Quadrilateral DEFG is a rectangle.
If the measure of Angle EDF = 5x-3 and the measure of Angle DFG = 3x+7,
find the measure of Angle EDF.

a. 20
b. 22
c. 18

Answer:

First option is correct. The location of point A after the transformations is (-5,1).

Step-by-step explanation:

It is given that the coordinates of point A are (-5,1).

Rectangle ABCD is reflected over the x-axis. then x-coordinate remains the same but the sign of y-coordinate is changed.

[tex](x,y)\rightarrow (x,-y)[/tex]

Coordinates of point A are

[tex](-5,1)\rightarrow (-5,-1)[/tex]

After that ABCD is reflected over the y-axis, then y-coordinate remains the same but the sign of x-coordinate is changed.

[tex](x,y)\rightarrow (-x,y)[/tex]

Coordinates of point A are

[tex](-5,-1)\rightarrow (5,-1)[/tex]

After that ABCD rotated 180 degrees about the origin, then the sign of both coordinates are changed.

[tex](x,y)\rightarrow (-x,-y)[/tex]

[tex](5,-1)\rightarrow (-5,1)[/tex]

Therefore option 1 is correct.

Answer with explanation:

The given fourth degree polynomial is:

  [tex]f(x)=x^4+6 x^3-3 x^2+17 x-15[/tex]

By rational root theorem , the possible Zeroes of the polynomial are factors of 15 .

Factors of 15 are

  [tex]=\pm 1, \pm 3, \pm 5, \pm 15[/tex]

There are 8 possible Zeroes of the given Polynomial expression.

First line because 4(2a+1)=8a+4 and 9a+4-(8a+4)= a

Second line because a times 2 =2a and 2a+1-2a=1

Although the second equality is more or less obvious since 2a+1 leaves a remainder of 1 when divided by a.

The area of the real rectangular dining hall (assuming that 5cm on paper is 6m in real) is 72m².

A rectangle is a quadrilateral whose opposite edges are of equal length and the opposite angles are of equal measurements.

Given that

the length of the dining hall, l = 10cm

the width of the dining hall, b = 5cm

As evident from the figure, the dining hall is rectangular, and since it is known that the area of a rectangle is the product of its length and width, therefore,

Area of the hall (on paper) = l × b

                          = 10 × 5

                          = 50 square cm.

Also, as it is given in the question that 5cm on paper represents 6m in real, therefore

[tex]Area\ of\ the\ hall\ (in\ real)= (\dfrac{10}{5}\times6)\times(\dfrac{5}{5}\times6)[/tex]

                                          [tex]\\= 2\times6\times6\\= 72\ m^2[/tex]

Hence, the area of the hall in real is 72m².

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Answer:

A.

Step-by-step explanation:

The equation 5x^2 + 5y^2 + 10x - y = 0 represents a circle. By completing the square, the equation is rewritten in the standard form as (x + 1)^2 + (y - 1/10)^2 = 101/100. The circle's center is at (-1, 1/10) with a radius of approximately 1.005.

To show that the equation 5x^2 + 5y^2 + 10x - y = 0 represents a circle and to rewrite it in standard form, we can complete the square for both x and y terms. First, factor out the coefficients of the squared terms:

Divide both sides by 5 to simplify the equation:

Now, complete the square by adding and subtracting the necessary constants inside each parenthesis:

By completing the square, the equation becomes:

This can be further simplified to:

The standard form of the equation for a circle is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius of the circle. Therefore, our circle's center is at (-1, 1/10) and its radius is the square root of 101/100, which simplifies to 10.05/10 or approximately 1.005.

Measure of angle TRV is [tex]130^{0}[/tex].

An angle is formed when two straight lines or rays meet at a common endpoint. The common point of contact is called the vertex of an angle.

According to the question

< TRS  + < TRV = [tex]180^{0}[/tex]

(x - 10) + (2x + 10) = [tex]180^{0}[/tex]

3x =  [tex]180^{0}[/tex]

x = [tex]\frac{180}{3}[/tex]

x = [tex]60^{0}[/tex]

< TRV = 2x + 10

= 2([tex]60^{0}[/tex])+10

= 120 + 10

= [tex]130^{0}[/tex]

Hence, measure of angle TRV is [tex]130^{0}[/tex].

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The value of x is 7/6.

To find the value of x, we can utilize the given information about the lengths of the medians and segments within triangle ABC.

Using medians CE and GC:

Since G is the centroid of triangle ABC, it divides each median into two segments in a ratio of 2:1. Therefore, we can set up two equations based on the given lengths:

CE = 2/3 * GC

6x = 2/3 * (3x + 7)

Solving for x, we get:

6x = 2x + 14/3

4x = 14/3

x = 7/6

Using medians BG and FG:

Similarly, we can set up two equations based on the given lengths of medians BG and FG:

FG = 2/3 * BG

x + 8 = 2/3 * (5x - 1)

Solving for x, we get:

x + 8 = 10x/3 - 2/3

8/3 = 9x/3

x = 8/9

Using medians DG and AG:

Following the same approach, we can set up two equations based on the given lengths of medians DG and AG:

AG = 2/3 * DG

9x - 6 = 2/3 * (4x - 5)

Solving for x, we get:

9x - 6 = 8x/3 - 10/3

x - 6 = 8x/3 - 10/3

-5 = 5x/3

x = -3

Comparing the values obtained from each set of equations, we find that x = 7/6 is consistent across all three sets. Therefore, the value of x is 7/6.

Answer:

Step-by-step explanation:

Simple interest formula is : [tex]p\times r\times t[/tex]

t is always in years here.

1.

p = 250

r = 2.7% or 0.027

t = 1

Simple interest earned = [tex]250\times0.027\times1[/tex] = $6.75

2.

p = 389.42

r = 3.2% or 0.032

t = 7

Simple interest earned = [tex]389.42\times0.032\times7[/tex] = $87.23

Amount after 7 years will be = [tex]389.42+87.23[/tex] = $476.65

3.

p = 1567.12

r = 1.9% or 0.019

t = [tex]9/12=0.75[/tex]

Simple interest earned = [tex]1567.12\times0.019\times0.75[/tex] = $22.33

Account balance after 9 months = [tex]1567.12+22.33[/tex] = $1589.45

Brandi rides her bike at a speed of 5 m/s from school to her house. She will take an additional 188 seconds to cover the remaining distance of 940 meters to arrive at her house.

Brandi's situation:

To find: Additional time needed to reach house.

Calculations:

The fifth term in the binomial expansion of [tex](x+5)^{8}[/tex] is [tex]\boxed{43,750\ x^{4}}[/tex].

Further explanation:

Given:

The binomial term is [tex](x+5)^{8}[/tex].

The expansion of [tex](x+5)^{8}[/tex] is as follows:

[tex]\boxed{{\left({a+b}\right)^n}=\sum\limits_{k=0}^n{{}^n{{\text{C}}_k}{a^{n - k}}{b^k}}}[/tex]

There are [tex]n+1[/tex] terms in the expansion of [tex](a+b)^{n}[/tex].

The sum of indices of [tex]a[/tex] and [tex]b[/tex] is equal to [tex]n[/tex] in every term of the expansion.

The general term [tex]T_{r+1}[/tex] of the binomial term [tex](a+b)^{n}[/tex] is as follows:

[tex]\boxed{{{\text{T}}_{r + 1}}={}^n{{\text{C}}_r}{a^{n - r}}{b^r}}[/tex]

For [tex]5^{th}[/tex] term the value of [tex]r[/tex] is calculated as follows:

[tex]\begin{aligned}r+1&=5\\r&=5-1\\r&=4\end{aligned}[/tex]

Now, the [tex]5^{th}[/tex] term of [tex](x+5)^{8}[/tex] is calculated as follows:

[tex]\begin{aligned}T_{5}&=T_{4+1}\\&=^8C_{4}\cdot x^{8-4}\cdot 5^{4}\\&=\dfrac{8\cdot 7\cdot 6\cdot 5}{4\cdot 3\cdot 2\cdot 1}\cdot x^{4}\cdot 625\\&=625\cdot 70x^{4}\\&=43,750x^{4}\end{aligned}[/tex]

Therefore, the fifth term of the binomial expansion [tex](x+5)^{8}[/tex] is [tex]\boxed{43,750\ x^{4}}[/tex].

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Answer details:

Grade: Senior school

Subject: Mathematics

Chapter: Binomial Theorem

Keywords: Binomial theorem, expansion, (x+5)^8, 175000x3, 43750x4, 3125x5, 7000x5, fifth term, binomial expansion, genral term, binomial, polynomial, indices.

The answer is:

[tex]\[\boxed{43750x^4}\][/tex]

To find the fifth term in the binomial expansion of [tex]\((x + 5)^8\)[/tex], we use the binomial theorem. The binomial theorem states that:

[tex]\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\][/tex]

For the expansion of [tex]\((x + 5)^8\)[/tex], we identify:

[tex]\[a = x, \quad b = 5, \quad n = 8\][/tex]

The general term in the expansion is given by:

[tex]\[T_{k+1} = \binom{n}{k} a^{n-k} b^k\][/tex]

We need to find the fifth term, which corresponds to  k = 4  (since k  starts from 0):

[tex]\[T_5 = \binom{8}{4} x^{8-4} 5^4\][/tex]

First, calculate the binomial coefficient [tex]\(\binom{8}{4}\):[/tex]

[tex]\[\binom{8}{4} = \frac{8!}{4! \cdot 4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70\][/tex]

Next, calculate [tex]\( x^{8-4} \)[/tex] and [tex]\( 5^4 \):[/tex]

[tex]\[x^{8-4} = x^4\]\[5^4 = 625\][/tex]

Now, combine these results:

[tex]\[T_5 = 70 \cdot x^4 \cdot 625\][/tex]

Finally, multiply the coefficients:

[tex]\[70 \times 625 = 43750\][/tex]

Thus, the fifth term is:

[tex]\[43750x^4\][/tex]

Therefore, the answer is:

[tex]\[\boxed{43750x^4}\][/tex]

The probability that exactly six children in the family are boys is 0.015625.

An arrangement of objects where the order in which the objects are selected does not matter.

The probability of having exactly six boys in a family with eight children can be calculated using the binomial distribution formula:

[tex]P(X = k) =^nC_{k} p^k(1-p)^(^n^-^k^)[/tex]

P(X = k) is the probability of having exactly k boys, n is the total number of children,  k is the number of boys we want to find , p is the probability of having a boy and  (1-p) is the probability of having a girl (also 1/2 in this case)

Substituting the values into the formula, we get:

P(X = 6) = ⁸C₆ (1/2)⁶ (1/2)⁸⁻⁶

= (8! / (6! ×2!)) (1/2)⁸

= (87/21) × (1/2)^8

= 0.015625

Therefore, the probability that exactly six children in the family are boys is 0.015625.

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Answer:

The answer is 5,5,5.

Step-by-step explanation:

Reflection mean same like a mirror, it says reflectional symmetry. So the sides are the same.

The function y=x-cosx has an extremum on the interval [0,2pi] at x=3pi/2, determined by setting the first derivative equal to zero. The second derivative test is inconclusive at this point, but the point (3pi/2, 3pi/2) is a minimum due to the nature of the function.

To find all points of extrema on the interval [0,2pi] for the function y=x-cosx, we first need to find the derivative of the function and determine where it is equal to zero. The first derivative of y=x-cosx is y' = 1 + sinx. The critical points occur where y' = 0, which translates to 1 + sinx = 0 or sinx = -1. The only point in the interval [0,2pi] where sinx = -1 is at x = 3pi/2. To determine if this critical point is a maximum or minimum, we apply the second derivative test.

The second derivative of the function is y'' = cosx. At x = 3pi/2, the second derivative y'' = cos(3pi/2) = 0; since the second derivative is zero, the second derivative test is inconclusive. However, we can analyze the function around the critical point to conclude or use other methods like analyzing the first derivative's sign change.

For y=x-cosx, since the original function is a combination of a linear increasing function and cosine function, the critical point at x = 3pi/2 will be a minimum as the function increases both before and after this point.

Therefore, the point of extrema on the interval [0,2pi] for the function y=x-cosx are at (3pi/2, 3pi/2 - cos(3pi/2)), which simplifies to (3pi/2, 3pi/2).

Answer:

(<2 and <4 are vert angles)     reason:given

(lines m and n intersect at p)   reason: def of vertical angles

(<2 and <3 are a linear pair)      reason: def of a linear pair

(<2 and <3 are a linear pair)      reason: def of a linear pair

(m<2+m<3=180)                          reason:angle addition postulate

(m<3+m<4=180)                          reason:angle addition postulate

(m<2+m<3=m<3+m<4                  reason:substitution property

(m<2 = m<4)                                 reason:subtraction property

<2 ~<4                                            reason:definition of ~ angles

Step-by-step explanation:

edgenuity 2020

The function f(x) is a piecewise function. To find the limit as x approaches 2, you must consider it from the left and right. From the left, the limit is ln(2), from the right, it's 4ln(2). Since these are not equal, the limit does not exist.

The function f(x) is a piecewise function where the first range covers 0 < x <= 2 and the second ranging from 2 < x <= 4. To find the limit of f(x) as x approaches 2, we need to find the limit from both directions, as x approaches 2 from the left (<) and the right (>).

For x approaches 2 from the left (<), f(x) boils down to ln(x), so the limit as x approaches 2 would be ln(2).

 For x approaching 2 from the right (>), f(x) translates to x^2 * ln(2). Substituting x = 2 in this function gives us (2)^2 * ln(2) which is 4ln(2)

The limit does not exist since ln(2) not equal to 4ln(2).

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The limit of the function f(x) as x approaches 2 does not exist, because the values obtained when approaching from the left (ln2) and from the right (4ln2) do not match.

The question is asking us to find the limit of the function f(x) as x approaches 2. This is a well-known concept in calculus, called the limit of a function. It is easy to solve using the rule that if f(x) and g(x) are two functions that agree at every point of a certain interval, except perhaps at one single point 'a', then their limits as x approaches 'a' are the same.

So let's find the limit as x approaches 2 by looking at each side independently. For x less than or equal to 2, the function is defined as ln(x). So when x approaches 2 from the left, we get ln(2). For x greater than or equal to 2, the function is defined as x^2ln(2). So when x approaches 2 from the right, we get 4ln(2).

Since the values obtained when approaching 2 from the left (ln2) and from the right (4ln2) do not match, we can conclude that the limit of f(x) as x approaches 2 does not exist.

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