The 4th time of me posting this question

Answer:

A).4m^(+5)

B).2m^ (+7)

Step-by-step explanation:

First let me convert the expression to words for easy simplification and evaluation.

2m squared times 2m to the power 3.

= 2m² * 2m³

m to the power 4 times 2m to the power 3

= m⁴ * 2m³

So to evaluate.

We just multiply and get our answer.

For 2m² * 2m³

= (2*2)(m²*m³)

= 4(m(²+³))

= 4m^(+5)

For m⁴ * 2m³

= 2(m⁴*m³)

= 2m(⁴+³)

= 2m^ (+7)

Answer:

Circle O is similar to Circle X because they have the same radius

The work done by the man against gravity in climbing to the top is 16740 lb-ft

The work done against gravity relies on the height of the object and the weight at which the object is changing.

From the given information:

Taking the vertical y-axis when y = 0, then:

w(0) = 20 lb

w(90) = 20 - 8 = 12 lb

Provided that the paint leaks steadily, the function of y i.e. w(y) can be expressed as a linear function in the form:

w(y) = a + by ---- (1)

Thus;

From equation (1)

w(y) = 20 - 4y/45

The total weight becomes;

w = w(y) + the man's weight

w = 20 - 4y/45 + 170

w = 190 - 4y/45

Therefore, the work done against gravity is computed as:

W = ∫ w dy

where;

[tex]\mathbf{W = \int ^{90}_{0}( 190 - \dfrac{4y}{45} )\ dy }[/tex]

W = 16740 lb-ft

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Answer:

  900

Step-by-step explanation:

We can write the proportion as ...

  calories/crackers = 60/12 = y/180

Multiplying by 180, we have ...

  180(5) = y = 900

There are 900 calories in 180 crackers.

Answer:

There are 24 pieces of fruit in the basket.

Step-by-step explanation:

There are 24  pieces of fruit in the basket because to find how many there are you would do [tex]\\ \frac{6}{5\\}\\ \\[/tex] · [tex]\frac{5}{6}[/tex]f = 20 · 6/5 times 5/6 cancels out, and 20 time 6/5 is 24. 20/24 can also simplify to 5/6.

Hope this helps!

(Make sure to stream ON, Black Swan. Boy With Luv, Mic Drop, and etc.!)

Answer:60

Step-by-step explanation:

⅟₂ of 30=15

¼ of 60=15

The required value is 60.

The size, number, or amount of one thing or group as compared to the size, number, or amount of another is considered as proportion.

Given that, 1/2 of 30 = 1/4 of ??

So, 1/2 of 30 = 15

a/b = c/d

ad = bc

Let the unknown value be x,

Therefore,

30 / 2 = x / 4

2x = 120

x = 60

Hence, the value of unknown element is 60.

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Answer: First side is 40.5 meters. Second side is 121.5 meters. The third side is 159.

Step-by-step explanation:

x + 3x + 4x -3 = 321

8x - 3 =321

      +3     +3

8x = 324

x = 40.5

Second Side:   3(40.5) = 121.5

Third side:  4(40.5) - 3  = 159

Check: 159 + 121.5 + 40.5 =  321    

Answer:

Expectation number of books a customer will​ purchase  =  1.3

Step-by-step explanation:

x     :​ 0      1              2

p(x) :0.2​,  0.3​, and 0.5​

Expectation

E(X) = ∑x P(x)

      = 0 X 0.2 +1 X 0.3+ 2 X 0.5

     = 1.3

Expectation  E(X) =  1.3

Expectation number of books a customer will​ purchase  =  1.3

Answer:

2x+1

Step-by-step explanation:

that's a hard question

we know that g(x)= x-3

so f(g(x))= f(x-3)

we put it in the equation :

f(x-3)= 2(x-3) +7 = 2x-6+7 = 2x +1

Answer:

n = 4

Step-by-step explanation:

These triangles are similar, so you can set up a proportion like this:

[tex]\frac{20}{n} =\frac{10}{2}[/tex]

→Cross multiply:

[tex]\frac{40}{10n}[/tex]

→Divide 40 by 10:

n = 4

Answer:

line AB is congruent to line CB . proof: given

line AE is congruent to line CD . proof: definition of a rectangle

angle EAB is congruent to angle BCD . proof: definition of a rectangle

triangle ABE is congruent to CBD . proof: SAS

line BE is congruent to line BD . proof: CPCTC

Step-by-step explanation:

Sorry if I'm wrong on this, but I hope you understand the general idea!

Corrected Question

The function is: [tex]R(x)=80+7.440\sqrt{x} , 0\leq x\leq 15000[/tex]

Answer:

(a)$48.02

(b)$656,299.92

Step-by-step explanation:

(a) We are to determine the change in monthly revenue if the monthly expenditure, x is raised from its current level of 6000 to 6001.

[tex]R(x)=80+7.440\sqrt{x}\\R'(x)=\dfrac{d}{dx} (80+7.440\cdot x^{1/2})\\=7.440 \times \frac{1}{2} \times x^{\frac{1}{2}-1}\\=3.72\times x^{-\frac{1}{2}}\\\\R'(x)=\dfrac{3.72}{\sqrt{x} }[/tex]

Therefore, the expected change in monthly revenue

[tex]R'(6000)=\dfrac{3.72}{\sqrt{6000} }=$0.04802 (thousands)[/tex]

=$48.02

(b)When the amount spent on advertising is $6000

[tex]Revenue, R(6000)=80+7.440\sqrt{6000}\\=\$656.29992$ (in thousands)\\=\$656,299.92[/tex]

Answer:

[tex] f(x) = x^2 -8x -20[/tex]

We know that one factor is (x-10). We know that we need to find two values that added we got -8 and multiplied we got -20. So then we can set the following options:

[tex] a(10) = -20[/tex]

[tex] a-10= -8[/tex]

And solving for a we got:

[tex] a = -8+10=-2[/tex]

And we got:

[tex] a = -2[/tex]

So then the other factor would [tex] x +2[/tex]

Step-by-step explanation:

For this case we have the following expression:

[tex] f(x) = x^2 -8x -20[/tex]

We know that one factor is (x-10). We know that we need to find two values that added we got -8 and multiplied we got -20. So then we can set the following options:

[tex] a(10) = -20[/tex]

[tex] a-10= -8[/tex]

And solving for a we got:

[tex] a = -8+10=-2[/tex]

And we got:

[tex] a = -2[/tex]

So then the other factor would [tex] x +2[/tex]

Answer:

Step-by-step explanation:

Answer:

D

Step-by-step explanation:

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

inverse ----> f(x) = y = x

x = [tex]y^{2}[/tex] - 1

[tex]y^{2} = x + 1[/tex]

y = ± [tex]\sqrt{x + 1}[/tex]

The question is incomplete. Here is the complete question.

The number of students, S, serviced by the school system in the town of Emor, t years from 2000 can be modeled by the function S(t) = 10000.[tex](1.1)^{t}[/tex]. The number of classrooms, C, in the town of Emor, t years from 2000 can be modeled by the function C(t) = 450 + 40t. Let D be the average number of students per classroom in Emor's school system t years from 2000.

Write a formula for D(t) in terms of S(t) and C(t).

Write a formula in terms of t.

Answer: D(t) = S(t) / C(t)

D(t) = [tex]\frac{10000.(1.1^{t} )}{450+40t}[/tex]

Step-by-step explanation: First, it is asked to write D(t), which is the average number of students per classroom in terms of students, S(t) and classroom, C(t).

Average is the total number of students divided by the total number of classrooms. Therefore:

D(t) = [tex]\frac{S(t)}{C(t)}[/tex]

Second, to write in term of t, which is time in years, for the average number of students per classroom:

D(t) = [tex]\frac{10000.(1.1)^{t} }{450+40t}[/tex]

In this formula it is clear that the average number of students per classroom is dependent of the growth factor of students each year represented by [tex]1.1^{t}[/tex] and the "growth factor" of classroom each year, represented by 40t.

Answer:

y= -5/24

Step-by-step explanation:

2y=50y+10

1: Subtract 50y from both sides.

2y−50y=50y+10−50y

−48y=10

2: Divide both sides by -48.

−48y /48y=10/-48

y= -5/24

Answer:

Step-by-step explanation:

y = -3x + 3

-2x + 2y = 6

-2x + 2(-3x + 3) = 6

-2x - 6x + 6 = 6

-8x + 6 = 6

-8x = 0

x = 0

y = -3(0) + 3

y = 0+ 3

y = 3

(0, 3)

Answer:

  see below

Step-by-step explanation:

The first answer choice is the only one that has the first sale correctly represented. Each term is the product of the number of pens of the type and the cost of pens of that type.

  12 dozen roller-ball pens will have a cost of 12x

  5 dozen ballpoint pens will have a cost of 5y

The total cost of these will be 116 (dollars).

The equation is written ...

  12x +5y = 116 . . . . . . . this equation is only found in choice A

Answer:

She liked the smell and was turned on slightly.

Step-by-step explanation:

Some people get aroused when smelling a fart. The teacher was most likely trying to get into your pants.

Answer:

Ok, i will answer in English.

The first step is to know that when we have a linear equation like:

y = a*x + b

any line with a slope equal to -1/a will be perpendicular to our line.

So first let's write our equation in the standard form:

4x + 6y - 6 = 0

6y = -4x + 6

y = (-2/3)*x + 1

then the slope is (-2/3)

then the slope of the perpendicular line will be:

-(1/(-2/3)) = 3/2

So the only thing any line with the equation:

y = (3/2)*x + c

where c can be any real number, is perpendicular to 4x+6y-6=0

Answer:

same problem with me.That's what happened

Null hypothesis (H0): The reaction time for the dominant hand is not faster than the non-dominant hand.

Alternative hypothesis (Ha): The reaction time for the dominant hand is faster than the non-dominant hand.

The null and alternative hypotheses for testing if the reaction time is faster for the dominant hand can be formulated as follows:

Null hypothesis (H0): The reaction time for the dominant hand is not faster than the non-dominant hand.

Alternative hypothesis (Ha): The reaction time for the dominant hand is faster than the non-dominant hand.

To test these hypotheses, we can perform a paired t-test because the same subjects are tested with both hands. The paired t-test compares the means of the paired observations and determines if there is a statistically significant difference between them.

Using a 5% level of significance (α = 0.05), we can calculate the test statistic and compare it to the critical value from the t-distribution with (n-1) degrees of freedom, where n is the number of subjects.

If the calculated test statistic falls in the critical region (i.e., the p-value is less than 0.05), we reject the null hypothesis and conclude that there is evidence to suggest that the reaction time for the dominant hand is indeed faster.

If the calculated test statistic does not fall in the critical region (i.e., the p-value is greater than or equal to 0.05), we fail to reject the null hypothesis and conclude that there is insufficient evidence to suggest a difference in reaction times between the dominant and non-dominant hand.

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The null and alternate hypothesis tests are:

H₀: μd = μnd

H₁: μd < μnd

Given data:

The hypotheses for testing if the reaction time is faster for the dominant hand can be stated as follows:

Null Hypothesis (H₀): The reaction time is the same for the dominant and non-dominant hand.

Alternative Hypothesis (H₁): The reaction time is faster for the dominant hand.

Mathematically, these can be represented as:

H₀: μd = μnd

H₁: μd < μnd

Where:

μd is the population mean reaction time for the dominant hand.

μnd is the population mean reaction time for the non-dominant hand.

To test these hypotheses, a statistical test such as a paired t-test can be used. The test will compare the means of the paired observations (reaction times for the dominant and non-dominant hand) and determine if there is enough evidence to reject the null hypothesis in favor of the alternative hypothesis.

The significance level of 5% (0.05) indicates that if the p-value obtained from the test is less than 0.05, we would reject the null hypothesis and conclude that there is evidence to support the claim that the reaction time is faster for the dominant hand.

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Answer: I dont have one for sure but I can explain why this is true.

Step by Step:

sin(x+y)−sin(x−y)=sinxcosy+sinycosx−(sinxcosy−sinycosx)

=sinxcosy+sinycosx−sinxcosy+sinycosx

=sinxcosy+sinycosx−sinxcosy+sinycosx

=sinycosx+sinycosx

=2sinycosx

henceforth proved.

Answer:

h = 76

Step-by-step explanation:

2 + h - 48 = 30

h + 2 - 48 = 30

h - 46 = 30

h - 46 + 46 = 30 + 46

h = 76

Answer:

2+h-48=30

2-48+h=30

-46+h=30

h=30+46

∴h=76

Answer:

On the first account, she invests $300, earning 0.02*300 = $6.

On the second acount, she invests 800 - 300 = $500, earning $25.

Step-by-step explanation:

This is a simple interest problem.

The simple interest formula is given by:

[tex]E = P*I*t[/tex]

In which E is the amount of interest earned, P is the principal(the initial amount of money), I is the interest rate(yearly, as a decimal) and t is the time.

In this question:

Two acounts, one with 2% interest(I = 0.02) and the other with 5% interest(I = 0.05).

Each account has two earnings, that i will call [tex]E_{1}[/tex] and [tex]E_{2}[/tex]

Two investments adding up to 800. I will call the first investment P and the second is 800 - P.

Time is not given, but for simplicity, i will use 1 year.

First investment:

I at 2%. The earnings are [tex]E_{1}[/tex].

[tex]E_{1} = 0.02P[/tex]

Second Investment:

Earnings [tex]E_{2}[/tex], at 5%. So

[tex]E_{2} = 0.05(800 - P)[/tex]

The earnings add to 31, so [tex]E_{1} + E_{2} = 31[/tex], then

[tex]E_{2} = 31 - E_{1}[/tex]

So

[tex]31 - E_{1} = 0.05(800 - P)[/tex]

[tex]31 - 0.02P = 0.05(800 - P)[/tex]

[tex]31 - 0.02P = 40 - 0.05P[/tex]

[tex]0.03P = 9[/tex]

[tex]P = \frac{9}{0.03}[/tex]

[tex]P = 300[/tex]

So:

On the first account, she invests $300, earning 0.02*300 = $6.

On the second acount, she invests 800 - 300 = $500, earning $25.

Answer:

(a)$4954

(b)$10

(c)The company should not produce the 101st item.

Step-by-step explanation:

(a)

[tex]C'(50)=\dfrac{C(52)-C(50)}{52-50} \\\\C(50) = 4900,C'(50) = 27\\\\$Therefore:\\27=\dfrac{C(52)-4900}{52-50}\\C(52)-4900=27*2\\C(52)=4900+54\\C(52)=\$4954[/tex]

(b)If C'(50) = 27 and R'(50) =37

Therefore, the profit earned on the 51st item

[tex]=R'(50)-C'(50)[/tex]

=37-27

=$10

(c)If C'(100) = 41 and R'(100) =37

Therefore, the profit earned on the 101st item

Profit [tex]=R'(100)-C'(100)[/tex]

=37-41

=-$4

The company should not produce the 101st item. It would lose $4 if it does.

Answer:

Step-by-step explanation:

Let the events be

W = Wednesday

T = Thursday

F = Friday

S = Saturday

The corresponding probability are

P (W) = 0.3

P (T) = 0.4

P (F) = 0.2

P (S) = 0.1

Let Y = number of days beyond wednesday that takes for both magazines to arrive

(so possible Y values are 0, 1 , 2 ,3)

The possible outcome are 4² = 16

(W,W) (W,T) (W,F) (W,S)

(T,W) (T,T) (T,F) (T,S)

(F,W) (F,T) (F,F) (F, S)

(S,W) (S,T) (S,F) (S,S)

The values associated for each of the following as follows

Y(W,W) =0, Y(W,T) =1, Y(W,F)=2, Y (W,S)=3

Y(T,W)=1, Y (T,T)=1, Y (T,F)=2, Y (T,S)=3

Y(F,W)=2, Y (F,T)=2, Y (F,F)=2, Y (F, S)=3

Y(S,W)=3, Y (S,T)=3, Y (S,F)=3, Y (S,S)=3

The probability mass function of Y is

P(Y=0)=0.3(0.3)=0.9

P(Y=1) = P[(W,T) or (T,W) or (T,T)]

= [0.3(0.4) + 0.3(0.4) + 0.4(0.4)]

=0.4

P(Y = 2) = P[(W,F) or (T,F) or (F,W) or (F,T) or (F,F)]

=0.3(0.2) + 0.4(0.2) + 0.2(0.3) + 0.2(0.4) + 0.2(0.2)]

=0.32

P(Y =3) = P[(W,S) or (T,S) or (F,S) or (S,W) or (S,T) or (S,F) or (S,S)]

= [0.3(0.1) + 0.4(0.1) + 0.2(0.1) + 0.1(0.3) + 0.1(0.4) + 0.1(0.2) + 0.1(0.1)]

= 0.19

Step-by-step explanation:

The surface area of a square pyramid is the area of the square base plus the area of the 4 lateral faces:

A = b² + 4 (½ bl)

A = b² + 2 bl

where b is the width of the base,

and l is the slant height.

Answe The surface area of a square pyramid is the area of the square base plus the area of the 4 lateral faces:

A = b² + 4 (½ bl)

A = b² + 2 bl

where b is the width of the base,

and l is the slant height.:

Step-by-step explanation:

Answer:

(a) [9, 19]

(b)  [tex]9 \leq x \leq 19[/tex]

(c)See attachment

Step-by-step explanation:

We want to choose a value of x at least 5 units away from 14.

(a)Now, 14-5=9 and 14+5=19

The possible values of x ranges is in the closed interval [9,19]

(b) Since x is at least 5 units away from 14, we have:

[tex]|14-x|\leq5[/tex]

Solving the absolute inequality

[tex]-5 \leq 14-x \leq 5\\$In $ -5 \leq 14-x\\ x \leq 14+5\\x \leq 19\\\\$In $ 14-x \leq 5\\ 14-5 \leq x\\9 \leq x\\$Therefore,an inequality that represents all values of x that meet this constraint is:$\\9 \leq x \leq 19[/tex]

(c)To draw the number line, we use a closed dot since we have a less than or equal to sign.

Inequalities are used to represent unequal expressions.

Represent the value with x.

A value that is at least 5 from 14 can take any of the following forms:

[tex]\mathbf{14 -x \ge 5}[/tex]

[tex]\mathbf{x-14 \ge 5}[/tex]

The inequalities can be combined as:

[tex]\mathbf{|x-14| \ge 5}[/tex]

The values of x are solved as follows:

[tex]\mathbf{14 -x \ge 5}[/tex]

[tex]\mathbf{14 - 5 \ge x}[/tex]

[tex]\mathbf{9\ge x}[/tex]

Rewrite as:

[tex]\mathbf{x \le 9}[/tex]

Similarly, we have:

[tex]\mathbf{x-14 \ge 5}[/tex]

[tex]\mathbf{x \ge 14 + 5}[/tex]

[tex]\mathbf{x \ge 19}[/tex]

So, we have:

[tex]\mathbf{x \le 9}[/tex] or [tex]\mathbf{x \ge 19}[/tex]

This can be combined as:

[tex]\mathbf{9 \ge x \ u\ x \ge 19}[/tex]

Some possible values of x are: 5 and 20

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