How use zero product property to solve equations

Final answer:

To find the derivative of a fraction without using the quotient rule, you can use the chain rule and the product rule.

Explanation:

To find the derivative of a fraction without using the quotient rule, you can use the chain rule and the product rule. Here's an example:

Let's differentiate the function f(x) = (3x + 2) / (2x + 1).

Simplifying further, the derivative of f(x) is (6x + 3 - 6x - 4) / ((2x + 1)^2) = -1 / ((2x + 1)^2).

In this set, the points reflect across the indicated lines of symmetry. Matches are as follows: 3 corresponds to X, 5 to F, 6 to B, 4 to Y, 1 to Q, and 2 to S.

R_I

This means that the line of symmetry is line I

R_m

This means that the line of symmetry is line m.

1. R_I:T

This matches with Q.

In this number, the line of symmetry is I. If you reflect T across line I, you will get point Q. This is because they are right across from each other with line I between them, and they are both the same distance from line I.

2. R_I:S

This matches with S.

Point S lies on the line of symmetry; thus, S reflects itself.

3. R_I:Z

This is similar to number 1. If you reflect Z across line I, you will get point X because they are directly across each other with line I between them and they have the same distance from the line of symmetry.

4. R_m:S

This time, S isn't on the line of symmetry. The line of symmetry is the vertical line m. Point Y matches with point S because they are directly across each other with line m between them. They are also the same distance apart from line m.

5. R_I:B

This is very similar to number 1. This matches with F. The point directly across the line of symmetry is F.

6. R_I:F

This is the exact same as number 5! With the same line of symmetry, B matched with F. Thus, F must match with B!

Here are the matches:

3-X

5-F

6-B

4-Y

1-Q

2-S

Answer:

Option A.(-5, 2).

Step-by-step explanation:

Picture attached in the question we know the coordinates of the point C are (5, -2). Now we rotate C with a rotation of 180° about the origin.

As we can see a line connecting C and a new point C'. This line shows the rotation of the poit by 180°.

Therefore the new coordinates of C will be C'(-5, 2).

Option A is the correct option.

Answer:

The last option is the correct answer: [tex]r=\sqrt{\frac{A}{\pi}}[/tex]

Step-by-step explanation:

The area of a circle is given by :

[tex]A=\pi r^{2}[/tex] ; where r is the radius.

Now, lets find the radius.

[tex]\frac{A}{\pi } =r^{2}[/tex]

[tex]\sqrt{\frac{A}{\pi}} =r[/tex]

or [tex]r=\sqrt{\frac{A}{\pi}}[/tex]

Therefore, the last option is the correct answer.

The lateral surface area of the equilateral triangular prism shaped shipping tube is 252 square inches.

The question pertains to the calculation of the lateral surface area of a triangular prism shaped shipping tube, more specifically an

equilateral triangular prism

. The lateral area of a triangular prism can be calculated using the formula: Perimeter of the base x length of the prism. Here, the base of the prism is an equilateral triangle, so the perimeter of the base will be 3 x the length of one edge, which in this case is 3 x 6 inches = 18 inches. The length of the prism is given to be 14 inches. Hence, by substituting these values into the formula, the lateral surface area = 18 inches x 14 inches =

252 square inches.

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Answer:

The locations of given number are shown on the below number line.

Step-by-step explanation:

The given numbers are

[tex]4\dfrac{1}{3} and -1\dfrac{1}{3}[/tex]

We need to plot these points on the given number line.

In the given number write the integers -4,-3,-2,-1,0,1,2,3,4 on points indicated and two sub point between two consecutive points divides the line in three equal parts.

[tex]4\dfrac{1}{3}=4+\dfrac{1}{3}[/tex]

It means [tex]4\dfrac{1}{3}[/tex] lies next to 4 on right side.

[tex]-1\dfrac{1}{3}=-(1+\dfrac{1}{3})=-1-\dfrac{1}{3}[/tex]

It means [tex]-1\dfrac{1}{3}[/tex] lies next to -1 on left side.

Therefore, the locations of given number are shown on the below number line.

The area of the base of the pyramid which is a hexagon with given parameters is; 96√3

We are told that the base of the pyramid is a hexagon.

Now, from the given image we can find the base length of the right angle triangle from trigonometric ratio which is;

x/8 = cos 60°

x = 8 cos 60°

x = 4

Thus, length of side of hexagon is;

s = 4 * 2

s = 8

Formula for area of hexagon is given by;

A = ((3√3) * s²)/2

Where;

s is length of a side of hexagon

A = ((3√3) * 8²)/2

A = 96√3

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The dimension of the rectangle will be 6 inches by 3 inches.

Assume L is the length and W is the width. Then the area is given as,

A = L × W

And the perimeter is given as,

P = 2(L + W)

A rectangular note card has an area of 18 square inches. Its perimeter is 18 inches. Then the equations are given as,

18 = 2(L + W)

9 = L + W             ...1

L x W = 18           ...2

From equations 1 and 2, then we have

L x (9 - L) = 18

L² - 9L + 18 = 0

L² - 6L - 3L + 18 = 0

L(L - 6) - 3(L - 6) = 0

(L - 6)(L - 3) = 0

L = 3, 6

Then the width of the rectangle will be calculated as,

W = 9 - 3 = 6

W = 9 - 6 = 3

Then the dimension of the rectangle will be 6 inches by 3 inches.

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Answer:

The correct answer is:  6.5%

Step-by-step explanation:

Answer:

hello the answer is choice D

Step-by-step explanation:

To find Coleman's test grade, set up the equation x/2 + 27 = 67, solve for x by first subtracting 27 and then multiplying by 2 to get x = 80.

The student is asking to find Coleman's test grade given that if you take Coleman's test grade, divide by 2, and add 27, the result is 67. To solve for Coleman's test grade, we can set up an equation where Coleman's test grade is represented by x. The equation based on the provided information would be x/2 + 27 = 67. We can solve for x by first subtracting 27 from both sides of the equation to get x/2 = 40, and then multiplying both sides by 2 to find x. Therefore, x = 80, which means Coleman's test grade is 80.

To find Coleman's test grade, we can set up an equation: (x/2) + 27 = 67. By solving this equation, we find that Coleman's test grade is 80.

To find Coleman's test grade, we can set up an equation using the information given. Let's represent Coleman's test grade as 'x'. We have the equation: (x/2) + 27 = 67.

To solve this equation, we need to isolate 'x'. We can start by subtracting 27 from both sides of the equation: (x/2) = 40.

Next, we need to get rid of the fraction by multiplying both sides of the equation by 2: x = 80.

So Coleman's test grade is 80.

If  cost of annual tution at a university increased from $10,500 to $11,300.Then 7.61%  increase in tution to the nearest tenth of a percent

Percentage, a relative value indicating hundredth parts of any quantity.

Given,

The cost of annual tution at a university increased from $10,500 to $11,300.

We need to find what percentage increase in tution.

Firstly we have to check how much increased.

11300-10500

800

Hence 800 is increase.

We have to check what is 800 percentage in 10500

percent increase=change/original

origianal=10500

change=800

800/10500=8/105=0.0761

By the definition of percentage we know percent means parts out of 100

0.0761×100=7.61%

Hence 7.61% is the  increase in tution to the nearest tenth of a percent.

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Answer:

11. Equation of hyperbola having  vertices at (0, ±6) and foci at (0, ±9).

is given by [tex]\frac{y^2}{a^2}- \frac{x^2}{b^2}=1[/tex]

also b²= c²-a²

 b²=81-36

b²=45

So, Equation becomes , [tex]\frac{y^2}{36}- \frac{x^2}{45}=1[/tex]

Option (D) is correct.

12.Vertices (0,  ± 4), Asymptotes = ±1/4.x

equation of asymptote is given by, [tex]x = \pm y \frac{b}{a}[/tex]

[tex]\frac{4}{b}=\frac{1}{4}[/tex]

So a=4 and b=16,

So , equation becomes [tex]\frac{y^2}{16}- \frac{x^2}{256}=1[/tex]

Option (B) is correct.

13.  x= t-3, and y = t²+ 5

Replace t by x+3, we get

y= (x+3)² +5

y=x²+ 6x +14

Option (A) is correct.

14. Polar coordinates is given by (r,∅)

Polar coordinates of point is (3, 2π/3)

So, r =3, ∅ =2π/3

x= r cos∅ and y = r sin∅

x=3 Cos (2π/3) and y= 3 Sin (2π/3)

x= -3/2 and y=3√3/2

Option (A) is correct.

14. Polar coordinate is given by (r,∅)

Here , r=1, and ∅ = -π/6

x= r Cos ∅ and y =r Sin∅

x= 1 Cos (-π/6) and  y= 1 Sin (-π/6)

x =√3/2 and y =1/2

Option (B) which is B) (1, negative pi divided by 6 + 2nπ) or (-1, negative pi divided by 6 + 2nπ) is correct.

16. Two pairs of polar coordinates for the point (4, 4) with 0° ≤ θ < 360°.

is  option (B) which is B) (4 square root 2 , 45°), (-4 square root 2 , 225°).

17. A circular graph with inner loop on the left of a limacon curve is given by

r = a + b Cos∅.

In this case a> b.

So , Option (D) r = 4 + cos θ as well as (A) r = 3 + 2 cos θ looks correct.

18. Equation of limacon curve is given by r = -2 + 3 cos θ, here , a<b

So it is symmetric about y-axis only.Option (B) is correct.