of the way this can be graphed.)

_____

If the comparison is ≤ instead of <, then the line is solid (not dashed), indicating it is part of the solution region. If the comparison is > or ≥, then the shaded region is above the line, where y-values are greater than those on the line.

The equation of the line through the given point parallel to the given line ...

The given line's equation is is slope-intercept form. The slope is -3, the coefficient of x.

In point-slope form, the equation of a line with slope m through point (h, k) is ...

... y -k = m(x -h)

For slope m=-3 and point (h, k) = (-4, -8) the equation of the line is ...

... y +8 = -3(x +4) . . . . equation in point-slope form

We can eliminate parentheses and add -8 to put this equation in slope-intercept form.

... y +8 = -3x -12

... y = -3x -20 . . . . equation in slope-intercept form

Answer:

Step-by-step explanation:

The answer is B

y=2x+11

Answer:

d. (1, 3) and (-3, -1)

Step-by-step explanation:

Equating the expressions for y, we have ...

... x² +3x -1 = y = x +2

Subtracting x+2 gives ...

... x² +2x -3 = 0

... (x +3)(x -1) = 0 . . . . . factored form

... x = -3, 1 . . . . . . . . . . .values that make the factors zero

The second equation tells us, y = x+2, so

... For x = -3, y = -3 +2 = -1. The solution is (-3, -1)

... For x = 1, y = 1 +2 = 3. The solution is (1, 3)

(- 3, - 1 ) or (1, 3 )

Since both equations express y in terms of x, equate both sides

x² + 3x - 1 = x + 2 ( subtract x + 2 from both sides )

x² + 2x - 3 = 0

(x + 3 )(x - 1 ) = 0

equate each factor to zero and solve for x

x + 3 = 0 ⇒ x = - 3

x - 1 = 0 ⇒ x = 1

Substitute these values into either of the 2 equations for y

y = - 3 : y = - 3 + 2 = - 1 ( using y = x + 2 )

x = 1 : y = 1 + 2 = 3

solutions are (- 3, - 1 ) or (1, 3 )

B

note that (f + g)(x) = f(x) + g(x)

f(x) + g(x) = -4x² - 6x - 1 - x² - 5x + 3 ( collect like terms )

               = - 5x² - 11x + 2

Answer:

The correct option is B.

Step-by-step explanation:

The given functions are

[tex]f(x)=-4x^2-6x-1[/tex]

[tex]g(x)=-x^2-5x+3[/tex]

Using the p addition property of functions (f + g)(x) can be written as

[tex](f+g)(x)=f(x)+g(x)[/tex]

Substitute the values of each function in the above equation.

[tex](f+g)(x)=-4x^2-6x-1-x^2-5x+3[/tex]

Combine like terms.

[tex](f+g)(x)=(-4x^2-x^2)+(-6x-5x)+(-1+2)[/tex]

[tex](f+g)(x)=-5x^2-11x+2[/tex]

Therefore the correct option is B.

Note that

[tex]A_{CMD}=\dfrac{1}{2}\cdot MC\cdot h=7\ sq. ft.[/tex]

Let H be the height of triangle ABC. Since [tex]\dfrac{MD}{DB}=\dfrac{1}{2},[/tex] then

[tex]\dfrac{H}{h}=\dfrac{5}{1}, \\ \\H=5h.[/tex]

1.

[tex]A_{BDC}=A_{MBC}-A_{CMD}=\dfrac{1}{2}\cdot MC\cdot H-\dfrac{1}{2}\cdot MC\cdot h=\dfrac{1}{2}\cdot MC\cdot (5h-h)=\\ \\=4\cdot \dfrac{1}{2}\cdot MC\cdot h=4\cdot 7=28 sq. ft.[/tex]

2. M is midpoint of AC, then AM=MC.

[tex]A_{AMB}=\dfrac{1}{2}\cdot AM\cdot H=\dfrac{1}{2}\cdot MC\cdot 5h=5\cdot \dfrac{1}{2}\cdot MC\cdot h=5\cdot 7=35\ sq. ft.[/tex]

3.

[tex]A_{ABC}=\dfrac{1}{2}\cdot AC\cdot H=\dfrac{1}{2}\cdot 2MC\cdot 5h=10\cdot \dfrac{1}{2}\cdot MC\cdot h=10\cdot 7=70\ sq. ft.[/tex]

Answer:

[tex]A_{BDC}=28\ sq. ft,\ A_{AMB}=35\ sq. ft,\ A_{ABC}=70\ sq. ft.[/tex]

Answer:

28,35,70

Step-by-step explanation:

The set of seashell weights that Mirna collected (12, 13, 16, 17, 18, 20) does not have a mode because all values appear only once.

The mode of a set of data refers to the number that appears most frequently. In the case of the weights of the seashells Mirna collected in Florida, the data set is: 12, 13, 16, 17, 18, 20. To find the mode, we look for the value that occurs the most:

Since all numbers occur only once, there is no number that appears more frequently than the others. Therefore, this set of data does not have a mode.

answer:

1. 24

2. -6

work:

1.

[tex]5n + 3n[/tex]

[tex]5 (3) + 3 (3)[/tex]

[tex]15 + 9[/tex]

24

2.

[tex]9 (x -7) - y[/tex]

[tex]9 ( (11) -7) - 19[/tex]

[tex]9 + 4 - 19[/tex]

[tex]13 - 19[/tex]

[tex]-6[/tex]

hope this helps! ❤ from peachimin

1.  How to get answer:

2. How to get answer:

It can't be proven because it isn't so.

You can show that A'B' = C'D' because each is half of AC (from the midsegment theorem).

slope = [tex]\frac{1}{12}[/tex]

the slope is the value of f' (36)

f(x) = √x = [tex]x^{\frac{1}{2} }[/tex]

f'(x) = [tex]\frac{1}{2}[/tex] [tex]x^{-\frac{1}{2} }[/tex] = [tex]\frac{1}{2\sqrt{x} }[/tex]

f'(36) = [tex]\frac{1}{2(6)}[/tex] = [tex]\frac{1}{12}[/tex]

Answer:

[tex]slope = \frac{1}{12}[/tex]

Step-by-step explanation:

Here is another method to solve your problem. I am showing this method because this is the first method normally taught and a student might not of had the chance yet to learn the other methods

We can solve this problem by using limits and the following function

[tex]\lim_{h\to 0} \frac{f(x+h) - x}{h}[/tex]

[tex]\lim_{h\to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}[/tex]

Next multiply by the conjugate of the numerator.

[tex]\lim_{h\to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} * \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}[/tex]

[tex]\lim_{h\to 0} \frac{x + h - x}{h(\sqrt{x+h} + \sqrt{x})}[/tex]

Cancel the x - x

[tex]\lim_{h\to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}[/tex]

Divide out the h

[tex]\lim_{h\to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}[/tex]

[tex]\lim_{h\to 0} \frac{1}{(\sqrt{x+h} + \sqrt{x})}[/tex]

Plugin 0 where h is located

[tex]\lim_{h\to 0} \frac{1}{(\sqrt{x+h} + \sqrt{x})}[/tex]

[tex]\lim_{h\to 0} \frac{1}{(\sqrt{x+0} + \sqrt{x})}[/tex]

[tex]\lim_{h\to 0} \frac{1}{(\sqrt{x} + \sqrt{x})}[/tex]

Combine Like terms in denominator

[tex]\lim_{h\to 0} \frac{1}{(\sqrt{x} + \sqrt{x})}[/tex]

[tex]\lim_{h\to 0} \frac{1}{2\sqrt{x}}[/tex]

Now lets use our derivative and plugin 36 where x is located and solve

[tex]\frac{1}{2\sqrt{x}}[/tex]

[tex]\frac{1}{2\sqrt{36}}[/tex]

[tex]\frac{1}{2(6)}[/tex]

[tex]\frac{1}{12}[/tex]

Note, this is a harder method but it is normally the first method taught in Calculus 1.

y = - [tex]\frac{2}{5}[/tex] x - 2

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y-intercept )

rearrange 2x + 5y = 10 into this form

subtract 2x from both sides

5y = - 2x + 10 ( divide all terms by 5 )

y = - [tex]\frac{2}{5}[/tex] x + 2 ← point- slope form with slope m = - [tex]\frac{2}{5}[/tex]

Parallel lines have equal slopes hence

y = - [tex]\frac{2}{5}[/tex] x + c is the partial equation of the parallel line

to find c, substitute ( 5, - 4 ) into the partial equation

- 4 = - 2 + c ⇒ c = - 4 + 2 = - 2

y = - [tex]\frac{2}{5}[/tex] x - 2 ← equation of parallel line

The equation of the line that is parallel to 2x + 5y = 10 and passes through the point (5, -4) is y + 4 = (-2/5)x + 2.

To find the equation of a line that is parallel to another line, we need to use the fact that parallel lines have the same slope. First, let's write the given equation in slope-intercept form (y = mx + b), where m represents the slope:

2x + 5y = 10

5y = -2x + 10

y = (-2/5)x + 2

Since the given line has a slope of -2/5, the parallel line will also have a slope of -2/5. Now, we can use the point-slope form of a line (y - y1 = m(x - x1)) to find the equation. Plugging in the coordinates of the point (5, -4):

y - (-4) = (-2/5)(x - 5)

y + 4 = (-2/5)(x - 5)

y + 4 = (-2/5)x + 2

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2 7/8 ft

We presume no length was lost in the cut, so that ...

... (remaining length) + (cut off length) = (original length)

Then ...

... (cut off length) = (original length) - (remaining length)

... = 12 3/4 - 9 7/8

... = (12 - 9) + (3/4 - 7/8)

... = 3 + (6/8 -7/8)

... = 3 - 1/8

... = 2 7/8

The cut off length was 2 7/8 feet.

25x²-16 is the difference of two squares, so can be written as the product ...

... (5x -4)(5x +4)

or

... (5x +4)(5x -4)

Answer:

[tex](5x+4)(5x-4)[/tex]

Step-by-step explanation:

25x^2- 16

To find out the product that is equivalent to the given expression we need to factor the given expression

we write the numbers in square form

25 = 5*5 = 5^2

16 = 4*4 = 4^2

5^2x^2 - 4^2

[tex](5x)^2- 4^2[/tex]

we apply difference in square formula

a^2 - b^2 = (a+b)(a-b)

[tex](5x)^2- 4^2=(5x+4)(5x-4)[/tex]

First, I should point out that g(x) should be written as g(x)=(x+2)/3, otherwise the problem is confusing.  

[tex]f(x)=3x-2 \enspace g(x)=\frac{x+2}{3}[/tex]

(A) [tex]f(g(x))=3(\frac{x+2}{3})-2=x\\g(f(x))=\frac{3x-2+2}{3}=x[/tex]

(B) Since [tex]f(g(x))=x[/tex] and [tex]g(f(x))=x[/tex], it holds that

[tex]f(g(x))=g(f(x))[/tex] for all x. This means the composed functions are *identical*

A

f(g(x)) = f([tex]\frac{x+2}{3}[/tex]) = 3([tex]\frac{x+2}{3}[/tex]) - 2 = x + 2 - 2 = x

g(f(x)) = g(3x - 2) = [tex]\frac{3x-2+2}{3}[/tex] = [tex]\frac{3x}{3}[/tex] = x

B

Since both composite functions f(g(x)) and g(f(x)) equal x

This indicates that the functions f(x) and g(x) are inverse functions

We won't have a function if for same value of x in (x,y) we get different values y.

So first step: figure out k so that the first coordinate (x) is the same:

3k-4=4k    | solve for k

k = -4

no check the values y for the elements of the relation

x = 3k-4 = -12-4=-16

so at -16 we get (-16,16) and (-16, 32), which mean for k=-4 the relation is not a function.

Let me know if you have any questions.

Answer:

K= -4

Step-by-step explanation:

Since it isn't possible that (3k-4, 16) is going to be equal to (4k, 32) in terms of positive numbers, you will have to go to the negative side of the number line.

K has to equal 4 because 3 · -4 = -12, and -12 minus 4 is equal to -16.

And since 4 · -4 = -16, K has to equal - 4.

Answer:

not a rectangle

Step-by-step explanation:

There are several ways to determine whether the quadrilateral is a rectangle. Computing slope is one of the more time-consuming. We can already learn that the figure is not a rectangle by seeing if the midpoint of AC is the same as that of BD. (It is not.) A+C = (-5+4, 5+2) = (-1, 7). B+D = (1-2, 8-2) = (-1, 6). (A+C)/2 ≠ (B+D)/2, so the midpoints of the diagonals are different points.

___

The slope of AB is ∆y/∆x, where the ∆y is the change in y-coordinates, and ∆x is the change in x-coordinates.

... AB slope = (8-5)/(1-(-5)) = 3/6 = 1/2

The slope of AD is computed in similar fashion.

... AD slope = (-2-5)/(-2-(-5)) = -7/3

The product of these slopes is (1/2)(-7/3) = -7/6 ≠ -1. Since the product is not -1, the segments AB and AD are not perpendicular to each other. Adjacent sides of a rectangle are perpendicular, so this figure is not a rectangle.

___

Our preliminary work with the diagonals showed us the figure was not a parallelogram (hence not a rectangle). For our slope calculation, we "magically" chose two sides that were not perpendicular. In fact, this choice was by "trial and error". Side BC is perpendicular to AB, so we needed to choose a different side to find one that wasn't. A graph of the points is informative, but we didn't start with that.

under a reflection in the y-axis

a point (x, y ) → (- x, y ), thus

R(6, 6 ) → R' (- 6, 6 )

S(3, - 6 ) → S'(- 3, - 6 )

T(0, 3 ) → T'(0, 3 )

Plot the sets of points and graph them

Answer:  The graph is attached below.

Step-by-step explanation:  Given that the co-ordinates of the vertices of ΔRST are R(6, 6), S(3, –6), and T(0, 3).

We are given to graph ΔRST and its image after a reflection over the Y-axis.

After reflection across Y-axis, the co-ordinates of the vertices of ΔRST will follow the following transformation :

(x, y)  ⇒  (-x, y), because the sign before the x-coordinates of the vertices will get reversed.

Therefore, the co-ordinates of the vertices of the image of ΔRST will be

R'(-6, 6), S'(-3, -6) and T'(0, 3).

The graphs of both the triangles, ΔRST and its image after reflection R'S'T' is drawn in the attached figure.

We see that the vertices T and T' coincide with each other.

Thus, the graph is shown below.

You need to pick from each column depending on what makes sense - they are mixed (some numbers are roots, some cubes, in the same column).

On the order:

3 - 27

4 - 64

5 - 125

512 - 8

8 - 2

729 - 9

216 - 6

1 - 1

Answer:

as>JDKFas;lkdjfsa;lkdfjas;kldfjlakdjfalsdkfjakdkdkdlddd

Step-by-step explanation:

Answer:

See the attached.

Step-by-step explanation:

A graph of f' is a graph of the slope of the function. Your function f(x) is piecewise linear, so different sections of its graph have different constant values of slope.

In the intervals (-5, -2) and (0, 2), the slope is -1. (The graph has a "rise" of -1 for each "run" of 1.) So, in those intervals, the graph of f' looks like a graph of y=-1.

In the interval (-2, 0), the rise is 2 for a run of 2, so the slope is 2/2 = 1. The graph of f' in that interval will look like a graph of y=1.

In the interval (2, 5), the rise of f(x) is 1 for a run of 3, so the slope in that interval is 1/3. There, the graph of f' will look like a graph of y=1/3.

If you want to get technical about it, the slope is undefined at x=-2, x=0, and x=2. Therefore, the line segments that make up the graph of f' ought to have open circles at those points, indicating that f' is not defined.

Step-by-step explanation:

6/8 / 3/4 = 1

The equation to describe the relationship between the number of lilies and roses is l/30 = r/78. This equation represents the proportion between the number of lilies and roses used in each arrangement.

The relationship between the number of lilies (l) and the number of roses (r) can be described by the equation l/30 = r/78. This equation represents the proportion between the number of lilies and roses used in each arrangement. By setting up this proportion, we can determine the ratio of lilies to roses in each arrangement.

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Hey mate!!

Answer⤵

The difference between 1,968 and 3,000 is 1,032.

3,000-1,968=1,032.

Answer confirmed= 1,032

Hope it helps you! ヅ

To find the difference between 1968 and 3000, you need to subtract the smaller number from the larger one.

Step 1: Identify the larger number. In this case, 3000 is larger than 1968.

Step 2: Subtract the smaller number from the larger number.
3000 - 1968 = 1032

So, the difference between 1968 and 3000 is 1032.

Answer: First option. The slope is -4 and the y-intercept is 2.

Solution:

y=-4x+2

When the equation is in the form:

y=mx+b (y isolated)

The coefficient of the variable "x" is the slope "m" of the right line. In this case the coefficient of "x" is -4, then the slope "m" is -4.

The indeoendent term is the y-intercept "b". In this case the independent term is +2, then the y-intercept "b" is 2.  

Solution:

Given equation of line [tex]y=-4x+2[/tex].

The given equation is in the form of y=mx+b, y  is isolated. So the coefficient of x is the slope of the line.

The slope of the line y=-4x+2 is -4

To find the y-intercept of the equation, substitute [tex]x=0[/tex] in the equation,

[tex]\Rightarrow y=-4(0)+2\\\Rightarrow y=2[/tex]

So,  y-intercept of the equation is [tex](0,2)[/tex].

Hence, the slope is -4 and y intercept is 2. (first option)