The given relationship using the Pythagorean theorem holds true for a right-angled triangle.
We are given a mathematical relationship using the Pythagorean theorem. The Pythagorean theorem, also known as Pythagoras' theorem, is a fundamental relationship between the three sides of a right triangle in Euclidean geometry. The equation is given below.
a² + b² = c²
We need to describe the situation when this relationship holds true. The variables "a", "b", and "c" represent the sides of a triangle. The Pythagorean theorem is applicable to a right-angled triangle. It states that the sum of the squares of the base and the perpendicular is equal to the square of the hypotenuse.
Here a, b, and c denote the lengths of the base, the perpendicular, and the hypotenuse of the triangle, respectively.
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This table represents the relationship between x and y described by the equation.y=-x1012141618SY6789Which list represents the dependent values in the table?5,6,7,8,95, 6, 7, 8, 9, 10, 12, 14, 16, 1810, 12, 14, 16, 181,2,3,4,5
ANSWER :
A. 5, 6, 7, 8, 9
EXPLANATION :
From the problem, we have the function :
[tex]y=\frac{1}{2}x[/tex]y is the dependent variable and
x is the independent variable.
So the dependent values are the y values.
That will be 5, 6, 7, 8, 9
April 25 ft long has got into three pieces. it's a first rope is 2x feet long, the second piece is 5X feet long, and the third piece is 4 ft long. A) Write an equation to find X.B) Find the length of the first and second pieces.
Given:
The length of the total rope = 25 ft
It is divided into three pieces
it's the first rope is 2x feet long, the second piece is 5X feet long, and the third piece is 4 ft long.
A) Write an equation for x.
The equation will be:
[tex]2x+5x+4=25[/tex]Which can be simplified to :
[tex]7x+4=25[/tex]so, the equation is 7x + 4 = 25
B) Find the length of the first and the second pieces
First, we will solve the equation to find x
[tex]\begin{gathered} 7x=25-4 \\ 7x=21 \\ \\ x=\frac{21}{7}=3 \end{gathered}[/tex]So, the length of the first piece = 2x = 6 ft
The length of the second piece = 5x = 15 ft
At time the position of a body moving along the s- axis is s = t ^ 3 - 6t ^ 2 + 9t m Find the body's acceleration each time the velocity is zero . Find the body's speed each time the acceleration is zero .
The body's acceleration each time the velocity is zero is 6 [tex]m/s^{2}[/tex] or -6 [tex]m/s^{2}[/tex] and the body's speed each time the acceleration is zero is -3m/s.
According to the question,
We have the following information:
s = [tex]t^{3} -6t^{2} +9t[/tex]
Velocity = ds/dt
Velocity = [tex]3t^{2} -12t+9[/tex]
Acceleration = dv/dt
Acceleration = 6t-12
When velocity is zero:
[tex]3t^{2} -12t+9= 0[/tex]
Taking 3 as a common factor:
[tex]t^{2} -4t+3=0\\t^{2} -3t-t+3=0[/tex] (Factorizing by splitting the middle term)
t(t-3)-1(t-3) = 0
(t-3)(t-1) = 0
t = 3 or t = 1
Now, putting these values of t in acceleration's equation:
When t =3:
A = 6*3-12
A = 18-12
A = 6 [tex]m/s^{2}[/tex]
When t = 1:
A = 6*1-12
A = 6-12
A = -6 [tex]m/s^{2}[/tex]
Now, when acceleration is zero:
6t-12 = 0
6t = 12
t = 2 s
Now, putting this value in velocity's equation:
[tex]3*2^{2} -12*2+9[/tex]
3*4-24+9
12-24+9
21-24
-3 m/s
Hence, the body's acceleration each time the velocity is zero is 6 [tex]m/s^{2}[/tex] or -6 [tex]m/s^{2}[/tex] and the body's speed each time the acceleration is zero is -3m/s.
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a carpet measures 7 feet long and has a diagonal measurement of (74) square root feet. find the width of the carpet
Let's use Pythagorean Theorem to solve this problem:
[tex]\sqrt[]{74}^2=w^2+7^2[/tex][tex]74=w^2\text{ + 49}[/tex]Solving for w:
[tex]\begin{gathered} w\text{ = }\sqrt[]{74\text{ - 49}} \\ w\text{ = 5} \end{gathered}[/tex]w = 5ft
Sketch a diagram of a 315° angle in standard position and indicate the measure of its reference angle.
Explanation:
The angle in the question is given below as
[tex]315^0[/tex]The angle is found the QUADRANT IV
Hence,
The sketch of the angle in standard positing will be given in the image below
[tex]360-315=45^0[/tex]Therefore,
The reference angle is
[tex]45^0[/tex]In the coordinate plane, three vertices of rectangle PQRS are P(0,0), Q(0,b), S(c,0). What are the coordinates of point R?Answers:A.(c,b)B.(b,c)C.(2b,2c)D.(2c,2b)
B.(b,c)
Explanation
Step 1
let's graph the rectangle
we know that in a rectangle the angles are rigth, so, we have vertical and horizontal lines
the missing point is the intersection of the lines
y=c
and
x=b
so
the answer is
B.(b,c)
I hope this helps you
Find the distance between the points. Round to the nearest tenth if necessary. (3, 7), (-5, -7) Distance?
To find the distance between both points you have to apply pythagoras theorem.
First draw both points and form a rigth triangle with the distance between them as the hypothenuse:
The length of the base of the triangle "a" is determined by the difference between the x-coordinates of both points:
[tex]a=x_2-x_1=3-(-5)=3+5=8[/tex]The heigth of the triangle "b" is determined by the difference between the y-coordinates of both points:
[tex]b=y_2-y_1=7-(-7)=7+7=14[/tex]Now using phytagoras theorem you can calculate the length of the hypotenuse as:
[tex]\begin{gathered} a^2+b^2=c^2 \\ (8)^2+(14)^2=c^2 \\ c^2=260 \\ c=\sqrt[]{260} \\ c=2\sqrt[]{65}=16.12 \end{gathered}[/tex]The distance between points (3,7) and (-5,-7) is 2√65
Model With Mathematics An archer shoots
an arrow to a height (meters) given by the
equation y = -5t2 + 18t - 0.25, where t is
the time in seconds. A target sits on a hill
represented by the equation y = 0.75x - 1.
At what height will the arrow strike the
target, and how long will it take?
The arrow strike the target within time of = 3.49s on height of = 1.62 m
What is quadratic equation?A quadratic equation is a quadratic algebraic expression of the form ax2 + bx + c = 0. The word quadratic comes from the word quad, which means square. There are many scenarios where quadratic equations are used. Did you know that when a rocket launches, its trajectory is described by a quadratic equation?In addition, quadratic equations have many uses in physics, engineering, astronomy, and more. 4A quadratic equation is a quadratic equation in x for which x has at most two answers. These two answers to x are also called the roots of the quadratic equation and are denoted by (α, β). In the next content, you will learn about roots of quadratic equations. The quadratic equation is his second algebraic equation of x. The first condition of the quadratic equation is that the coefficient of the x2 term is nonzero (a ≠ 0).Calculationy1 = -5t² + 18t -0.25
y2 = 0.75t - 1
the arrow will strike the target when y1 = y2
-5t² + 18t -0.25 = 0.75t - 1
t = 3.49s(approx.)
the height y = 1.62m
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Find g(1) and find one value of x for which g(x)=-1.
To solve g(1) = ? we must do a vertical line at x = 1, it goes DOWN! because the graph is below the x-axis, if we do the line we will see that it will stop at y = -4, therefore, g(1) = -4
[tex]g(1)=-4[/tex]To find out the value of x for which g(x) = -1 we will start the process by doing a horizontal line at y = -1, if we do it we will see two possible values: -2 and 2, they're both correct! So you can choose which one you will put as your answer.
help me please I want to learn how to solve this
SOLUTION
In this question, we are meant to find the slope of the line represented
by 5x - 12 y = 24.
Re-arranging the equation, we have: 12 y = 5x - 24
Dividing both sides by 12,
[tex]\begin{gathered} y\text{ = }\frac{1}{12}(\text{ 5x -24 )} \\ y\text{ = }\frac{5}{12}x\text{ - 2} \\ \text{CONCUSION: The slope of the line is }\frac{5}{12}\text{ ------OPTION J} \end{gathered}[/tex]Explain how you know that the function represented by the data in the given table is quadratic.XV0-17123-13-313
Given:
Here a table of equation is given
Required:
How to know that the function represented by the data in the given table is quadratic.
Explanation:
here the first differences of y values are as below
[tex]\begin{gathered} -13-(-17)=-13+17=4 \\ -3-(-13)=-3+13=10 \\ 13-(-3)=13+3=16 \\ 35-13=22 \\ 63-35=28 \end{gathered}[/tex]now again take difference of the first difference which is called as second difference.
[tex]\begin{gathered} 10-4=6 \\ 16-10=6 \\ 22-16=6 \\ 28-22=6 \end{gathered}[/tex]so here we can see that the second difference is same which is 6
now if second difference of any table is equal we can say that the given table is the table of quadratic equation.
Final answer:
The second differences are all 6
F(x) = log10 X
The question is which answer represents the domain of the logarithmic function below?
Answer:hi
Step-by-step explanation:1+1
Joe goes running in the park. He runs 3 miles and does it in 42 minutes. How many minutes doe it take him to run a mile? This topic is distance = rate x time
You must use this formula:
[tex]d=rt[/tex]Where "d" is the distance, "r" is the rate and "t" is time.
If you solve for "r":
[tex]r=\frac{d}{t}[/tex]If you solve for "t":
[tex]t=\frac{d}{r}[/tex]Knowing that Joe runs 3 mile in 42 minutes, you can find "r". Notice that:
[tex]\begin{gathered} d=3mi \\ t=42\min \end{gathered}[/tex]Then:
[tex]r=\frac{3mi}{42\min}=0.0714\frac{mi}{\min}[/tex]Knowing the rate, you can set up the following in order to find the time in minutes it takes Joe to run a mile:
[tex]\begin{gathered} d=1mi \\ r=0.071\frac{mi}{\min} \end{gathered}[/tex]Substituting values into the formula for calculate the time, you get:
[tex]t=\frac{1\min}{0.0714\frac{mi}{\min}}=14\min [/tex]The answer is: It takes him 14 minutes to run a mile.
Solving equations using quadratic formula m² -5m - 14 = 0
Given:
an equation is given as m² -5m - 14 = 0
Find:
we have to solve the given quadratic equation.
Explanation:
Compare the given equation with am² + bm + c = 0, we get
a = 1, b = -5, c = -14
we will solve the given equation as following
[tex]\begin{gathered} ()=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-14)}}{2(1)} \\ ()=\frac{5\pm\sqrt{25+56}}{2}=\frac{5\pm\sqrt{81}}{2} \\ ()=\frac{5\pm9}{2} \\ ()=\frac{5+9}{2},\frac{5-9}{2} \\ ()=\frac{14}{2},-\frac{4}{2} \\ ()=7,-2 \end{gathered}[/tex]Therefore, the solution of given equation is m = 7, -2
Answer:
x = 7 ; -2
Step-by-step explanation:
Solving equations using quadratic formula:[tex]\sf \boxed{\bf x = \dfrac{-b \± \sqrt{b^2 - 4ac}}{2a}}[/tex]
m² - 5m - 14 = 0
a = 1 ; b = -5 ; c = -14
b² - 4ac = (-5)² - 4 *(1)*(-14)
= 25 + 56
= 81
[tex]\sf x = \dfrac{-(-5) \± \sqrt{81}}{2*1}\\\\x = \dfrac{5 \± 9}{2}\\\\\\x = \dfrac{5+9}{2} \ ; x =\dfrac{5-9}{2}\\\\\\x = \dfrac{14}{2} \ ; x =\dfrac{-4}{2}\\\\[/tex]
x = 7 or -2
what is the explicit rule of 4, -16, 64, -256
Given sequence is
4, -16, 64, -256
If we have a look closely, we can see a common ratio between the consecutive terms. For example
-16/4 = -4
64/-16 = -4
-256/64 = -4
If there is a common ratio (r) between the consecutive terms of a sequence, it is called a geometric sequence. The explicit rule for such a sequence is:
[tex]a_n=a_1\cdot r^{n-1}[/tex]
Here, r is the common ratio, that is -4 in this case.
a1 is the first term, that is 4.
Now, put the values of a and r in the equation to get the explicit formula
[tex]a_n=4_{}\cdot(-4)^{n-1}[/tex]You can verify the sequence by placing different values of n.
The minimum of a parabola is located at (–1, –3). The point (0, 1) is also on the graph. Which equation can be solved to determine the a value in the function representing the parabola?1 = a(0 + 1)^2 – 31 = a(0 – 1)^2 + 30 = a(1 + 1)^2 – 30 = a(1 – 1)^2 + 3
Given:
The minimum of a parabola is located at (–1, –3).
The general equation of the parabola will be as follows:
[tex]y=a(x-h)^2+k[/tex]Where (h,k) is the vertex of the parabola
given the vertex is the minimum point (-1, -3)
So, h = -1, k = -3
substitute into the general form, so, the equation of the parabola will be:
[tex]y=a(x+1)^2-3[/tex]The point (0, 1) is also on the graph.
So, when x = 0, y = 1
substitute with the given point to determine the value of (a)
So, the equation will be:
[tex]1=a(0+1)^2-3[/tex]So, the answer will be the first option:
1 = a(0 + 1)^2 – 3
A length measure can never be more than one half unit in error. why is this the case?can someone please answer this question.
Answer:
This is because the degree of accuracy is half a unit each side of the unit of measure
[tex]\text{When an instrument measures in '1' s any value betwe}en\text{ 6}\frac{1}{2}\text{ and 7}\frac{1}{2\text{ }}\text{ is measured as 7}[/tex]
Solve each system by graphing. Check your solution. (I'll send the photo)
The equations in the system are equal and therefore the graph results in one over the other.
could you please help me out with a question
At figure , Diameter = √ 24^2 + 7^2 = √ 625 = 25
Then ANSWERS ARE
Circumference = π• D = 3.14 x 25 = 78.5 cm
Area of circle = π• D^2/4 = 3.14 x 25^2/4 = 490.6 cm2
A man wishes to invest $3500. He can buy savings bond which pay Simple Interest at the rate of 12% per annum or he can start a savings account which pays Compound Interest at the same rate.
Calculate the difference in the amount of the two investments at the end of 3 years.
let f(x)= - |x-3|+4 what interval describes when f is decreasing
Answer:
(3, ∞)
Explanation:
Given the function:
[tex]f\mleft(x\mright)=-|x-3|+4[/tex]The graph of the function is attached below:
The interval when f(x) is decreasing is therefore:
[tex](3,\infty)[/tex]39An amusement park issued a coupon to increase the number of visitors to the park each week. The function below representsthe number of visitors at the amusement park x weeks after the issuance of the couponVx) = 500(1.5)What is the approximate average rate of change over the interval [2,6]?OA 949 visitors per weekB 281 visitors per weekC1,143 visitors per weekD. 762 visitors per weekResetSubmitCrved12-39
The Solution.
Given the exponential function below:
[tex]V(x)=500(1.5)^x[/tex]The average rate of change over the interval [2,6] is given as below:
[tex]\text{Average rate of change =}\frac{V(6)-V(2)}{6-2}[/tex]To find V(6):
[tex]V(6)=500(1.5)^6=500\times11.3906=5695.313[/tex]To find V(2):
[tex]V(2)=500(1.5)^2=500\times2.25=1125[/tex]So, substituting for the values of V(6) and V(2) in the above formula, we get
[tex]\begin{gathered} \text{Average rate of change over \lbrack{}2,6\rbrack =}\frac{5695.313-1125}{6-2} \\ \\ \text{ = }\frac{4570313}{4}=1142.578\approx1143\text{ visitors per week} \end{gathered}[/tex]Thus, the correct answer is 1143 visitors p
Alex and George are each charged a constant rate for every text they send on their cell phones, shown in the table below:Number ofTexts (0)AmountCharged (a)$0.20$0.40$0.6023At the end of the month, Alex learned that he had sent 150 texts and paid $30 for his bill. George learned that he had sent 125 texts and paid $6.25.Who paid the correct amount? How do you know?
the payment rate is $0.2 per texts. So Alex should pay
[tex]150\cdot0.2=30[/tex]so the paid is correct for Alex. And we know because it follows the payment rate
just need help understanding how to do these step by step explanation please
Solution:
Given the simultaneous equations:
[tex]\begin{gathered} 4x+3y=15\text{ --- equation 1} \\ 5x-2y=13\text{ ---- equation 2} \end{gathered}[/tex]To solve for x and y, using the elimination method, we have
[tex]\begin{gathered} 2\times(4x+3y=15)\Rightarrow8x+6y=30\text{ --- equation 3} \\ 3\times(5x-2y=13)\Rightarrow15x-6y=39\text{ --- equation 4} \end{gathered}[/tex]Add up equations 1 and 2.
thus, this gives
[tex]\begin{gathered} 8x+15x+6y-6y=30+39 \\ \Rightarrow23x=69 \\ divide\text{ both sides by the coefficient of x, which is 23} \\ \frac{23x}{23}=\frac{69}{23} \\ \Rightarrow x=3 \end{gathered}[/tex]To solve for y, substitute the value of 3 for x into equation 1.
thus, from equation 1
[tex]\begin{gathered} 4x+3y=15 \\ when\text{ x = 3,} \\ 4(3)+3y=15 \\ \Rightarrow12+3y=15 \\ add\text{ -12 to both sides,} \\ -12+12+3y=-12+15 \\ 3y=3 \\ divide\text{ both sides by the coefficient of y, which is 3} \\ \frac{3y}{3}=\frac{3}{3} \\ \Rightarrow y=1 \end{gathered}[/tex]Hence, the solution to the equation is
[tex]\begin{gathered} x=3 \\ y=1 \end{gathered}[/tex]Write an equation for the line that contains (-81, 17) and is perpendicularto the graph 9(2x - 4) - 6(2y - 3) = 4y +2Help please! Big test tomorrow
To write the equation of a paerpendicular line that cross a given point we first need the slope of the given line, then we transform into the spole of the perpendicular line and find the intercept using the given point.
So, we want an equation like this:
[tex]y=ax+b[/tex]And we need "a" and "b". First, let's rewrite the given equation in the slope-interscept form:
[tex]\begin{gathered} 9(2x-4)-6(2y-3)=4y+2 \\ 18x-36-12y+18=4y+2 \\ 18x-12y-18=4y+2 \\ 18x-18-2=4y+12y \\ 18x-20=16y \\ y=\frac{18}{16}x-\frac{20}{16} \\ y=\frac{9}{8}x-\frac{5}{4} \end{gathered}[/tex]This is equivalent to the given graph. 9/8 is the slope. To get the slope of the perpendicular line, we invert it and change its sign. So "a" (the slope of the perpendicular line) is:
[tex]a=-\frac{1}{\frac{9}{8}}=-\frac{8}{9}[/tex]Now we got:
[tex]y=-\frac{8}{9}x+b[/tex]To find "b", we input the values of the point we want it to contain, which is (-81,17):
[tex]17=-\frac{8}{9}(-81)+b[/tex]And we solve for b:
[tex]\begin{gathered} 17=\frac{8\cdot81}{9}+b \\ 17=8\cdot9+b \\ b=17-8\cdot9 \\ b=17-72 \\ b=-55 \end{gathered}[/tex]So, the equation is:
[tex]y=-\frac{8}{9}x-55[/tex]4) Which of the following could represent the lengths of the sides of a right triangle? Hint: Remember Pythagorean Triple :a) 3,4,5b) 5,12,12c) 15,30,45d) 24,32,40
We have to find which of the following could represent the lengths of the sides of a right triangle.
To be a right triangle, the lengths a, b and c have to satisfy the Pithagorean theorem:
[tex]a^2+b^2=c^2[/tex]Of course, c has to be the largest of the sides.
We can write for the first option:
[tex]\begin{gathered} 3^2+4^2=5^2 \\ 9+16=25 \\ 25=25 \end{gathered}[/tex]As the expression is satisfied, we can conclude that the triangles with sides 3, 4 and 5 is a right triangle.
Option B (5,12,12) can not be a right triangle, as it has 2 largest sides. It can only have one, that is the hypothenuse. NOTE: it can have two equal smallest sides, but no two largest.
Option C is 15, 30 and 45. We test the equation:
[tex]undefined[/tex]Jerry's Paint Service use 3 gallons ofpaint in 2 hours. At this rate howmany hours will it take them to use 14 gallons of paint?
Jerry's Paint Service uses 3 gallons of
paint in 2 hours. At this rate how
many hours will it take them to use 14 gallons of paint?
Apply proportion
2/3=x/14
solve for x
x=14*2/3
x=9.33 hours or 9 1/3 hours
How to make the proportion
2 ways
First
2 hours/3 gallons=x hours/14 gallons
solve for x
multiply in cross
14*2=3x
x=28/3
second way
3 gallons/2 hours=14 gallons/x hours
solve for x
multiply in cross
3x=14*2
x=28/3
the result is the same both ways
I definitely absolutely recommend this needed a tutor for it can one help me out if your available
The given coordinates : ( 5, 5 ) & ( 11, 3 )
The expression for the mid point is :
[tex]x=\frac{x_1+x_2}{2},\text{ y=}\frac{y_1+y_2}{2}[/tex]Substitute the value of coordinates as :
[tex]\begin{gathered} x_1=5,y_1=5,x_2=11,y_2=3 \\ x=\frac{5+11}{2} \\ x=\frac{16}{2} \\ x=8 \\ y=\frac{5+3}{2} \\ y=\frac{8}{2} \\ y=4 \end{gathered}[/tex]So, the mid point between (5, 5) & (11, 3) is ( 8, 4)
Factor 4a²x - 4ax - 8x.
Answer:
4x(a+1)(a-2)
Explanation:
Given the polynomial:
[tex]4a^2x-4ax-8x[/tex]First, factor out 4x in all the terms:
[tex]=4x(a^2-a-2)[/tex]Next, factor the expression in the parenthesis:
[tex]\begin{gathered} =4x(a^2-a-2) \\ =4x(a^2-2a+a-2) \\ =4x[a(a-2)+1(a-2)] \\ =4x(a+1)(a-2) \end{gathered}[/tex]The factored form of the polynomial is 4x(a+1)(a-2).
4. Each month for 2 months, Kyle buys a pack of 8 replacement tires for his remote-control car. At the end of each month, he has 1 tire left. Explain how the numerical expression for the number of tires Kyle uses in 2 months compares to the numerical expression for the number of tires he uses in 1 month.
Each month Kyler buys a pack of 8 replacement tires.
He does this for 2 months.
At the end of each month he has 1 tire left.
So the numerical expression for the number of tires he uses per month will be;
Number of tires used in 1st month = 8-1 =7
Number of tires used in the 2nd month = 8 - 1= 7
Total number of tires used in two months = 7*2 = 14
Tires left in two months = 1+ 1 = 2
Comparison : The number of tires used in two months is twice that used in one month.