–5(–2) also i checked and apparently its not -7

Answer:

d.

Step-by-step explanation:

Given a length [tex]l[/tex] and a width [tex]w[/tex], we can express the perimeter of a rectangle as [tex]2(l+w)[/tex].

Therefore the perimeter [tex]P[/tex] can be expressed as

[tex]2\times[(x-8)+(x-9)]\\=2\times(2x-17)\\=4x-34[/tex]

It is true teachers can increase spatial awareness as they use position words or implement mapping and shape manipulation.

This refers to the ability to understand the position of one's body in relation to space and other objects as well as understanding how location and movement work.

In children spatial awareness can be increased through:

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14.11 is the price of one share in the company TYK.

Given:

Find:

Solution:

The dividend is the amount paid to the shareholders for their investment in the company.

Dividend at end of 7th year = $1 per share (given)

dividend expected growth rate = 6%  (given)

the dividend will be paid from the end of seven years.

So find the price of the share at the end of the 6th year.

price at the end of the 6th year

= Dividend at the 7th year/discount rate growth rate

= 1/(10%-6%) = 1/4% = 25

price of the share at the end of the 6th year is $25

discount price 10% (given)

current price share = 25/(1+10%)^6

                              = 25/(1.1)^6

                              = 25/1.77

                             = 14.11

The current price of the share is $ 14.11

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The graph of function f is shifted right 3 units and up 4 units.

Let f(x) be the parent function

Now use the Rule : f(x)→f(x-b) to find the shifting of graph

The graph shifts right by b units

So, shift the graph by 3 unit

So, f(x)→f(x-3)

So, the graph shifted right by 3 units

2nd rule used is Rule : f(x)→f(x)+b to find the another shifting of graph

The graph shifts up by b units

Now shift the obtained graph up by 4 units

So,  f(x-3)→f(x-3)+4

So, the graph shifts up by 4 units

Thus, The graph of function f is shifted right 3 units and up 4 units.

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Answer:

1. x=2       2.

y=1

Step-by-step explanation:

Im sorry I TRIED :(

let's firstly convert the mixed fraction to improper fraction and then take it from there, keeping in mind that the whole is "x".

[tex]\stackrel{mixed}{5\frac{5}{6}}\implies \cfrac{5\cdot 6+5}{6}\implies \stackrel{improper}{\cfrac{35}{6}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{7}{3}x~~ = ~~5\frac{5}{6}\implies \cfrac{7}{3}x~~ = ~~\cfrac{35}{6}\implies 42x=105\implies x=\cfrac{105}{42} \\\\\\ x=\cfrac{21\cdot 5}{21\cdot 2}\implies x=\cfrac{21}{21}\cdot \cfrac{5}{2}\implies x=1\cdot \cfrac{5}{2}\implies x=2\frac{1}{2}[/tex]

In around 6.35 years, the population will be 1 million.

The population is modeled by the exponential equation:

[tex]P(t) = 232,012*e^{0.23*t}[/tex]

Then we just need to solve the equation for t:

[tex]P(t) = 232,012*e^{0.23*t} = 1,000,000[/tex]

Let's solve that:

[tex]232,012*e^{0.23*t} = 1,000,000\\\\e^{0.23*t} = 1,000,000/232,012 = 4.31\\[/tex]

If we apply the natural logarithm to both sides:

[tex]ln(e^{0.23*t}) = ln(4.31)\\\\0.23*t = ln(4.31)\\\\t = ln(4.31)/0.23 = 6.35[/tex]

So in around 6.35 years, the population will be 1 million.

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The walkway should be about 2.57feet.

What is the total surface area?

The total surface area of the building is the  sum of the total faces of an object contains.

The formula is expressed as:

 TSA = 2(lw + wh + lh)

where,

l is the length ,w is the width , h is the height

Calculation

Given the following parameters

TSA = 168 square feet

length = 8 feet and  Height = 6 feet

Required

Width of the walkway, Substitute the given values in the formula

168 = 2(lw + wh + lh)

84 = (8(6) + 6w + 8w)

84 = 48 + 14w

4w = 84 - 48

14w = 36

w = 2.57feet

The walkway should be about 2.57feet.

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The correct steps to transform x^2 + 6x +8=0 into the form (x-p)² =q are;

x^2 + 6x +8 +1  = 0 + 1

x^2 + 6x +8 +1  = 1

x^2 + 6x + 9 = 1

(x+3)² = 1

The vertex form of a quadratic equation is expressed as:

a(x-h)² + k

Given the quadratic equation

x^2 + 6x +8=0

Subtract 8 from both sides

x^2 + 6x +8 - 8 = 0 - 8

x^2 + 6x  = -8

Complete the square

(x^2+6x+3^2) = -8 + 9

(x+3)² = -8 + 9

(x+3)² = 1

Hence the correct steps to transform x^2 + 6x +8=0 into the form (x-p)² =q are;

x^2 + 6x +8 +1  = 0 + 1

x^2 + 6x +8 +1  = 1

x^2 + 6x + 9 = 1

(x+3)² = 1

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[tex]\\ \rm\Rrightarrow \dfrac{x^2+3x+2}{x^2-4x-5}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{x^2+2x+x+2}{x^2-5x+x-5}[/tex]

[tex]\\ \rm\Rrightarrow {(x+2)(x+1)}{(x-5)(x+1)}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{x+2}{x-5}[/tex]

And

[tex]\\ \rm\Rrightarrow \dfrac{x^2+7x+12}{x^2-3x-28}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{x^2+4x+3x+12}{x^2-7x+4x-28}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{(x+4)(x+3)}{(x-7)(x+4)}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{x+3}{x-7}[/tex]

Answer:

A∩B = { 1 }

Step-by-step explanation:

Consider the sets :

A = {1, 2, 6}

B = {x  ;where x is an odd whole number less than 8}

Suppose the numbers of B are natural numbers (not integers).

Then

B = {1 , 3 , 5 , 7}

We obtain :

A = {1, 2, 6}  and  B = {1 , 3 , 5 , 7}

The intersection of A and B or the set A∩B :

is the set of the common numbers of A and B.

Then

A∩B = { 1 }

The set A intersection B(or, A∩B) for the given set A and set B exists {1}.

"The set A intersection B(or, A∩B) exists described as the set composed of all elements that belong to both A and B".

The intersection of a set A with a B exists the set of elements that exist in both sets A and B. The intersection exists represented as A∩B.

The intersection of two mathematical objects exists where they overlap. For geometric objects, the intersection exists as a point or group of points where the objects cross.

For numerical or non-numerical sets, the intersection exists every number or object the two sets contain in common.

Given: A = {1, 2, 6}

B = {x | x is an odd whole number less than 8}.

From the above information, we get

⇒ B = {1, 3, 5, 7}

If A = {1, 2, 6} and B = {1, 3, 5, 7}

then A∩B = {1, 2, 6} ∩ {1, 3, 5, 7}

A∩B = {1}

Therefore, the correct answer A∩B exists {1}.

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Answer:

84 flowers.

Step-by-step explanation:

We must multiply 90 by 1/15 to get 6 bulbs that will not flower. 90-6 is 84.

It's 4 buns & 5 cakes

Hope my explanation to the other one helps, if not, you can draw them; while following the steps & you may get it better.

E.g draw a box for the bun & a circle for the cake, then, adding them up to get the 95p. Also, you keep taking away from 435p

Answer:

[tex]\frac{15}{2}[/tex]

Step-by-step explanation:

The absolute value of a number is just its distance from zero. In other words, if it's negative, the absolute value would make the number positive. This makes the absolute value [tex]\frac{45}{6}[/tex]. Now, reduce to get[tex]\frac{15}{2}[/tex].

The range of [tex]y =\log_8(x)[/tex] is the set of all real values

The function is given as:

[tex]y =\log_8(x)[/tex]

The above function is a parent logarithmic function.

The parent logarithmic function have a range of the set of all real values

Hence, the range of [tex]y =\log_8(x)[/tex] is the set of all real values

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Answer:

D

Step-by-step explanation:

on edg

The inequality graphed is the one in the first option.

On the graph we can see a solid parabola, the parabola opens downwards, and we can see that the vertex of the parabola is on the point (-1, 5).

And the shaded region is above the parabola.

Then the inequality is of the form:

y  ≥  parabola.

By knowing where is the vertex, we know that:

parabola = -a*(x + 1)^2 + 5

Then the correct option is the first one.

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Answer:

Discount: $30 off

New Sale Price: $270

Step-by-step explanation:

$300*10% =

30

$300-30=

$270

or

$300*.1=

30

$300-30=

$270

The domain of the composite function is given as follows:

[–3, 6) ∪ (6, ∞)

The composite function of f(x) and g(x) is given as follows:

[tex](f \circ g)(x) = f(g(x))[/tex]

In this problem, the functions are:

The composite function is of the given functions f(x) and g(x) is:

[tex]f(g(x)) = f(\sqrt{x + 3}) = \frac{1}{\sqrt{x + 3} - 3}[/tex]

The square root has to be non-negative, hence the restriction relative to the square root is found as follows:

[tex]x + 3 \geq 0[/tex]

[tex]x \geq -3[/tex]

The denominator cannot be zero, hence the restriction relative to the denominator is found as follows:

[tex]\sqrt{x + 3} - 3 \neq 0[/tex]

[tex]\sqrt{x + 3} \neq 3[/tex]

[tex](\sqrt{x + 3})^2 \neq 3^2[/tex]

[tex]x + 3 \neq 9[/tex]

[tex]x \neq 6[/tex]

Hence, from the restrictions above, of functions f(x), g(x) and the composite function, the domain is:

[–3, 6) ∪ (6, ∞)

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Based on the student's grades, and the credit hours, their grade point average is 2.60.

First, find the grade points earned:

= (4 x 3) + (1 x 3) + (3 x 3) + (2 x 1)

= 12 + 3 + 9 + 2

= 26

The GPA is:

=Grade points earned / total number of credits

= 26 / (3 + 3 + 3 + 1)

= 26 / 10

= 2.6

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Answer:

telephone was $45, utility was $135

Step-by-step explanation:

Let the telephone bill be x, then the utility bill is 3x

3x+x = 180

4x = 180

x = 45

3x = 135

When [tex]n=0[/tex], we have

[tex]5^{3^0} + 1 = 5^1 + 1 = 6[/tex]

[tex]3^{0 + 1} = 3^1 = 3[/tex]

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for [tex]n=k[/tex], that

[tex]3^{k + 1} \mid 5^{3^k} + 1[/tex]

Now for [tex]n=k+1[/tex], we have

[tex]5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)[/tex]

so we know the left side is at least divisible by [tex]3^{k+1}[/tex] by our assumption.

It remains to show that

[tex]3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1[/tex]

which is easily done with Fermat's little theorem. It says

[tex]a^p \equiv a \pmod p[/tex]

where [tex]p[/tex] is prime and [tex]a[/tex] is any integer. Then for any positive integer [tex]x[/tex],

[tex]5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3[/tex]

Furthermore,

[tex]5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3[/tex]

which goes all the way down to

[tex]5^{3^k} \equiv 5 \pmod 3[/tex]

So, we find that

[tex]\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3[/tex]

QED

The mean and the median based on the information given will be 166.08 and 195.

It should be noted that the mean is also the average. It's gotten by adding all the numbers and dividing by 13. This will be 166.08.

The median is simply the middle number. This will be 195.

The range will be:

= 301 - 7

= 294

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The path of the given Hamiltonian diagram is A, E, B, C, D, F. See explanation below.

A Hamiltonian path, also known as a traceable path, is one that visits each vertex of the graph precisely once.

A traceable graph is one that contains a Hamiltonian path.

Hence, the path is given as;
A, E, B, C, D, F

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Answer:

20

Step-by-step explanation:

20+20=40

It is false that x is a function of y

The equation is given as:

x = y^2

Set y = 2

x = (2)^2

x = 4

Set y = -2

x = (-2)^2

x = 4

Different values of y give the same x value.

This means that x is not a function of y

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The probability of selecting a respondent who does not prefer Cell C is 0.775

The complete question is in the attached image

From the attachment, we have

Prefer cell C = 90

Total = 400

So, the number of respondents that do not prefer cell C is

Do not = 400 - 90

Do not = 310

The probability is then calculated as:

P = 310/400

Evaluate

P = 0.775

Hence, the probability of selecting a respondent who does not prefer Cell C is 0.775

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Answer:

all real values of x where x < -1

Step-by-step explanation:

See attached graph.