Can y’all help with this?

The 5th term of the sequence is 27.98

The given parameters are:

The nth term of a geometric sequence is calculated using:

[tex]T_n = ar^{n-1[/tex]

So, we have:

[tex]T_5 = 1 * 2.3^{5-1[/tex]

Evaluate the difference

[tex]T_5 = 1 * 2.3^4[/tex]

Evaluate the product

[tex]T_5 = 27.98[/tex]

Hence, the 5th term of the sequence is 27.98

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Answer:  p = 2 3/7

Step-by-step explanation:

7p+13 = 30

7p = 17

p = 17/7

p = 2 3/7

A fraction is written in the form of a numerator and a denominator where the denominator is greater than the numerator.

The value of p as a fraction in its simplest form is 17/7

A fraction is written in the form of a numerator and a denominator where the denominator is greater than the numerator.

Example:

1/2, 1/3 is a fraction.

2/4 is a fraction but we can simplify further since there is a common factor of 2 with 2 and 4.

2/4 = 1/2 is a fraction.

4/16 = 4/(4x4) = 1/4 is a fraction.

We have,

7p + 13 = 30

Solve for p.

Subtract 13 on both sides.

7p + 13 - 13 = 30 -13

7p = 17

Divide both sides by 7.

p = 17/7

We can not further simplify the fraction as there are no common factors between 17 and 7.

Thus,

The value of p as a fraction in its simplest form is 17/7

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The sample space S =  {1, 2, 3, 4, 5, 6} describing all possible outcomes and outcomes for the event of rolling the number 2 is E = {2}

It is defined as the space in which all possible outcomes are represented in a list form for a particular event.

We have:

A number cube with faces labeled from 1 to 6 will be rolled once.

Let S is the sample space:

As we know, the sample space is the set of all possible outcomes:

S = {1, 2, 3, 4, 5, 6}

The total number of outcomes:

n(S) = 6

The outcomes for the event of rolling the number 2.

E = {2}

n(E) = 1

Thus, the sample space S =  {1, 2, 3, 4, 5, 6} describing all possible outcomes and outcomes for the event of rolling the number 2 is E = {2}

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The ratio in the form of 1 :n is 1:300.

Ratio of two number A and B is the comparison of A with respect to B and when two ratios are equal they are said to be in proportion.

It is given that a ratio is

6mm : 180cm

It has to be simplified in the form of

1:n

1 cm = 10mm

therefore

180 cm = 1800 mm

6mm:180cm = 6/ 1800

= 1/300

Therefore the ratio in the form of 1 :n is 1:300.

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[tex]\text{L.H.S}\\\\=\begin{vmatrix} 1 & bc& b+c\\ 1 & ca& c+a\\ 1 &ab&a+b\end{vmatrix}\\\\\\=\begin{vmatrix} 1 & bc& b+c-(a+b+c)\\ 1 & ca& c+a-(a+b+c)\\ 1 &ab&a+b-(a+b+c)\end{vmatrix}~~~~~~~~~~~~~~~;c_3\rightarrow c_3 -(a+b+c)c_1\\\\\\=\begin{vmatrix} 1 & bc& -a\\ 1 & ca& -b\\ 1 &ab&-c\end{vmatrix}\\\\\\=\dfrac 1{abc}\begin{vmatrix} a & abc& -a^2\\ b & abc& -b^2\\ c &abc&-c^2\end{vmatrix}\\\\\\=-\dfrac {abc}{abc}\begin{vmatrix} a & 1& a^2\\ b & 1& b^2\\ c &1&c^2\end{vmatrix}\\[/tex]

[tex]=-1\begin{vmatrix} a & 1& a^2\\ b & 1& b^2\\ c &1&c^2\end{vmatrix}\\\\\\=-1 \times -1 \begin{vmatrix} 1 & a& a^2\\ 1 & b& b^2\\ 1 &c&c^2\end{vmatrix}\\\\\\=\begin{vmatrix} 1 & a& a^2\\ 1 & b& b^2\\ 1 &c&c^2\end{vmatrix}\\\\\\=\text{R.H.S}\\\\\text{Showed.}[/tex]

Answer:

200 KM/H

Step-by-step explanation:

A plane has a speed of 840 km/h in still air. It can travel 3120 km with the wind in the same time it would take to travel 1920 km against the wind. Find the speed of the wind.

let x=speed of wind

(840+x)=speed of plane with wind

(840-x)=speed of plane against wind

Travel time = distance/speed (same with and against the wind)

3120/(840+x)=1920/(840-x)

cross-multiply

3120*840-3120x=1920*840+1920x

3120*840-1920*840=1920x+3120x

5040x=1008000

x=200 km/hr

answer

Speed of wind=200 km/hr

You should must check for an extraneous solution when  the variable appears both inside and outside the absolute value expression

An extraneous solution is a solution that in obtained after completely solving an equation but it does not  work in the original given equation.

You should must check for an extraneous solution when  the variable appears both inside and outside the absolute value expression (Option D).

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The income that's made based on the equation when 1 comb is sold is $0.5.

From the information given, the equation is given as y = x/2. In this case, it was illustrated that the income is depicted based on the combs sold.

Therefore, the income by selling 1 comb will be:

y = 1/2

y = $0.5

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Answer:

$113.64

Step-by-step explanation:

0.88 x $113.636364 = 100 euros

= $113.64

(Hope this helps ^-^)

Answer:

its 150

Step-by-step explanation:

the "ten" is 5 to round it it will have to be the number to the right, which is 0 if its below 4 its changed to 0 and nothing happens to the ten

Step-by-step explanation:

1) The width of the rectangle is 10.2 in.

2) The perimeter of the rectangle is 71.4 in.

3) The area of the rectangle is 260.1 sq.in

By finding the proportional relation between x and y, we will see that when x = 16, we have y = 32.

If y varies directly with x, then we can write the relation as:

y = k*x

Where k is the constant of proportionality.

First, we know that when x = 4, we have y = 8, replacing that we get:

8 = k*4

Solving for k, we get:

k = 8/4 = 2.

Then the relation between x and y is:

y = 2*x

When x = 16, we have:

y = 2*16 = 32

Then the correct option is a.

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Answer:

ehhhh probably C cuz it makes more sense

Answer:

230??

Step-by-step explanation:

If C =270°

And b=40°

270-40= 230

Actually belive the one on top

Answer:

5/6

Step-by-step explanation:

i'm not rly confident,, but i used a calculator and solved it,,

so im sry if im wrong. Good luck!!

Answer:

32 sq inches

Step-by-step explanation:

Answer:

Part A)

[tex]\displaystyle z_6 = 9609600000x^{10}[/tex]

Part B)

[tex]z_4 = 10500 x^9[/tex]

Step-by-step explanation:

Recall the binomial expansion theorem:

[tex]\displaystyle (x+y)^n = \sum_{k=0}^{n}{n \choose k} x^{n-k} y^k[/tex]

Part A)

Our expression is equivalent to:

[tex]\displaystyle (2x+5)^{15} = \sum_{k = 0}^{15} {15 \choose k} (2x)^{15-k}\cdot 5^k[/tex]

To find the sixth term, let k  = 5. Therefore, the sixth term is:

[tex]\displaystyle \begin{aligned} z_6 &= {15\choose 5} (2x)^{15-5}\cdot 5^5 \\ \\ & = {15\choose 5}x^{10} \cdot (2)^{10}\cdot 5^5 \\ \\ &= 9609600000x^{10}\end{aligned}[/tex]

Part B)

Likewise:

[tex]\displaystyle \begin{aligned} \left(x^2 + \frac{5}{x}\right)^9 = \sum_{k=0}^9 {9\choose k}(x^2)^{9-k}\left(\frac{5}{x}\right)^{k}\end{aligned}[/tex]

To find the fourth term, let k = 3. Therefore, the fourth term is:

[tex]\displaystyle \begin{aligned} z_4 & = {9\choose 3}\left(x^2\right)^{9-3} \left(\frac{5}{x}\right)^{3} \\ \\ & = {9\choose 3}x^{12} \cdot \frac{5^3}{x^3} \\ \\ & = 10500x^9\end{aligned}[/tex]

Answer: y > -1/2x + 2

Step-by-step explanation: first, in order to find the inequalities you should find the gradient by choosing two points from the line and you should you the formula m=y2-y1/x2-x1 to find the gradient.

Next, you should find the y-intercept in order to complete the inequality it can be easily found as the y-intercept is the place where the line crosses the y axis

Then you create your equation { y = -1/2x + 2 } and then if above the line is shaded then it is {> greater than} and if below the line is shaded then it should be {< less than}

(so you should replace the equation with the lesser or greater sign according to the way the graph is shaded)

Answer:

C

Step-by-step explanation:

The y-intercept if the graph is 2 and the slope is 2 so the rule is y=2x+2

Answer:

yes of course it was a question about the same as you can use it for a while now I am with you forever

Given the amount of water needed to fill the cement dam and the water pump rate, the time needed to fill the dam is 400 minutes or 6hours 40minutes.

Option B)6h 40min is the correct answer.

Given that;

To determine how long it take to fill the dam, we say;

Time need = Amount of water needed ÷ Pump rate

T = A ÷ r

T = 30000 litres ÷ 75 litres/minute

T = 400 minutes

Note that; 60min = 1hrs

Hence,

T = 6hours 40minutes

Given the amount of water needed to fill the cement dam and the water pump rate, the time needed to fill the dam is 400 minutes or 6hours 40minutes.

Option B)6h 40min is the correct answer.

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Answer:

Karen expects the vehicle to depreciate by 600 each year.

Step-by-step explanation:

Given:

- initial value: 4400

- salvage value: 800

- another six years

It is expected to depreciate from 4400 to 800 in 6 years.

The depreciation per year is the ratio

depreciation per year = (total depreciation) / (number of years)

Total depreciation is the change in value, so the depreciation per year is

 (4400 - 800)/6 = 600

Answer: x = 13

Step-by-step explanation: First, we simplify the equation to get 4x + 7 = 5x - 6. Then, from there, we subtract 7 on both sides to get 4x = 5x - 13. Then we subtract 5x from both sides, which would give us -x = -13. Finally, we can divide both sides by -1 (because remember, x = 1) to get that x = 13. Hope this helps!

Answer:

a=

b=8

Step-by-step explanation:

20-5=15 (admittance fee)

15-3=12 (food and drink)

12÷1.5=8 rides because 1.50 is price for every ride

Answer:

infinite number of solutions

Step-by-step explanation:

x + y = 1 → (1)

3y = - 3x + 3 ( add 3x to both sides )

3x + 3y = 3 ( divide through by 3 )

x + y = 1 → (2)

since the 2 equations are the same, this indicates the system has an infinite number of solutions

Answer:

x=30

Step-by-step explanation:

You have to use intercept theorem, also known as Thales's theorem to slvoe this.

Answer:

[tex]\normalsize \textsf{$x = -\dfrac{8}{3}$}[/tex]

Step-by-step explanation:

Given: [tex]\normalsize \textsf{$3(2x - 9) - 4(3x + 4) = 2(x - 12) + \dfrac{7}{3}$}[/tex]

1. Distribute

For 3:

[tex]\normalsize \textsf{$3(2x - 9)$}\\\normalsize \textsf{$3(2x) + 3(- 9)$}\\\normalsize \boxed{\textsf{$6x -27$}}\\[/tex]

For -4:

[tex]\normalsize \textsf{$-4(3x + 4)$}\\\normalsize \textsf{$-4(3x) + -4(4)$}\\\normalsize \boxed{\textsf{$-12x -16$}}\\[/tex]

For 2:

[tex]\normalsize \textsf{$2(x - 12)$}\\\normalsize \textsf{$2(x) + 2(-12)$}\\\normalsize \boxed{\textsf{$2x -24$}}\\[/tex]

Now we have: [tex]\normalsize \textsf{$6x - 27 - 12x -16 = 2x-24 + \dfrac{7}{3}$}[/tex]

2. Combine like terms:

[tex]\normalsize \textsf{$-6x -43 = 2x -\dfrac{72}{3} + \dfrac{7}{3}$}\\\\\normalsize \textsf{$-6x -43 = 2x -\dfrac{65}{3}$}\\\\[/tex]

3. Subtract 2x from both sides:

[tex]\normalsize \textsf{$-6x-2x -43 = 2x -2x -\dfrac{65}{3}$}\\\\\normalsize \textsf{$-8x -43 = -\dfrac{65}{3}$}\\\\[/tex]

4. Add 43 to both sides:

[tex]\normalsize \textsf{$-8x -43 +43 = -\dfrac{65}{3} + 43$}\\\\\normalsize \textsf{$-8x = -\dfrac{65}{3} + \dfrac{129}{3}$}\\\\\normalsize \textsf{$-8x = \dfrac{64}{3}$}[/tex]

5. Divide both sides by -8

[tex]\normalsize \textsf{$-8x = \dfrac{64}{3}$}\\\\\normalsize \textsf{$-8x \div 8 = \dfrac{64}{3} \div -\dfrac{8}{1}$}\\\\\normalsize \boxed{\textsf{$x = -\dfrac{8}{3}$}}[/tex]

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