Answer:
The answer of 2 x 3 would be 6
Step-by-step explanation:
The reason why 2 x 3 or 2 times 3 would be six/6. Is because if you had 2 pairs of 3 items, you would get a total of 6 items. And if you had 3 pairs of 2 items, you would still have a total of 6 items.
Please help I'll mark brainliest
✿————✦————✿————✦————✿————✦————✿
So, to find the mean they added up the scores from 20 seasons and divided by 20, the number of seasons to get an average score of 10.4.
10.4 = (x1 +x2 +x3 ... +x20 )/20
Then;
10.4*(20) = (x1 +x2 +x3 ... +x20 )
208 = sum of the first 20 seasons
We want to add one more number in there, the 21st season with a score of 14. Add the 14 to the sum of the first 20 seasons and then divide by 21, the new number of seasons you're averaging.
Mean = (208+14)/21
✿————✦————✿————✦————✿————✦————✿
Answer
= 10.6
✿————✦————✿————✦————✿————✦————✿
#carryonlearning
Luke's house is due west of Toronto and due south of Barrie. Toronto is 16 kilometres from Luke's house and 20 kilometres from Barrie. How far is Barrie from Luke's house, measured in a straight line?
Using the Pythagorean Theorem, it is found that Barrie is 12 miles from Luke's house, in a straight line.
What is the Pythagorean Theorem?The Pythagorean Theorem relates the length of the legs [tex]l_1[/tex] and [tex]l_2[/tex] of a right triangle with the length of the hypotenuse h, according to the following equation:
[tex]h^2 = l_1^2 + l_2^2[/tex]
This problem can be modeled by a right triangle, with [tex]l_1 = 16, l_2 = d, h = 20[/tex], hence:
[tex]h^2 = l_1^2 + l_2^2[/tex]
[tex]16^2 + d^2 = 20^2[/tex]
[tex]d^2 = 144[/tex]
[tex]d = 12[/tex]
Barrie is 12 miles from Luke's house, in a straight line.
More can be learned about the Pythagorean Theorem at https://brainly.com/question/654982
g[f(x)] if g(x) = x2 and f(x) = x + 3.
Choose the expression that represents 3 less than 7 times a number
Answer:
7x - 3, x being the number
Step-by-step explanation:
number = x
7x - 3
Answer:
Three less than 7 times a number is 39.
Step-by-step explanation:
A dinner plate has a circumference of 113.04 cm. What is the area of the dinner plate? (Use 3.14 for .)
A.
226.08 cm2
B.
2,034.72 cm2
C.
1,017.36 cm2
D.
56.52 cm2
option C
hope it helps...!!!
Answer:
[tex]\fbox{Area = 1017.36 sq.cm.}[/tex]
Step-by-step explanation:
Given:
circumference of circular dinner plate= 113.04 cm
To find:
Area of dinner plate = ?
Solution:
circumference = 113.04 cm.
we know that circumference = 2πr
[tex]2 \pi r = 113.04 \: cm \\ r = \frac{113.04 }{2 \pi} \\ r = \frac{113.04 }{2 \times 3.14} \\r = \frac{113.04 }{6.28} \\ \fbox{r = 18 \: cm}[/tex]
We know that,
Area of circle = πr²
now substituting value of r in above formula,
[tex] \pi {r}^{2} = 3.14 \times 18 \times 18 = 1017.36 \: {cm}^{2} [/tex]
[tex] \fbox{Area = 1017.36 sq.cm.}[/tex]
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URGENT I NEED IT NOW PLEASE HELP ME!!!!
A High School’s football Stadium holds a total of 7,000 fans. A survey shows that of the total number of fans, 68% root for the home team. How many fans attending a football game are rooting for the home team? Please help!
Find the slope of the line that passes through the two given points:
(-1, 4) and (5, 3)
A line starting in the third quadrant, crossed the x axis twice, and then ending in the first quadrant
Answer:
[tex] \frac{ - 1}{6} [/tex]
Step-by-step explanation:
[tex]slope \: = \frac{y2 - y2}{x2 - x1} \\ slope = \frac{3 - 4}{5 - - 1} \\ sope = \frac{ - 1}{ \: \: \: 6} [/tex]
How do you write 7,100 in scientific notation
Answer:
7.1 x 10³
Step-by-step explanation:
Move your decimal so that it follows your first whole number greater than zero. The number of places you moved the decimal is the same as the exponent. Since the number 7.1 needs to INCREASE to be 7,100, then the exponent is positive.
Answer:
7.1 x 10[tex]^{3}[/tex] (note the 3 is an exponent)
Step-by-step explanation:
1. Count how many places there are to the right of 7
2. There are 3
3. Apply the formula of scientific notation a x 10[tex]^{b}[/tex]
4. 7.1 x 10[tex]^{3}[/tex] (remove all digits after 7.1
Hope this helps :)
I’m not sure the steps to figure out the problem. X = 2 to x =4
Answer:
you have to put the Y in the formula or make a comparison
Describe how to find the sale price of an item that has been discounted 15%
Alice is making bracelets to sell. She plans to put 12 beads on each bracelet. Beads are sold in packages of 20.
#1. What is the least number of packages she can buy to make bracelets and have no beads left over?
#2 If each package of beads costs $5.50, how much will the beads cost for her project?
Show and Explain your work
Hint: the “T” Method works well for show and explain. Make sure you have all of the details explained. If someone only read your explanation they should know… what the question is, all of the steps needed to answer the question and what your final answer is.
Answer:
question 1 is 60 beads and question 2 is 16 dollars and 50 cents.
Step-by-step explanation:
for question 1 the are asking for the lowest common denomonator (60)
for question 2 you devide 60/20 and multiply by 5.50
3 x 5.50 = 16.5$
Find the measure of ∠ABC.
Answer:
170°
Step-by-step explanation:
There is a total of 180° and the angle is 10° less...
So, 180°-10° = 170°
Please 5 stars if correct!!!
I hope this helps!!!
General solution of: (1-xy)^-2 dx + [y^2 + x^2 (1-xy)^-2]dy = 0
show two solution on your answer
nonsense answer deleted
[tex] \Large \bold{SOLUTION\ 1:} [/tex]
[tex] \small \begin{array}{l} \text{First, we need to check if the given differential} \\ \text{equation is exact.} \\ \\ (1-xy)^{-2} dx + \big[y^2 + x^2 (1-xy)^{-2}\big]dy = 0 \\ \\ \dfrac{dx}{(1-xy)^2} + \left[y^2 + \dfrac{x^2}{(1-xy)^2}\right]dy = 0 \\ \\ \quad M(x, y) dx + N(x, y) dy = 0 \end{array} [/tex]
[tex] \small \begin{array}{l l}\tt\: M(x,y) = \dfrac{1}{(1 - xy)^2}, & N(x,y) = y^2 + \dfrac{x^2}{(1-xy)^2}\\ \\\tt \dfrac{\partial M}{\partial y} = \dfrac{-2(-x)}{(1 - xy)^3}, & \dfrac{\partial N}{\partial x} = \dfrac{2x}{(1 - xy)^2} + \dfrac{-2(-y)x^2}{(1 - xy)^3} \\ \\\tt \dfrac{\partial M}{\partial y} = \dfrac{2x}{(1 - xy)^3}, & \dfrac{\partial N}{\partial x} = \dfrac{2x(1 - xy)+2x^2y}{(1 - xy)^3} \\ \\\tt \: & \dfrac{\partial N}{\partial x} = \dfrac{2x}{(1 - xy)^3} \end{array} [/tex]
[tex] \small \begin{array}{l} \tt\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x} \implies \text{Differential equation is exact.} \\ \\\tt \dfrac{\partial F}{\partial x} = M(x, y) = \dfrac{1}{(1 - xy)^2} \\ \tt\displaystyle F(x, y) = \int \dfrac{1}{(1 - xy)^2} \partial x = -\dfrac{1}{y} \int \dfrac{1}{(1 - xy)^2}(-y)\partial x \\ \\ \tt\:F(x, y) = \dfrac{1}{y(1 - xy)} + h(y) \\ \\ \tt\dfrac{\partial F}{\partial y} = N(x, y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\\tt \dfrac{\partial}{\partial y}\left[\dfrac{1}{y(1 - xy)} + h(y)\right] = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ \tt-\dfrac{1 - xy + y(-x)}{y^2(1 - xy)^2} + h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ \tt-\dfrac{1 - 2xy}{y^2(1 - xy)^2} + h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} + \dfrac{1 - 2xy}{y^2(1 - xy)^2} \\ \\ \tt\:h'(y) = y^2 + \dfrac{x^2y^2 - 2xy + 1}{y^2(1-xy)^2} = y^2 + \dfrac{1}{y^2} \\ \\ h(y) = \dfrac{y^3}{3} - \dfrac{1}{y} + C \\ \\ \tt\text{Substituting to }F(x,y),\text{we get} \\ \\ \dfrac{1}{y(1 - xy)} + \dfrac{y^3}{3} - \dfrac{1}{y} = C \\ \\ \quad \quad \text{or} \\ \\ \tt\red{\boxed{\dfrac{x}{1 - xy} + \dfrac{y^3}{3} = C} \Longleftarrow \textit{Answer}} \end{array} [/tex]
[tex] \Large \bold{SOLUTION\ 2:} [/tex]
[tex] \small \begin{array}{l} \tt\text{Since we already know that the equation is exact,} \\ \text{we can then continue solving for the solution by} \\ \text{inspection method or by algebraic manipulation.} \\ \\ \tt(1-xy)^{-2} dx + \big[y^2 + x^2 (1-xy)^{-2}\big]dy = 0 \\ \\ \tt\dfrac{dx}{(1-xy)^2} + \left[y^2 + \dfrac{x^2}{(1-xy)^2}\right]dy = 0 \\ \\ \tt\dfrac{dx}{(1-xy)^2} + y^2 dy + \dfrac{x^2}{(1-xy)^2} dy = 0 \\ \\ \tt\dfrac{dx + x^2dy}{(1-xy)^2} + y^2 dy = 0 \\ \\ \tt\text{Divide both numerator and denominator of the} \\ \tt\text{fraction by }x^2. \end{array} [/tex]
[tex] \small \begin{array}{c}\tt \dfrac{\dfrac{1}{x^2}dx + dy}{\dfrac{(1-xy)^2}{x^2}} + y^2 dy = 0 \\ \tt\\ \tt\dfrac{\dfrac{1}{x^2}dx + dy}{\left(\dfrac{1}{x}-y\right)^2} + y^2 dy = 0 \\ \\ \tt-\dfrac{\left(-\dfrac{1}{x^2}dx - dy\right)}{\left(\dfrac{1}{x}-y\right)^2} + y^2 dy = 0 \\ \\ \tt\displaystyle {\large{\int}} -\frac{d\left(\dfrac{1}{x}-y\right)}{\left(\dfrac{1}{x}-y\right)^2} + \int y^2 dy = \int 0 \\ \\ \tt\implies\tt \dfrac{1}{\dfrac{1}{x}-y} + \dfrac{y^3}{3} = C \\ \\\text{or} \\ \\ \tt\red{\boxed{\dfrac{x}{1 - xy} + \dfrac{y^3}{3} = C} \Longleftarrow \textit{Answer}} \end{array} [/tex]
#CarryOnLearning
#BrainlyMathKnower
Shoe Size
2
4
6
8
10
12
At soccer practice, Mike asked each of his teammates to write down their shoe size.
What is the interquartile range for Mike's teammates' shoe sizes?
Answer:
2
Step-by-step explanation:
To find the interquartile range, subtract the lower quartile from the upper quartile.
The upper quartile is 7.
The lower quartile is 5.
Subtract.7–5=2
The interquartile range for Mike's teammates' shoe sizes is 2 sizes.
The interquartile range for Mike's teammates' shoe sizes is R = 2
What in Interquartile Range ( IQR )?The distance between the upper and lower quartiles is known as the interquartile range. Half of the interquartile range corresponds to the semi-interquartile range. Finding the values of the quartiles in a small data set is straightforward.
The spread of the data, or statistical dispersion, is measured by the interquartile range. The middle 50%, fourth spread, or H-spread are further names for the IQR. It is described as the spread between the data's 75th and 25th percentiles.
Given data ,
Let the interquartile range be represented as R
Now , the equation will be
Let the shoe sizes be represented as set A
A = { 2 , 4 , 6 , 8 , 10 }
Now , the upper range of the quartile is = 7
The lower range of the quartile is = 5
And , the interquartile range R = upper quartile - lower quartile
On simplifying , we get
The interquartile range R = 7 - 5 = 2
Hence , the interquartile range R = 2
To learn more about interquartile range click :
https://brainly.com/question/15686744
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Manu had ₹ 120. He bought one book for ₹ 25.75 and a pen for ₹ 12.50. How much money does he have now?
Answer:
₹ 81.75
Step-by-step explanation:
Total = ₹ 120.
He bought one book for ₹ 25.75 and a pen for ₹ 12.50.
₹ 25.75 + ₹ 12.50 = ₹ 38.25
How much money does he have now?
₹ 120 - ₹ 38.25 = ₹ 81.75
₹81.75
you just minus 120 from 25.75 and 12.50
(2)Please help with both questions
Answer:
See below for answers and explanations
Step-by-step explanation:
Problem 1
Let's think of the boat and wind as vectors:
Boat Vector --> [tex]\langle28cos36^\circ,28sin36^\circ\rangle[/tex]
Wind Vector --> [tex]\langle10cos22^\circ,10sin22^\circ\rangle[/tex]
Now, let's add the vectors:
[tex]\langle28cos36^\circ+10cos22^\circ,28sin36^\circ+10sin22^\circ\rangle[/tex]
Find the magnitude (the true velocity):
[tex]\sqrt{(28cos36^\circ+10cos22^\circ)^2+(28sin36^\circ+10sin22^\circ)}\approx37.78\approx38[/tex]
Find the direction (angle):
[tex]\theta=tan^{-1}(\frac{28sin36^\circ+10sin22^\circ}{28cos36^\circ+10cos22^\circ})\approx32.32^\circ\approx32^\circ[/tex]
Thus, D is the best answer
Problem 2
Recall that the angle between two vectors is [tex]\theta=cos^{-1}(\frac{u\cdot v}{||u||*||v||})[/tex] where [tex]u\cdot v[/tex] is the dot product of the vectors and [tex]||u||*||v||[/tex] is the product of each vector's magnitude:
[tex]\theta=cos^{-1}(\frac{u\cdot v}{||u||*||v||})\\\\\theta=cos^{-1}(\frac{\langle-82,47\rangle\cdot\langle92,80\rangle}{\sqrt{(-82)^2+(47)^2}*\sqrt{(92)^2+(80)^2}})\\\\\theta=cos^{-1}(\frac{(-82)(92)+(47)(80)}{\sqrt{6724+2209}*\sqrt{8464+6400}})\\\\\theta=cos^{-1}(\frac{(-7544)+3760}{\sqrt{8933}*\sqrt{14864}})\\\\\theta=cos^{-1}(\frac{-3784}{\sqrt{132780112}})\\\\\theta\approx109.17^\circ\approx109^\circ[/tex]
Therefore, C is the best answer
A rule for creating a pattern is given below. The rule begins with a number called the input and creates a number called the output.
Rule: Multiply the input by 5. Then subtract 4 from the result to get the output.
Which input and output table works for the rule?
Choose 1 answer:
(Choice A)
Input: 5 Output: 5
(Choice B)
Input: 3 Output: 7
(Choice C)
Input: 2 Output: 6
Answer:
Choice C - Input: 2 Output: 6
Answer:
C
Step-by-step explanation:
evaluate the output for the given input values using the rule
choice A
5 × 5 - 4 = 25 - 4 = 21 ≠ 5
choice B
3 × 5 - 4 = 15 - 4 = 11 ≠ 7
choice C
2 × 5 - 4 = 10 - 4 = 6 ← equals the output value
How many solutions does this nonlinear system of equations have?
Answer:
ONE “1”
Step by step:
Taking into account the definition of a system of equations, this nonlinear system of equations has one solution.
A system of equations is a set of two or more equations that share two or more unknowns. The solutions of a system of equations are all the values that are valid for all equations.
That is, to find the solution to a system of equations, you must find a value (or range of values) that satisfies all the equations in the system.
Graphically, the points where the graphs of the equations intersect will be the solutions to the system.
A system of equations with a linear equation and a quadratic equation can have two, one, or no solutions. To find its quantity, we must look for the intersection between both graphs:
If the graphs of the equations do not intersect, then there are no solutions for both equations. If the parabola and the line touch at a single point, then there is a solution for both equations. If the line intersects the parabola in two places, then there are two solutions to both equations.
In this case, you can see that the parabola and the line touch at a single point. Then there is a solution for both equations.
In summary, this nonlinear system of equations has one solution.
The giraffes at the Liberia zoo eat 75 pounds of food per day, and 52.5 pounds per day comes from acacia leaves. How much of their diet comes from other types of leaves?
Julia, the zookeeper, used the equation 52.5 + x = 75 to find the answer. Which equation is an equivalent equation that can be used to solve the problem?
Answer:
Step-by-step explanation:
75-52.5=x
3. The object has the shape of a rectangular prism, but part of a face is missing.
Find the surface area of the object and show your work.
a)
Find the surface area of the complete prism.
5 cm
30 cm
20 cm
10 cm
15 cm
I need an answer
A. 140°
B. 90°
C. 70°
D. 50°
Help please
Answer:
i think the answer is d which is 50°
As Juan continued his hitting practice showing off his abilities, one of the balls flew over the center field stands and into the parking lot. “Did you see that shot”, he yelled at the girls. “The ball hung in the air for at least 10 seconds”, he exclaimed.
The formula for this hit is:
h(x−) = 16+x2 8+3x 4 where h is the height of the ball and x is the number of seconds the ball is in the air.
I need help with part B
Answer:
A. Find h(10) = 0 . . . (it is not)
B. h(5.235) = 0; hang time was about 5.235 seconds
Step-by-step explanation:
The hang time of the ball is the value of x in h(x) such that h(x) = 0.
__
A.Juan can show the hang time is 10 seconds by finding h(10) = 0. The value of h(10) is ...
h(10) = -16(10²) +83(10) +4 = -1600 +834 = -766 ≠ 0
Juan cannot prove the hang time was 10 seconds.
__
B.The hang time is the solution to ...
h(x) = 0
-16x² +83x +4 = 0 . . . . . use the given expression for h(x)
x² -5.1875x = 0.25 . . . . divide by -16
x² -5.1875x +6.7275390625 = 6.9775390625 . . . . add (5.1875/2)²
(x -2.59375)² = 6.9775390625 . . . . . . . . . . rewrite as a square
x = 2.59375 ±√6.9775390625 ≈ {-0.048, 5.235} . . . . square root, find x
The hang time of the ball was about 5.235 seconds.
__
We like a graphing calculator for finding a quick solution to questions like this.
What is the product of 3/4 and -6/7?
Answer:
-9/14
Step-by-step explanation:
Hey there!
Guide:
• Difference means subtract/subtraction
• Product means multiply/multiplication
• Sum means add/addition
• Quotient means divide/division
• Now that we know what “product” means… we can make the question/equation easier to solve.
3/4 × -6/7
= 3(-6) / 4(7)
= -18 / 28
= -18 ÷ 2 / 28 ÷ 2
= -9 / 14
Therefore, your answer is: -9/14
Good luck on assignment & enjoy your day!
~Amphitrite1040:)
a) the cost of. kg oranges is rs 350 how many kg of oranges can be bought for rs 665
b.). A car travel for 50 km.with the help of 5 litres of petrol how many litres of petrol is reauired to travel 120 km
A: 1.9 Kilograms of oranges for rs 665.
B: 12 liters of petrol will make a car travel 120 km.
Find the value of X and perimeter.
Find the factorization of the polynomial below.
81x^2 - 18x + 1
A. (9x + 1)^2
B. (18x + 1)^2
C. (18x - 1)^2
D. (9x - 1)^2
Answer:
[tex](9x-1)^2[/tex]
Step by step explanation:
[tex]81x^2-18x+1\\\\=81x^2 - 9x-9x+1\\\\=9x(9x-1)-(9x-1)\\\\=(9x-1)(9x-1)\\\\=(9x-1)^2[/tex]
f ( 1 ) =
solve f ( x ) =1:
x =
Answer:
If b=5:7, find the value of B in terms of c
Can somebody help me pls!
Answer: C
Step-by-step explanation:
Just look at a z-score table and multiply by 100.
-> (0.308538)(100) is about 30.85%
Solve the system of equations:
4x-5y=21
3x+10y= -53
Step-by-step explanation:
2× (4x-5y) = 21
3x+10y= -53
-------------------------
8x-10y = 42
3x+10y=-53
------------------------
11x= -11
{x=-1}
-----------------
3x+10y=-53
x=-1
3(-1)+10y=-53
-3+10y=-53
10y= -50
{y=5}