Answer:
A
Step-by-step explanation:
:)
hope it helps
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MATH
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Model the product on the grid. Record the product.
1. 3 X 13 =
2. 5 >
Find the product.
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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For the school play, tickets cost $9.50 for adults and $3 for kids under 12. What would be the total cost for 8 adult tickets and 21 kids tickets? What would be the total cost for aa adult tickets and kk kids tickets?
Answer:
$1399.5aa +3kkStep-by-step explanation:
a)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.
__
b)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.)
In ΔABC, m∠1=70° and m∠4=35°. What is m∠3?
Answer:
75
Step-by-step explanation:
A triangle can only have 3 angles that add up to 180 degrees. Subtract 70 and 35 from 180 to get 75.
180=x+70+35
180-(70+35)=x
Find all real values of $x$ such that $x^2-5x + 4=0$.
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.
What is the equation?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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need help solving this
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]
Q10. Five t-shirts and a hat cost £83.00. Two t-shirts and a hat cost £38.00. How much does one t-shirt cost?
No spam links and plz write down the answer.
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 :)
MAX POINTS!!!! Please help me solve the problem in the picture and tell me what I’m doing also explain each step.
I know myself that the answer is 0.4
Help help help help math
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
Chang works as a tutor for $11 an hour and as a waiter for $13 an hour. This month, he worked a combined total of 90 hours at his two jobs. Let t be the number of hours Chang worked as a tutor this month. Write an expression for the combined total dollar amount he earned this month.
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
a natural number multiplied by itself. is it possible for the result to have 66 as the last 2 digits.
yes or no that is all i need
Third-degree, with zeros of -4, -2, and 1, and a y-intercept of -11.
[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]
I want a way to solve such a problem
========================================================
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.
What is the slope of the line?
slope:change in x\change in y
Which value of x is a solution of the equation 4(x + 3) - 10 =6 (x -2)
[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]
HELP ANSWER PART A AND B
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.
please help quickly!!!!!!!
If y varies directly with x, determine an equation for the given direct variation. For this question, find the value of the given variable.
Find y when x = 10 if y = 8 when x = 20.
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
Which of the following rational functions is graphed below
Answer:
C
Step-by-step explanation:
Ti-83 graph proves it.
Simplify -3(-4x + 2)
Step-by-step explanation:
12x - 6 is your answer in simplified form
The sum of two angles of a triangle is
110°. Calculate the third angle in
degrees.
Answer:
70
Step-by-step explanation:
the sum of all angles is 180°
so 180-110= 70
1. Gavin buys 89 blue pansies and
86 yellow pansies. He will plant the flowers
in 5 rows with an equal number of plants
in each row. Draw a bar model to help you
find how many plants will be in each row,
O
17
Spiral Review
nose nilo que dise perdoooooooon xd
QUESTION 17 PLEASE HELP ME ASAPP
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:)
help me plssssssssssssss
Answer:
C .......
Step-by-step explanation:
Please help!!!!!!!!!!!!!!
Answer:
The answer is A.
Step-by-step explanation:
1 plus 2x (276-5y) + 69= A.
4.
The expression V-18 can be rewritten as ✓-1. 9. 2. What is a simplified version of the
expression?
А 3iv2
8 312
C -3iv2
D-3V2
Tropical dried fruit costs $1.50 per pound and regular dried fruit costs $0.90 per pound. You want to create a mixed bag of tropical dried fruit and regular dried fruit that is worth $1.30 per pound. How many pounds of tropical dried fruit do you need if you have already purchased 50 pounds of regular dried fruit?
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]
PLEASE HELP -7y + 11 = 75 + y
Answer:
y = -8
Step-by-step explanation:
-7y + 11 = 75 + y
Bring the "y" variable on one side, and rest on the other.
Rearranging, we get,
-7y - y = 75 - 11
-8y = 64
-y = 8
y = -8
Hope it helps :)
Lisa is earning 7.50 dollars per hour she worked for 33 hours last week. how much money did she earn?
Answer:
Lisa earned $247.50 for working 33 hours.
Step-by-step explanation:
WAY 1: proportional reasoning
7.50 x
------ = ----
1 h 33h
33 / 1 = 33
7.50 x 33 = 247.5
WAY 2: unit rate reasoning
7.50 per hour so x value of money for 33 h
7.50 x 33 = 247.5
Answer:
247.50 Dollars
Step-by-step explanation:
If Lisa earns 7.50$ per hour and she has worked for 33 hours, and then if the question asks how much money she earned it'd have to be 7.50*33. Which is 247.50, and its asking how much money so then you'd have to write the dollars next to the answer, so the answer is,
247.50 Dollars
A farmer's total payment for tools and compost over the course of two years is $27,894. Write and
solve an equation to find the payment for the second year if the first year’s payment is $10,205.
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
If ABCD is a parallelogram, m
Answer:
ABCD are the first 4 letters ina alphabet sooooo brooo wheres the parallelogram??
Step-by-step explanation:
The function v(t) is the velocity in m/sec of a particle moving along the x-axis. Use analytic methods to do each of the following: (a) Determine when the particle is moving to the right, to the left, and stopped. (b) Find the particle's displacement for the given time interval. If s(0) = 3, what is the particle's final position? (c) Find the total distance traveled by the particle. v(t) = 5 (sint)^2(cost); 0 ≤ t ≤ 2π
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].