(b) Four friends buy cinema tickets using this offer.
Cinema tickets
Buy 3 tickets and get a ticket free
They each pay £6.45.
How much does a ticket cost?
Answer:
The cost of one ticket will be £8.20
Every month, Helen budgets $110 for coffee from the coffee bar located in the lobby of her apartment building. If she stops by the bar every morning and her average order costs $3.88, how much money will she have left after 16 days?
answer:
$47.92
explanation:
$3.88 × 16=$62.08
$110-$62.08=$47.92
the 16 is the days
110is the budget
3.88 is the cost
hope that helps
The total money Helen have left after 16 days will be $47.92
What is the unitary method?The unitary method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value. If an event can occur in m different ways and if following it, a second event can occur in n different ways, then the two events in succession can occur in m × n different ways.
Given that Helen budgets for coffee from the coffee bar located in the lobby of her apartment building was $110. And she stops by the bar every morning and her average order costs $3.88.
The budget = 110 dollars
The average cost = 3.88 dollars
Then the average order costs for 16 days;
$3.88 × 16 = $62.08
Therefore, the money she have left after 16 days;
$110-$62.08=$47.92
Hence, The total money she have left after 16 days will be $47.92
Learn more about the unitary method, please visit the link given below;
https://brainly.com/question/23423168
#SPJ2
Nonsense will be reported!!
[tex]\qquad \qquad\huge \underline{\boxed{\sf Answer}}[/tex]
For the first figure ~
The diagonals of a kite intersect each other at 90°
So, we can apply Pythagoras theorem here :
[tex]\qquad \sf \dashrightarrow \: CD² = OC² + OD²[/tex]
[tex]\qquad \sf \dashrightarrow \: CD² = {7}^{2} + {9}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \: CD² = 49 + 81[/tex]
[tex]\qquad \sf \dashrightarrow \: CD² = 130[/tex]
[tex]\qquad \sf \dashrightarrow \: CD=x = \sqrt{ 130}[/tex]
For the second figure ;
we have same concept of kite, and use of Pythagoras theorem !
Also, the diagonal QS bisects diagonal PR
Hence,
[tex]\qquad \sf \dashrightarrow \: PR = 2 \times OR [/tex]
[tex]\qquad \sf \dashrightarrow \: 10 = 2 \times OR [/tex]
[tex]\qquad \sf \dashrightarrow \: OR = 10 \div 2[/tex]
[tex]\qquad \sf \dashrightarrow \: OR = 5 \: mm[/tex]
now, apply pythagoras theorem ~
[tex]\qquad \sf \dashrightarrow \: QR² = OR² + OQ²[/tex]
[tex]\qquad \sf \dashrightarrow \: QR² = {5}^{2} + {6}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \: QR² = 25 + 36[/tex]
[tex]\qquad \sf \dashrightarrow \: QR² = 61[/tex]
[tex]\qquad \sf \dashrightarrow \: QR=x = \sqrt{61} \: mm[/tex]
here, 2 OR = 2 OP = PR
so, similarly OP = 5 mm
Applying pythagoras theorem again ;
[tex]\qquad \sf \dashrightarrow \: SP² = OS² + OP²[/tex]
[tex]\qquad \sf \dashrightarrow \: SP² = {10}^{2} + {5}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \: SP² = {100}^{} + 25[/tex]
[tex]\qquad \sf \dashrightarrow \: SP² = 125[/tex]
[tex]\qquad \sf \dashrightarrow \: SP = \sqrt{125}[/tex]
[tex]\qquad \sf \dashrightarrow \: SP = y = 5\sqrt{5} \: mm[/tex]
Jane has to stickers /
or Janes stickers
aquals or
Andy stickers
aj How much sticker
der andy have
Answer:
Andy basically has two stickers
Find the value of x in the triangle shown below.
Answer:
√12
Step-by-step explanation:
Use the Phythagorean Theorem.
2² + b² = 4²
= 4 + b² = 16
-4 -4
= b² = 12
√12
Answer:
[tex]\sqrt{12}[/tex] or [tex]2\sqrt{3}[/tex] (simplified)
Step-by-step explanation:
Pythagorean theorem is
α²+b²=c²
a is the leg, which is 2 in this triangle
b is the base, which x in this triangle
c is the hypotenuse, which is 4 in this triangle
So, we can plug in the values:
2²+x²=4²
4+x²=16
-4 -4
x²=12
x=[tex]\sqrt{12}[/tex], or if simplified: [tex]2\sqrt{3}[/tex]
Hope this helps!
I need the correct answers help pls
PLS HELP ME PLSS!!!!! ; _ ;
Answer:
a) P(not red) = 37/49
b) P(purple) = 17/49
Step-by-step explanation:
Prob. of not red = (20+17) / (12+20+17) = 37/49
Solve for c (1.1+1).2=c
Answer is 0.42. Hope that helps.
Can you please answer comment i place on your question? I will give you a full description and answer for the work if you do.
Help and hurry pls!what is the perimeter of this figure?
PLEASE HELP ME I ASAP
I WILL MARK YOU BRAINLIEST
Observe the abscicca and ordinates
*Note that:- y-coordinate is ordinate and x-coordinate is abscicca.
The ordinate having 0 as abscicca in function 1 is (0,1), Thus.. The y-intercept is 1Function 2 :Observe the graph and mark the point where function meets y-axis
*Note that:- The point of the graph where the function meets y-axis is called y-intercept.
The point where the function meets is (0,1). Therefore, The y-intercept of function 2 is also 1[tex]\red{ \rule{35pt}{2pt}} \orange{ \rule{35pt}{2pt}} \color{yellow}{ \rule{35pt} {2pt}} \green{ \rule{35pt} {2pt}} \blue{ \rule{35pt} {2pt}} \purple{ \rule{35pt} {2pt}}[/tex]
Thus, Option C is the correct choice!!~
 prove that a/b x b/a = 1
The equation a/b x b/a = 1 is a product equation
It is true that a/b x b/a = 1
How to prove the product expression?The product expression is given as:
a/b x b/a = 1
Rewrite the product expression properly as follows:
[tex]\frac ab * \frac ba = 1[/tex]
Multiply the numerator of the fractions
[tex]\frac {ab}b * \frac 1a = 1[/tex]
Multiply the denominator of the fractions
[tex]\frac {ab}{ab} = 1[/tex]
Evaluate the quotient
[tex]1= 1[/tex]
Hence, it is true that a/b x b/a = 1
Read more about mathematical proofs at:
https://brainly.com/question/1788884
Determine whether the following value could be a probability
0.12
Answer:
Yes, it could be a probability
Step-by-step explanation:
The probability of an event HAS to be between the numbers 0 and 1. Not less than 0, not greater than 1. The number 0.12 is between 0 and 1, thus, making it a valid probability.
Which is the equation of the line that passes through the points (-4, 8) and (1, 3)?
A. Y=x+4
B. Y=-x+12
C. Y=-x+4
D. Y=x+12
Answer:
Therefore the equation is y=-x+4 (correct optlon: C)
Step-by-step explanation:
In order to find the equation that passes through both points, we can use the slope-intercept form of the linear equation:
y=mx+b
Where m is the slope and b is the y-intercept.
Using the given points on this equation, we have:
(-4,8):
(-4,8):8=m*(-4)+b
(-4,8):8=m*(-4)+bb=8+4m
(-4,8):8=m*(-4)+bb=8+4m(1,3):
(-4,8):8=m*(-4)+bb=8+4m(1,3):3=m+b
(-4,8):8=m*(-4)+bb=8+4m(1,3):3=m+b3=m+8+4m
(-4,8):8=m*(-4)+bb=8+4m(1,3):3=m+b3=m+8+4m5m=3-8
(-4,8):8=m*(-4)+bb=8+4m(1,3):3=m+b3=m+8+4m5m=3-85m=-5
(-4,8):8=m*(-4)+bb=8+4m(1,3):3=m+b3=m+8+4m5m=3-85m=-5m=-1
(-4,8):8=m*(-4)+bb=8+4m(1,3):3=m+b3=m+8+4m5m=3-85m=-5m=-1b=8+4*(-1)=8-4=4
Therefore the equation is y=-x+4 (correct optlon: C)
Please help Me I Make You an Brainalist
Answer:
1.) 12
2.) 36
3.) 36
4.) 36
5.) 72
6.) 72
7.) 120
8.) 30
Step-by-step explanation:
You're welcome make me brainliest :)
Answer: LCM=least common multiple so
Step-by-step explanation: 1)12
2)18 3)36 4)36 5)72 6)72 7)120 8)30
A rectangular prism has a length of 20 inches, a width of 11 inches, and a height of
13 inches. What is the volume in cubic inches of this rectangular prism?
Answer:
V =2860 in ^3
Step-by-step explanation:
The volume of a rectangular prism is given by
V = l*w*h where l is the length, w is the width and h is the height.
V = 20 * 11 * 13
V =2860 in ^3
help me with this question pleasee :)))
Answer:
C+3
Step-by-step explanation:
Please Someone help me i got a question and i have a exams please help! i have attached the picture help me!
Will recommend to keep in touch with picture as name of angles are based on it!
[tex] \\ [/tex]
So as you can see BA is parallel to DC as the arrow signs are made on it. And the figure has four sides , therefore we get to know that the above picture is of parallelogram.
[tex] \\ [/tex]
[tex] \rm\angle B = \angle D [/tex]
How ?
As we know it's a parallelogram, so there's one property of parallelogram that opposite sides of Angle are equal.
[tex] \\ [/tex]
[tex] \hookrightarrow \sf2x =130 \degree [/tex]
[tex] \\ [/tex]
[tex] \hookrightarrow \sf x =130 \degree\div 2 \degree [/tex]
[tex] \\ [/tex]
[tex] \hookrightarrow \sf x = \dfrac{130}{2} [/tex]
[tex] \\ [/tex]
[tex] \hookrightarrow \sf x = \dfrac{13 \times 10}{2} [/tex]
[tex] \\ [/tex]
[tex] \hookrightarrow \sf x = \dfrac{13 \times\cancel{ 10}}{\cancel2} [/tex]
[tex] \\ [/tex]
[tex] \hookrightarrow \sf x = \dfrac{13 \times5}{1} [/tex]
[tex] \\ [/tex]
[tex] \hookrightarrow \sf x =13 \times5[/tex]
[tex] \\ [/tex]
[tex] \hookrightarrow \sf x =65 \degree[/tex]
[tex] \\ [/tex]
Now Let's find value of y.
[tex] \rm \angle DAB =\angle DCB[/tex]
How ?
As we know it's a parallelogram, so there's one property of parallelogram that opposite sides of Angle are equal.
[tex] \\ [/tex]
[tex] \dashrightarrow \rm x = y[/tex]
[tex] \\ [/tex]
[tex] \dashrightarrow \rm 65 = y[/tex]
[tex] \\ [/tex]
[tex] \dashrightarrow \bf y = 65 \degree[/tex]
[tex] \\ [/tex]
y + z + 60° = 180°angles on same line
[tex] \\ [/tex]
65 + 60 + z = 180°125° + z = 180°z = 180° - 125 °z = 55°At last :-
x = 65°
y = 65°
z = 55°
The values of the unknown x, y and z in the figure are 65°, 35° and 85° respectively.
Data
x = ?y = ?z = ?Sum of Angles in a ParallelogramThe sum of angles in a parallelogram is equal to 360°. To solve for all the missing numbers, we have to use some properties of parallelogram.
2x = 130
Reason: Opposite angles in a parallelogram are equal.
let's solve for x
[tex]2x = 130 \\\frac{2x}{2} = \frac{130}{2}\\x = 65^0[/tex]
The value of x is 65°
Let's solve for y
[tex]x+ 130 + y + 2x= 360\\x = 65\\65 + 130 + y + 2(65) = 360\\325 + y = 360\\y = 360 - 325\\y = 35[/tex]
The value of y is 35°
Let's solve for z
[tex]60 + y + z = 180[/tex]
reason: sum of angles on a straight line is equal to 180°
[tex]60 + y + z = 180\\60 + 35 + z = 180\\95 + z = 180\\z = 180 - 95\\z = 85^0[/tex]
The values of the unknown x, y and z in the figure are 65°, 35° and 85° respectively.
Learn more on parallelograms here;
https://brainly.com/question/24056495
A golfer hits an errant tee shot that lands in the rough. A marker in the center of the fairway is 150 yards from the center of the green. While
standing on the marker and facing the green, the golfer turns 100° towards his ball. He then paces off 30 yards to his ball. How far is the
ball from the center of the green?
150 yd
[tex]\bold{\huge{\underline{ Solution}}}[/tex]
Given :-A marker in the center of the fairway is 150 yards away from the centre of the green While standing on the marker and facing the green, the golfer turns 100° towards his ball Then he peces off 30 yards to his ball To Find :-We have to find the distance between the golf ball and the center of the green .Let's Begin :-Let assume that the distance between the golf ball and central of green is x
Here,
Distance between marker and centre of green is 150 yards That is, Height = 150 yards For facing the green , The golfer turns 100° towards his ball That is, Angle = 100° The golfer peces off 30 yards to his ball That is, Base = 30 yardsAccording to the law of cosine :-
[tex]\bold{\red{ a^{2} = b^{2} + c^{2} - 2ABcos}}{\bold{\red{\theta}}}[/tex]
Here, a = perpendicular height b = base c = hypotenuse cos theta = Angle of cosineSo, For Hypotenuse law of cosine will be :-
[tex]\sf{ c^{2} = a^{2} + b^{2} - 2ABcos}{\sf{\theta}}[/tex]
Subsitute the required values,
[tex]\sf{ x^{2} = (150)^{2} + (30)^{2} - 2(150)(30)cos}{\sf{100°}}[/tex]
[tex]\sf{ x^{2} = 22500 + 900 - 900cos}{\sf{\times{\dfrac{5π}{9}}}}[/tex]
[tex]\sf{ x^{2} = 22500 + 900 - 900( - 0.174)}[/tex]
[tex]\sf{ x^{2} = 22500 + 900 + 156.6}[/tex]
[tex]\sf{ x^{2} = 23556.6}[/tex]
[tex]\bold{ x = 153.48\: yards }[/tex]
Hence, The distance between the ball and the center of green is 153.48 or 153.5 yards
A cylinder has a cross-section circumference of 56cm. The height of the cylinder is 3.2cm. Calculate the volume and surface area of the cylinder.
Solve trigonometric function:
csc0 × tan0
Answer:
[tex]\sec(\theta)[/tex]
Step-by-step explanation:
[tex]\csc(\theta)=\dfrac{1}{\sin(\theta)}\\\\\\\tan(\theta)=\dfrac{\sin(\theta)}{\cos(\theta)}\\[/tex]
Therefore,
[tex]\csc(\theta) \times \tan(\theta)=\dfrac{1}{\sin(\theta)}\times\dfrac{\sin(\theta)}{\cos(\theta)}=\dfrac{1}{\cos(\theta)}=\sec(\theta)\\[/tex]
can I have help with the problem in the picture pls
Use the drop-down menus below to state the sequence of transformations that maps Figure W onto Figure X in the animation below. Then use those transformations to determine: are the two figures congruent? Use the drop-down menus to explain why or why not.
Answer:
Dilate, Reflect
Not congruent, dilations are used
Step-by-step explanation:
Dilate the figure by a scale factor of 1/3
Reflecting the resulting shape will map the figure
Step 1: Dilate by 1/3
Step 2: Reflect over the y-axis
Since dilations are used, the figures are not the same, as it changes area and side lengths.
-Chetan K
You have started manufacturing tote bags from vintage fabric. It costs you
$12 to have each tote bag manufactured. You decide that you are going to sell them with a 50% markup. What will your retail price be?
[7/8]2 evaluate in simplest form as a fraction
SOLUTION:
=) 49/64 is the answer I think
Simplify the expression:
2a+4b=9c
Answer:
[tex] a=\frac{9c}{2}-b [/tex]
Step-by-step explanation:
Isolate the variable by dividing each side by factors that don't contain the variable.
Answer:
[tex]a=\frac{9c}{2}-b[/tex]
Step-by-step explanation:
Isoate the variable to solve
What is the equation of a parabola with a focus of (5,3) and a directrix of y=-3
The area of a rectangle is 96 cm 2. if the breath of the rectangleis 8 cm, find its lenght and perimeter.
Answer:
length= 12cm
perimeter= 40cm
Step-by-step explanation:
[tex]area = {96 \: cm}^{2} \\ breadth = \: 8cm \\ therefore = \: length \: \\ = 96 \div 8 \: = 12cm[/tex]
perimeter= (length+breadth)2
length= 12cm
breadth= 8cm
perimeter = (12+8)2
20×2= 40
Find the indefinite integral using the substitution x = 3 sin(θ). (Use C for the constant of integration.) 1 (9 − x2)3/2 dx
It looks like the integral might be
[tex]\displaystyle \int (9 - x^2)^{3/2} \, dx[/tex]
or perhaps
[tex]\displaystyle \int \frac1{(9 - x^2)^{3/2}} \, dx[/tex]
Take note of the fact that both integrands are defined only over the interval -3 < x < 3.
For either integral, we substitute x = 3 sin(θ) and dx = 3 cos(θ) dθ.
Note that we want this substitution to be reversible, so we must restrict -π/2 ≤ θ ≤ π/2, an interval over which sine has an inverse. Then θ = arcsin(x/3).
The first case then reduces to
[tex]\displaystyle \int (9 - (3\sin(\theta))^2)^{3/2} (3 \cos(\theta) \, d\theta) = 3 \times 9^{3/2} \int (1 - \sin^2(\theta))^{3/2} \cos(\theta) \, d\theta \\\\ = 81 \int (\cos^2(\theta))^{3/2} \cos(\theta) \, d\theta \\\\ = 81 \int |\cos^3(\theta)| \cos(\theta) \, d\theta[/tex]
By definition of absolute value,
[tex]\displaystyle 81 \int |\cos^3(\theta)| \cos(\theta) \, d\theta = \begin{cases}\displaystyle 81 \int \cos^4(\theta) \, d\theta & \text{if }\cos(\theta) \ge 0 \\ \displaystyle -81 \int \cos^4(\theta) \, d\theta & \text{if }\cos(\theta) < 0\end{cases}[/tex]
and these cases correspond to 0 ≤ θ < π/2 and π/2 < θ ≤ π, respectively. But we are assuming -π/2 ≤ θ ≤ π/2, so the negative case doesn't matter to us.
You can compute the remaining antiderivative by exploiting the half-angle identity for cosine,
[tex]\cos^2(\theta) = \dfrac{1 + \cos(2\theta)}2[/tex]
Then
[tex]\cos^4(\theta) = \left(\cos^2(\theta)\right)^2 = \dfrac{1 + 2\cos(2\theta) + \cos^2(2\theta)}4 = \dfrac{3 + 4\cos(2\theta) + \cos(4\theta)}8[/tex]
and so
[tex]\displaystyle \int \cos^4(\theta) \, d\theta = \dfrac{12\theta + 8\sin(2\theta) + \sin(4\theta)}{32} + C[/tex]
We can simplify this using the double angle identity for (co)sine,
sin(2θ) = 2 sin(θ) cos(θ)
cos(2θ) = 1 - 2 sin²(θ)
as well as the relations,
sin(arcsin(x/3)) = x/3
cos(arcsin(x/3)) = √(9 - x²)/3
which gives us
[tex]\displaystyle \int \cos^4(\theta) \, d\theta = \dfrac{12\theta + 16 \sin(\theta) \cos(\theta) + 4 \sin(\theta) \cos(\theta) (1 - 2\sin^2(\theta))}{32} + C[/tex]
Putting this in terms of x, we get
[tex]\displaystyle \int (9 - x^2)^{3/2} \, dx \\ = 81 \times \dfrac{12\arcsin\left(\frac x3\right) + 16 \times \frac x3 \times \frac{\sqrt{9-x^2}}3 + 4\times\frac x3\times\frac{\sqrt{9-x^2}}3 \left(1 - 2\left(\frac x3\right)^2\right)}{32} + C[/tex]
[tex]\displaystyle \int (9 - x^2)^{3/2} \, dx = 81 \times \dfrac{12\arcsin\left(\frac x3\right) + \frac{16x\sqrt{9-x^2}}9 + \frac{4x(9-2x^2)\sqrt{9-x^2}}{81}}{32} + C[/tex]
[tex]\boxed{\displaystyle \int (9 - x^2)^{3/2} \, dx = \dfrac{12\arcsin\left(\frac x3\right) + (180x-8x^3)\sqrt{9-x^2}}{32} + C}[/tex]
If you were asking about the other integral, the first few steps are similar and you end up with the far more trivial integral and antiderivative
[tex]\displaystyle \frac19 \int \frac{d\theta}{\cos^2(\theta)} = \frac19 \int \sec^2(\theta) \, d\theta = \frac19 \tan(\theta) + C[/tex]
Putting it back in terms of x, we get
[tex]\displaystyle \int \frac1{(9 - x^2)^{3/2}} \, dx = \frac19 \tan\left(\arcsin\left(\frac x3\right)\right) + C[/tex]
Recall that tan(θ) = sin(θ)/cos(θ), so
[tex]\displaystyle \int \frac1{(9 - x^2)^{3/2}} \, dx = \frac19 \times \frac{\frac x3}{\frac{\sqrt{9-x^2}}3} + C[/tex]
[tex]\boxed{\displaystyle \int \frac1{(9 - x^2)^{3/2}} \, dx = \frac{x}{9\sqrt{9-x^2}} + C}[/tex]
What is the area of a triangle with a height of 3/4 km and a base of 1 km
Answer:
0.38km²
Step-by-step explanation:
A=h and b/2
= 1·0.75/2=
0.375km²
A store in Iowa advertises that during their Labor Day sale, everything is 25% off, plus they will pay the sales tax. Mike buys shoes for $80 and socks for $5. Since there will be no tax added, what is the final price for Mike’s purchase?
$85.00
$63.75
$65.00
$68.00
Answer:
(c) $63.75
Step-by-step explanation:
The discounted price will be 100% -25% = 75% of the marked price. Mike's final price will be ...
($80 +5)×0.75 = $63.75