Hey mathematicians ! Can anyone help me with these problems please ! Will give brainly !

Answer:

Step-by-step explanation:

8 adult tickets cost 8($9.50) = $76

21 kids tickets cost 21($3) = $63

The total cost of these tickets is $76 +63 = $139.

__

aa adult tickets cost 9.50aa

kk kids tickets cost 3.00kk

The total cost of these tickets is 9.50aa +3.00kk.

_____

Additional comment

We ordinarily expect a variable to be a single letter. This problem statement defines each variable using a double letter, so that is what we have used. (Sometimes Brainly doubles up variable content in copied text, so we suspect that may have happened here.)

Answer:

60

Step-by-step explanation:

To begin solving think of just the pants and the shirts. Each of the 3 pair of pants can be paired with one of the 5 shirts giving you. 5x3=15 combinations

Therefore to apply this type of situation to all 5 shirts 3 pairs of pants and 4 ties

(5)(3)(4)=60

I know myself that the answer is 0.4

Using the grid model, the product of 3 and 13 is 39

Product refers to the result of multiplying two or more numbers or quantities together. It is a fundamental operation in arithmetic and algebra. When multiplying two numbers, the product is obtained by adding together a number of copies of one of the numbers, equal to the value of the other number.

Given, two numbers 3 and 9, and a grid.

To determine the product of 3 and 13 just count all the grids covered in 3 rows up to 13 columns or adding 3 for 13 times will result in 39.

Count the grids covered in a red box attached in the image below.

3+3+3+3+3+3+3+3+3+3+3+3+3=39

So, the product of 3 and 13 is 39.

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t = total amount of lbs of Tropical dried fruit

m = total amount of lbs of mixed bag

[tex]\begin{array}{lcccl} &\stackrel{fruit}{lbs}&\stackrel{per~lb}{price}&\stackrel{total}{price}\\ \cline{2-4}&\\ \textit{tropical fruit}&t&1.50&1.5t\\ \textit{regular fruit}&50&0.90&45\\ \cline{2-4}&\\ \textit{mix bag}&m&1.30&1.3m \end{array}~\hfill \begin{cases} t+50=m\\ 1.5t+45=1.3m \end{cases} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\stackrel{\textit{substituting on the 1st equation}}{1.5t+45=1.3\stackrel{m}{(t+50)}}\implies 1.5t+45=1.3t+65\implies 0.2t+45=65 \\\\\\ 0.2t=20\implies t=\cfrac{20}{0.2}\implies \boxed{t=100}[/tex]

Answer:

1.5

Step-by-step explanation:

1.47826086

rounded to the nearest tenth is 1.5

nose nilo que dise perdoooooooon xd

Step-by-step explanation:

Two t-shirts that are(31.80) plus a hat(6.20) cost £38.00.

which means that one t-shirt costs £15.90 :)

Step-by-step explanation:

12x - 6 is your answer in simplified form

Can you show more information please? I need more.

T = 90t + 90w

Step-by-step explanation:

90t = 90x11 = 990

90w = 90 x 13 = 1170

T = 90t + 90w

T = 990 + 1170

T = 2160

slope:change in x\change in y

[tex]\huge\text{Hey there!}[/tex]

[tex]\large\textsf{4(x + 3) - 10 = 6(x -2)}\\\large\textsf{4(x) + 4(3) - 10 = 6(x) + 6(-2)}\\\large\textsf{4x + 12 - 10 = 6x - 12}\\\large\textsf{4x + 2 = 6x - 12}\\\large\text{SUBTRACT 6x to BOTH SIDES}\\\large\textsf{4x + 2 - 6x = 6x - 12 - 6x}\\\large\text{SIMPLIFY IT!}\\\large\text{NEW EQUATION: \textsf{-2x + 2 = -12}}\\\large\text{SUBTRACT 2 to BOTH SIDES}\\\large\textsf{-2x + 2 - 2 = 6x - 12 - 6x}\\\large\text{SIMPLIFY IT!}\\\large\textsf{-2x = -14}\\\large\text{DIVIDE -2 to BOTH SIDES}[/tex]

[tex]\mathsf{\dfrac{-2x}{-2}= \dfrac{-14}{-2}}\\\large\text{CANCEL out: }\rm{\dfrac{-2}{-2}}\large\text{ because it gives you 1}\\\large\text{KEEP: }\rm{\dfrac{-14}{-2}}\large\text{ because it helps solve for the x-value}\\\large\text{NEW EQUATION: }\mathsf{x = \dfrac{-14}{-2}}\\\large\text{SIMPLIFY IT!}\\\large\textsf{x = 7}\\\\\\\huge\boxed{\text{Therefore, your answer is: \boxed{\textsf{x = 7}}}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]

Answer:

C .......

Step-by-step explanation:

[tex]\begin{cases} x = -4\implies &x+4=0\\ x=-2\implies &x+2=0\\ x=1\implies &x-1=0 \end{cases}\qquad \implies (x+4)(x+2)(x-1)=\stackrel{y}{0}[/tex]

now, that's the equation or polynomial in factored form, hmmm we also know that it has a y-intercept of -11, namely, when x = 0  y = -11, well let's plug in a factor to it, that will reflect those values, namely say hmmm factor "a", so

[tex](x+4)(x+2)(x-1)=y\qquad \stackrel{\textit{adding "a" factor for vertical shift}}{a(x+4)(x+2)(x-1)}=y \\\\\\ \stackrel{\textit{we know that when x = 0, y = -11}}{a(0+4)(0+2)(0-1)=-11}\implies -8a=-11\implies a=\cfrac{-11}{-8}\implies a = \cfrac{11}{8} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\mathbb{FOIL}}{\cfrac{11}{8}(x^2+6x+8)}(x-1)=y\implies \cfrac{11}{8}(x^3+6x^2+8x-x^2-6x-8)=y \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{11}{8}(x^3+5x^2+2x-8)=y~\hfill[/tex]

Hey there!

• If it CHANGES it means your temperature/time changes a bit overtime

• If it CONSTANT your temperature/time remains the same

So, basically to this question it was INCREASED then it was DECREASED

Therefore, your answer is: “Option A. “

Because it didn’t stay same

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

Answer:

The second years payment is 17689.

Step-by-step explanation:

If the first year is $10,205 then you subtract 10205 from 27894 which is 17689.

Equation: $27894-$10,205= $17689

Answer:

x=4 && x=1

Step-by-step explanation:

x^2 - x - 4x + 4=0

x(x-1) - 4(x - 1)=0

(x-4)(x-1)=0

X=4 and x=1

All real values of $x$ such that $x²-5x + 4=0$ will be x=4 and x=1.

An equation is a statement that two expressions, which include variables and/or numbers, are equal. In essence, equations are questions, and efforts to systematically find solutions to these questions have been the driving forces behind the creation of mathematics.

It is given that, the quadratic equation is, x²-5x+4=0

Any equation of form ax²+bx+c=0  where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.

We have to find the value of x as,

x²-5x + 4=0

x² - x - 4x + 4=0

x(x-1) - 4(x - 1)=0

(x-4)(x-1)=0

X=4 and x=1

Thus, all real values of $x$ such that $x²-5x + 4=0$ will be x=4 and x=1.

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Answer:

(a) The particle is moving to the right in the interval  [tex](0 \ , \ \displaystyle\frac{\pi}{2}) \ \cup \ (\displaystyle\frac{3\pi}{2} \ , \ 2\pi)[/tex] , to the left in the interval [tex](\displaystyle\frac{\pi}{2}\ , \ \displaystyle\frac{3\pi}{2})[/tex], and stops when t = 0, [tex]\displaystyle\frac{\pi}{2}[/tex], [tex]\displaystyle\frac{3\pi}{2}[/tex] and [tex]2\pi[/tex].

(b) The equation of the particle's displacement is [tex]\mathrm{s(t)} \ = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(t)} \ + \ 3[/tex]; Final position of the particle [tex]\mathrm{s(2\pi)} \ = \ 3[/tex].

(c)  The total distance traveled by the particle is 9.67 (2 d.p.)

Step-by-step explanation:

(a) The particle is moving towards the right direction when v(t) > 0 and to the left direction when v(t) < 0. It stops when v(t) = 0 (no velocity).

Situation 1: When the particle stops.

[tex]\-\hspace{1.7cm} v(t) \ = \ 0 \\ \\ 5 \ \mathrm{sin^{2}(t)} \ \mathrm{cos(t)} \ = \ 0 \\ \\ \-\hspace{0.3cm} \mathrm{sin^{2}(t) \ cos(t)} \ = \ 0 \\ \\ \mathrm{sin^{2}(t)} \ = \ 0 \ \ \ \mathrm{or} \ \ \ \mathrm{cos(t)} \ = \ 0 \\ \\ \-\hspace{0.85cm} t \ = \ 0, \ \displaystyle\frac{\pi}{2}, \ \displaystyle\frac{3\pi}{2} \ \ \mathrm{and} \ \ 2\pi[/tex].

Situation 2: When the particle moves to the right.

[tex]\-\hspace{1.67cm} v(t) \ > \ 0 \\ \\ 5 \ \mathrm{sin^2(t) \ cos(t)} \ > \ 0[/tex]

Since the term [tex]5 \ \mathrm{sin^{2}(t)}[/tex] is always positive for all value of t of the interval [tex]0 \ \leq \mathrm{t} \leq \ 2\pi[/tex], hence the determining factor is cos(t). Then, the question becomes of when is cos(t) positive? The term cos(t) is positive in the first and third quadrant or when [tex]\mathrm{t} \ \epsilon \ (0, \ \displaystyle\frac{\pi}{2}) \ \cup \ (\displaystyle\frac{3\pi}{2}, \ 2\pi)[/tex] .  

*Note that parentheses are used to demonstrate the interval of t in which cos(t) is strictly positive, implying that the endpoints of the interval are non-inclusive for the set of values for t.

Situation 3: When the particle moves to the left.

[tex]\-\hspace{1.67cm} v(t) \ < \ 0 \\ \\ 5 \ \mathrm{sin^2(t) \ cos(t)} \ < \ 0[/tex]

Similarly, the term [tex]5 \ \mathrm{sin^{2}(t)}[/tex] is always positive for all value of t of the interval [tex]0 \ \leq \mathrm{t} \leq \ 2\pi[/tex], hence the determining factor is cos(t). Then, the question becomes of when is cos(t) positive? The term cos(t) is negative in the second and third quadrant or  [tex]\mathrm{t} \ \epsilon \ (\displaystyle\frac{\pi}{2}, \ \displaystyle\frac{3\pi}{2})[/tex].

(b) The equation of the particle's displacement can be evaluated by integrating the equation of the particle's velocity.

[tex]s(t) \ = \ \displaystyle\int\ {5 \ \mathrm{sin^{2}(t) \ cos(t)}} \, dx \ \\ \\ \-\hspace{0.69cm} = \ 5 \ \displaystyle\int\ \mathrm{sin^{2}(t) \ cos(t)} \, dx[/tex]

To integrate the expression [tex]\mathrm{sin^{2}(t) \ cos(t)}[/tex], u-substitution is performed where

[tex]u \ = \ \mathrm{sin(t)} \ , \ \ du \ = \ \mathrm{cos(t)} \, dx[/tex].

[tex]s(t) \ = \ 5 \ \displaystyle\int\ \mathrm{sin^{2}(t) \ cos(t)} \, dx \\ \\ \-\hspace{0.7cm} = \ 5 \ \displaystyle\int\ \ \mathrm{sin^{2}(t)} \, du \\ \\ \-\hspace{0.7cm} = \ 5 \ \displaystyle\int\ \ u^{2} \, du \\ \\ \-\hspace{0.7cm} = \ \displaystyle\frac{5u^{3}}{3} \ + \ C \\ \\ \-\hspace{0.7cm} = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(t)} \ + \ C \\ \\ s(0) \ = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(0)} \ + \ C \\ \\ \-\hspace{0.48cm} 3 \ = \ 0 \ + \ C \\ \\ \-\hspace{0.4cm} C \ = \ 3.[/tex]

Therefore, [tex]s(t) \ = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(t)} \ + \ 3[/tex].

The final position of the particle is [tex]s(2\pi) \ = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(2\pi)} \ + \ 3 \ = \ 3[/tex].

(c)

[tex]s(\displaystyle\frac{\pi}{2}) \ = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(\frac{\pi}{2})} \ + \ 3 \\ \\ \-\hspace{0.85cm} \ = \ \displaystyle\frac{14}{3} \qquad (\mathrm{The \ distance \ traveled \ initially \ when \ moving \ to \ the \ right})[/tex]

[tex]|s(\displaystyle\frac{3\pi}{2}) - s(\displatstyle\frac{\pi}{2})| \ = \ |\displaystyle\frac{5}{3} \ (\mathrm{sin^{3}(\frac{3\pi}{2})} \ - \ \mathrm{sin^{3}(\displaystyle\frac{\pi}{2})})| \ \\ \\ \-\hspace{2.28cm} \ = \ \displaystyle\frac{5}{3} | (-1) \ - \ 1| \\ \\ \-\hspace{2.42cm} = \displaystyle\frac{10}{3} \\ \\ (\mathrm{The \ distance \ traveled \ when \ moving \ to \ the \ left})[/tex]

[tex]|s(2\pi) - s(\displaystyle\frac{3\pi}{2})| \ = \ |\displaystyle\frac{5}{3} \ (\mathrm{sin^{3}(2\pi})} \ - \ \mathrm{sin^{3}(\displaystyle\frac{3\pi}{2})})| \ \\ \\ \-\hspace{2.28cm} \ = \ \displaystyle\frac{5}{3} | 0 \ - \ 1| \\ \\ \-\hspace{2.42cm} = \displaystyle\frac{5}{3} \\ \\ (\mathrm{The \ distance \ traveled \ finally \ when \ moving \ to \ the \ right})[/tex].

The total distance traveled by the particle in the given time interval is[tex]\displaystyle\frac{14}{3} \ + \ \displaystyle\frac{5}{3} \ + \ \displaystyle\frac{10}{3} \ = \ \displaystyle\frac{29}{3}[/tex].

Answer:

part A: median is 12

part B: interquartile rage is 10

Step-by-step explanation:

Part A: to find your median on a box and whisker plot, you just need to look at your middle line in your box. In this case it's 12.

Part B: to find your interquartile range on a box and whisker plot, you subtract your Q1 and Q3. Your Q1 and Q3 will be the two outside lines that make up the rest of your box, in this case would be 6 and 16. To find your Q1 if you don't already have your plot set up, you separate the numbers that are given to you in half and find the median of each side. The first half will be your Q1 and the second half your Q3.

Answer:

  x = (√21)/2

Step-by-step explanation:

The sides of a 30°-60°-90° triangle have the ratios 1 : √3 : 2. That means the middle-length side is (√3)/2 times the length of the longest side.

  x = (√3)/2·√7

  [tex]\boxed{x=\dfrac{\sqrt{21}}{2}}[/tex]

========================================================

Explanation:

Refer to the table below.

The first step is to find the midpoint of each interval.

To do that, we add up the endpoints and divide by 2. The midpoint of the interval from 2 to 6 is 4 because (2+6)/2 = 8/2 = 4. Repeat these steps for the other intervals to find that the remaining midpoints are: 8, 12, 16, 20.

Overall, the midpoints from top to bottom are 4, 8, 12, 16, 20

Note how the midpoints are incrementing by 4 each time, which is the same distance between each successive left endpoint (and right endpoint endpoints as well).

Let's say m represents the set of midpoints which I'll form in the third column. Multiply the frequency column f by each corresponding midpoint m. The result fm forms the fourth column.

Add up everything in that fourth column to get 1140. We then divide that by the sum of the frequencies (100) to get 1140/100 = 11.4 which is the final answer. The average weight is 11.4 pounds.

Answer:

The answer is A.

Step-by-step explanation:

1 plus 2x (276-5y) + 69= A.

Answer:

y=K*x equation

k=constant of variation

y= 4 when x = 10

Step-by-step explanation:

8=k*20

divide both sides by 20 to solve for k

2/5=k

y=2/5 *10

y= 4 when x = 10

Answer:

70

Step-by-step explanation:

the sum of all angles is 180°

so 180-110= 70

Answer:

C

Step-by-step explanation:

Ti-83 graph proves it.

Answer:

ABCD are the first 4 letters ina alphabet sooooo brooo wheres the parallelogram??

Step-by-step explanation:

Answer:

x=-5, y=-4

Step-by-step explanation:

to solve this system of equations, first multiply the second equation by 7 to get -21x-7y=133. this allows for the y terms in both equations to cancel once you add the two equations together.

adding these together, the y term cancels and you are left with -26x=130. so dividing both sides by -26 will get x=-5. then plug this x value back into either original equation (i chose -3x-y=19) so solve for y.

-3(-5)-y=19 so -y=4 so y=4

so the answer to the system of equations is x=-5 and y=-4