Total bill = $29
Tip left = 15%
The valu of the tip = 15% of $29
[tex]\begin{gathered} \frac{15}{100}\text{ x 29} \\ \\ =\text{ \$4.35} \end{gathered}[/tex]The value of the tip = $4.35
Let us consider the options given
$4.35 - $3 = $1.35
$4.35 - $3.5 = $0.85
$4.35 - $4 = $0.35
$4.5 - $4.35 = $0.15
Looking t the deviation from the real value of the tip ($4.35), the least deviation is $0.15. Hence, we can conclude that $4.5 is the closest to the tip.
The answer is option D ($4.50)
Todd forgot the first two numbers of his locker combination.The number can be any number 1 through 6. What is the probability that he will guess the first number correctly and the second number incorrectly
Todd forgot the first two numbers of his locker combination. The number can be any number 1 through 6. What is the probability that he will guess the first number correctly and the second number incorrectly?
______________________________________________
Please, give me some minutes to take over your question
________________________________________
The probability that he will guess the first number correctly and the second number incorrectly
1/6 (the first number correctly)
5/6 (the second number incorrectly)
1/6 * 5/6 = 5/36
_________________________________________
Answer
The probability that he will guess the first number correctly and the second number incorrectly is 5/36 = 0.1389 = 13. 89%.
Identify the quadrant or ask is that the following points lie on if the point lies on an axis specify which part positive or negative of which axis X or Y
ANSWER
Quadrant II
EXPLANATION
There are four (4) quadrants on the coordinate plane:
Let us now plot the point:
Therefore, the point (-1, 9) lies on quadrant II.
How much will the account be worth in 46 months?
In the question we are given the following parameters
Principal = $5100
Rate = 16.87% compounded semi-annually
Time = 46 months = 3yrs 10 months = 3 5/6 years
Explanation
We can solve the question using the formula below
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]"nt" is the number of months the principal accrues interest twice a year.
Therefore we have;
[tex]\begin{gathered} A=5100(1+\frac{16.87\div100}{2})^{\frac{23}{6}\times2} \\ A=5100(1+0.08435)^{\frac{23}{3}} \\ A=5100(1.08435)^{\frac{23}{3}} \\ A=9488.62 \end{gathered}[/tex]Answer:$9488.62
plssss help!!!! Right triangle RST is drawn below. A square is drawn on to each leg of the triangle, and a square is drawn onto the hypotenuse of the triangle. The area of square A is 25 cm² . The area of square B is 144 cm². Determine the length in centimeters of x, the hypotenuse of the right triangle.O 13 cmO 17 cm O 169 cmO 119 cm
the hypotenuse of the right angled triiangle = x = 13cm (option A)
Explanation:
Area of B = 144cm²
Area of a square = length²
length = √Area of a square
shape of B is a square, hence the length of B:
the length of one of the side of B = √144 = 12cm
Area of A = 25cm²
shape of A is a square
the length of one of the side of A = √25 = 5cm
Triangle is a right angled-triangle
From the diagram, the length of the side of B = opposite = 12cm
The length of the side of A = adjacent = 5cm
The length of the third square marked x = hypotenuse
Using pythagoras' theorem:
Hypotenuse² = opposite² + adjacent²
x² = 12² + 5²
x² = 144 + 25
x² = 169
x = √169
x = 13cm
Hence, the hypotenuse of the right angled triiangle = x = 13cm (option A)
Johnny is going to use ASA to prove that VWX=YZX.Which of these is a necessary step in Johnny's proof? A. Prove that WX= XZby CPCTC.B. Prove that VW=YZ by CPCTC.C. Prove that VWX=YZX by alternate interior angles.D. Prove that VXW = YXZ by vertical angles.
In order to prove that two triangles are congruent using the ASA theorem, it is necessary to show two pairs of congruent angles and one pair of congruent sides of the triangles.
In the given diagram, there is already a pair of congruent triangles and a pair of congruent sides shown.
Then, in order to use the ASA theorem, we need to prove that there is another pair of congruent angles, one from each triangle.
Since the side VX must be adjacent to both angles involved in the procedure, then we need to show that the angle VXW is congruent to the angle YXZ, which is adjacent to the side YX.
This can be proven using the fact that VXW and YXZ are vertical angles.
Therefore, the necessary step in Johnny's proof is shown in option D)
[tex]\text{Prove that }\angle VXW\cong\angle YXZ\text{ by vertical angles}[/tex]I’m not sure how to graph the equation and not sure what it means by “interpret”
(a) Graphing the equation
(i) let x = 0 ; then
y = -0.05 (0) +16
∴ y = 16
Point 1 = ( 0;16 )
(ii) let y = 0 , then
0 = -0.05x +16
0.05x = 16
x = 16 /0.05
∴ x = (320 )
point 2 = ( 320; 0 )
The graph of the line ( y = -0.05x+16) will then be as follows :
(b) Interpret the x and y intercept :{To interpret means to explain in details or translate in writing the meaning of the values of x and y . }
• x represents the number of miles travelled
,• y represents gasoline used i gallons
Interpretation:
• when ,x is 0 miles, , the ,gasoline ,is sitting at, 16 gallons.,( this might be the initial stage of travelling)
,• however, when the, person has travelled 320 miles,, all gasoline is ,completly used up and sits at 0 gallons, .( this might be the end stage of travelling)
need help. first correct answer gets brainliest plus 15 pts
We are given that lines V and 0 and lines C and E are parallel.
We are asked to prove that ∠15 and ∠3 are congruent (equal)
In the given figure, angles ∠3 and ∠7 are "corresponding angles" and they are equal.
[tex]\angle3=\angle7[/tex]In the given figure, angles ∠7 and ∠6 are "Vertically opposite angles" and they are equal.
[tex]\angle7=\angle6[/tex]Angles ∠6 and ∠14 are "corresponding angles" and they are equal.
[tex]\angle6=\angle14[/tex]Angles ∠14 and ∠15 are "Vertically opposite angles" and they are equal.
[tex]\angle14=\angle15[/tex]Therefore, the angles ∠15 and ∠3 are equal.
[tex]\angle3=\angle7=\angle6=\angle14=\angle15[/tex]Solve 5x² + 25 = 0Ox= -5x = -5 and x = 5Ox=5No Real Solutions
Solve for x:
Subtract 25 from both sides:
[tex]\begin{gathered} 5x^2+25-25=-25 \\ 5x^2=-25 \end{gathered}[/tex]Divide both sides by 5:
[tex]\begin{gathered} \frac{5x^2}{5}=-\frac{25}{5} \\ x^2=-5 \end{gathered}[/tex]Take the square root of both sides:
[tex]\begin{gathered} x=\pm\sqrt{-5} \\ x=\pm\sqrt{5}i \end{gathered}[/tex]Therefore, there are no real solutions
Answer:
No Real Solutions
Find the surface area to the nearest tenth.19 m4536.5 m22268.2 m2O 238.8 m2477.5 m2
Answer:
Explanation:
The given solid is a sphere of radius 19m.
The surface area of a sphere is calculated using the formula:
[tex]A=4\pi r^2[/tex]Substitute 19 for r:
[tex]\begin{gathered} A=4\times\pi\times19^2 \\ =4536.46m^2 \\ \approx4536.5\; m^2 \end{gathered}[/tex]The surface area of the sphere to the nearest tenth is 4536.5 square mete.
Given the diagram below which could be used to calculate AC
Cos a = adjacent side / hypotenuse
Where:
a= angle = 37°
adjacent side = 20
Hypotenuse = x (the longest side , AC)
Replacing:
Cos (37)=20/ x (option B)
A construction worker dropped a hammer while building the Grand Canyon skywalk, 8100 feetabove the Colorado River. Use the formula t=(square root of h)/4 to find how many seconds it took for thehammer to reach the river.
Given:
[tex]t=\frac{\sqrt[]{h}}{4}[/tex]To find the time when the height h=8100 feet:
Substitute h=8100 in the given function.
[tex]\begin{gathered} t=\frac{\sqrt[]{8100}}{4} \\ t=\frac{90}{4} \\ t=22.5\text{ seconds} \end{gathered}[/tex]Thus, the time required for the hammer to reach the river is 22.5 seconds.
(3,-8),(-2,5) write an equation for the line in point slope form .Then rewrite the equation in slope intercept form
The equation for the line in point-slope form is:
[tex]y-y_1=m(x-x_1)[/tex]Where m is the slope and (x1, y1) is a point of the line. If we have two points (x1,y1) and (x2, y2), the slope is equal to:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]So, replacing (3, -8) and (-2, 5), we get that the slope and the equation of the line are:
[tex]m=\frac{5-(-8)}{-2-3}=\frac{5+8}{-5}=\frac{-13}{5}[/tex][tex]\begin{gathered} y-(-8)=\frac{-13}{5}(x-3) \\ y+8=-\frac{13}{5}(x-3) \end{gathered}[/tex]Therefore, the equation in slope-intercept form is calculated as:
[tex]\begin{gathered} y+8=-\frac{13}{5}x-\frac{13}{5}\cdot(-3) \\ y+8=-\frac{13}{5}x+\frac{39}{5} \\ y=-\frac{13}{5}x+\frac{39}{5}-8 \\ y=-\frac{13}{5}x-\frac{1}{5} \end{gathered}[/tex]Answer: Point-slope form:
[tex]y+8=-\frac{13}{5}(x-3)[/tex]slope-intercept form:
[tex]y=-\frac{13}{5}x-\frac{1}{5}[/tex]Are y = 3x +7 and y = 3x - 8 parallel to each other?
Answer:
They are parallel to each other
Explanation:
Two lines are parallel if they have the same slope.
Additionally, in an equation with the following form:
y = mx + b
The number m beside the x, is the slope
So, in this case, both equations have a 3 besides the x, then, they are parallels
Determine which point is the solution to the given system. y= -7/2 x + 32 y= 4/5 x -11
Answer:
(10, -3)
Step-by-step explanation:
[tex]-\frac{7}{2}x+32=\frac{4}{5}x-11 \\ \\ -35x+320=8x-110 \\ \\ -43x=-430 \\ \\ x=10 \\ \\ \therefore y=-\frac{7}{2}(10)+32=-3[/tex]
2000.5 - 351.748 +62.1
Given the expression :
[tex]2000.5-351.748+62.1[/tex]At first make all the decimal digits equally for all terms
The maximum decimal is 3 so, add 00 to the first and the last terms
So,
[tex]\begin{gathered} 2000.5-351.748+62.1 \\ =2000.500-351.748+62.100 \\ =1710.852 \end{gathered}[/tex]So, the answer is : 1,710.852
Cobalt-60 has a half-life of about 5 years. After 20 years, how many grams of a2,076 gram sample will remain? Round to the hundredths place, if answer doesn'thave a tenths place then use a zero so the answer does.
Solution:
The formula for half-life is given below as
[tex]N(t)=N_0(\frac{1}{2})^{\frac{t}{\frac{t1}{2}}}[/tex]Where the given values are
[tex]\begin{gathered} N_0=2076g \\ t=20years \\ t^{\frac{1}{2}}=5years \end{gathered}[/tex]By substituting the values, we will have
[tex]\begin{gathered} N(t)=N_{0}(\frac{1}{2})^{\frac{t}{\frac{t\times1}{2}}} \\ N(t)=2076\times(\frac{1}{2})^{\frac{20}{5}} \\ N(t)=2076\times(\frac{1}{2})^4 \\ N(t)=\frac{2076}{16} \\ N(t)=129.75g \end{gathered}[/tex]Hence,
The final answer is
[tex]\Rightarrow129.75g[/tex]Which of the equations below could be the equation of this parabola? A. y = 1/2 x² B. x-1/2 y2 c. y = -1/2 x² D. x = 1/2 y2SUBMIT
The equation of the parabola is given as;
[tex]y=\frac{1}{2}x^2[/tex]The correct answer is option A.
write a quadratic equation in the form of ax²bx+c=0
The general form of a quadratic equation is expressed as
ax^2 + bx + c = 0
In order to write the equation, we would substitute values for a, b and c. If a = 3, b = 8, c = 25, the equation would be
3x^2
Use < or > to write a true sentence. Show your work in the lining up decimals and adding zeroes8.41 8.051
8.41 > 8.051
the digit after the decimal point is greater on 8.41 (4) than on 8.051 (0)
Solving a Debra, Ravi, and Ahmad sent a total of 76 text messages during the weekend. Ahmad sent 2 times as many messages as Ravi. Debra sent 8 more messages than Ravi. How many messages did they each send?
Let the number of messages sent by Ravi be x.
Ahmad sent 2 times as many messages as Ravi. Therefore, Ahmad sent 2x messages.
Debra sent 8 more messages than Ravi. Therefore, Debra sent (x + 8) messages.
The sum of all the messages is 76:
[tex]x+2x+x+8=76[/tex]Solving for x, we have:
[tex]\begin{gathered} 4x+8=76 \\ 4x=76-8 \\ 4x=68 \\ x=\frac{68}{4} \\ x=17 \end{gathered}[/tex]The number of messages Ahmad sent will be:
[tex]2(17)=34[/tex]The number of messages Debra sent will be:
[tex]17+8=25[/tex]ANSWER
[tex]\begin{gathered} Debra\to25\text{ }messages \\ Ravi\to17\text{ }messages \\ Ahmad\to34\text{ }messages \end{gathered}[/tex]Find the first five terms in sequences with the following 3n+2
To determine the first five terms of the sequence we substitute n by 1, 2, 3, 4, and 5.
For n=1, we get:
[tex]3(1)+2=3+2=5.[/tex]For n=2, we get:
[tex]3(2)+2=6+2=8.[/tex]For n=3, we get:
[tex]3(3)+2=9+2=11.[/tex]For n=4, we get:
[tex]3(4)+2=12+2=14.[/tex]For n=5, we get:
[tex]3(5)+2=15+2=17.[/tex]Answer: The first five terms of the sequence are:
[tex]5,\text{ 8, 11, 14, 17.}[/tex]A tank is in the shape of a cylinder of radius 15 cm and height 50 cm.Work out the volume of the tank.
Answer: [tex]11250\pi \\[/tex] cm^3
Step-by-step explanation:
This could be solved with integral calculus or simple arithmetic.
If you need to show the work in calculus, let me know, otherwise, here's the easiest way to reach the answer:
Volume of a solid is equal to the area of its 2D projection multiplied by its height, assuming that it's uniform throughout its entire height. Fortunately, a cylinder is uniform throughout its height.
What is a cylinder's 2D projection? A circle!
Area of a circle = [tex]\pi r^{2}[/tex]
r = 15
Area = 225pi cm^2
Now, we multiply the area of the 2D projection by the height of the cylinder.
225pi * 50 = 11250pi cm^3
find the equation of the line?
Let's calculate the straight line equation
To do this we will take two points from the graph
A = (0,3)
B= (2,0)
For them we will first calculate the slope of the curve
[tex]m=\frac{y2-y1}{x2-x1}[/tex][tex]\begin{gathered} m=\frac{0-3}{2-0} \\ m=\frac{-3}{2} \end{gathered}[/tex]Now let's calculate the y-axis intersection
[tex]\begin{gathered} b=y-mx \\ b=3-m\cdot0 \\ b=3 \end{gathered}[/tex]The equation of the line in the slope-intercept form is
[tex]y=-\frac{3}{2}x+3[/tex]Question 6 of 25Simplify the radical expression below.이히O A.A.v28O B.9O c.NIC3
We need to simplify the next given expression:
[tex]\sqrt{\frac{2}{9}}[/tex]We can rewrite it as:
[tex]\sqrt{\frac{2}{9}}=\frac{\sqrt{2}}{\sqrt{9}}[/tex]Solve each square root:
√2 =√2
√9 = 3
Then, the result is:
[tex]=\frac{\sqrt{2}}{3}[/tex]Hence, the correct answer is option A.
This is lines, functions and systems. Graph the line with slope 2/3 passing through the point (2, 1).
Note that the slope is expressed as :
[tex]\text{slope}=\frac{\text{rise}}{\text{run}}[/tex]From the given, the slope is 2/3
[tex]\text{slope}=\frac{\text{rise}}{\text{run}}=\frac{2}{3}[/tex]So it means that from the point (2,1)
You need to rise 2 units upward and run 3 units to the right
It will be look like this :
Next step is to connect these two points by drawing a line.
That's it, the line is in blue line.
Why might you use a different power of 10 instead of leaving bothnumbers in scientific notation?
Why might you use a different power of 10 instead of leaving both
numbers in scientific notation?
Because scintific notation is a simplification for large or small numbers, so a big number with a lot of zeros can be writen by a power of 10 multiplied by a number. But we cannot make substractions using this, because the powers ten represents very different values, for instance
[tex]\begin{gathered} 1\times10^1=10 \\ 1\times10^2=100 \end{gathered}[/tex][tex]1\times10^2-1\times10^1=100-10=90,[/tex]But if they have the same power, we can use the distibutive law to make the substraction
[tex]1\times10^2-1\times10^1=10\times10^1-1\times10^1=(10-1)\times10^1=9\times10^1^{}[/tex]which is the same thing. No matter the power of 10, if the power is the same you can use the same argument I've made before.
Completely the instructions to move from one point to another along the line y = 2/3x+1. Down 4 units then. Units.
The parent function given is,
[tex]y=\frac{2}{3}x+1[/tex]We were told the parent function was translated 4 units down, which means
[tex]\begin{gathered} y=\frac{2}{3}x+(1-4) \\ y=\frac{2}{3}x-3 \end{gathered}[/tex]Hence, the transformed function would be,
[tex]y=\frac{2}{3}x-3[/tex]Let us now plot the graph of the parent function and the transformed function in order to compare the two graphs.
From the graph above, the parent function is represented in the green line while the transformed function is represented in the black line.
Therefore, the answer is
Down 4 units, then left 6 units.
The vertices of a figure are A(1, -1), B(5, -6), and C(1, - 6). Rotate the figure 90° counterclockwise about the origin. Find the coordinates of the image. Polygon
A'(1,1)
B' (6,5)
C' (6,1)
Explanation
Step 1
Let
A(1,-1)
B(5,-6)
C(1,-6)
Step 2
find the image (A'B'C')
When rotating a point 90 degrees counterclockwise about the origin our point A(x,y) becomes A'(-y,x). In other words, switch x and y and make y negative.
Hence
[tex]\begin{gathered} A\mleft(1,-1\mright)\rightarrow A^{\prime}(1,1) \\ B(5,-6)\rightarrow B^{\prime}(6,5) \\ C(1,-6)\rightarrow C^{\prime}(6,1) \end{gathered}[/tex]so, the coordinates of the image are
A'(1,1)
B' (6,5)
C' (6,1)
I hope this helps you
i need help, i already did the first part but i don’t understand the second part.
a) To convert to radical form, we follow this:
[tex]m^{\frac{a}{b}}=\sqrt[b]{m^{a}}[/tex]So:
[tex]R=73.3m^{\frac{3}{4}}=73.3\sqrt[4]{m^{3}}[/tex]b) The formula we have is for mass in Kilograms, so the first step is to convert the mass stated from lbs to kg.
1 lb -- 0.454 kg
160 lb -- m
[tex]m=0.454\cdot160=72.64\operatorname{kg}[/tex]Now, we can use this value in the formula:
[tex]R=73.3m^{\frac{3}{4}}=73.3\cdot(72.64)^{\frac{3}{4}}=1823.84[/tex]a) To convert to radical form, we follow this:
[tex]m^{\frac{a}{b}}=\sqrt[b]{m^{a}}[/tex]So:
[tex]R=73.3m^{\frac{3}{4}}=73.3\sqrt[4]{m^{3}}[/tex]b) The formula we have is for mass in Kilograms, so the first step is to convert the mass stated from lbs to kg.
1 lb -- 0.454 kg
160 lb -- m
[tex]m=0.454\cdot160=72.64\operatorname{kg}[/tex]Now, we can use this value in the formula:
[tex]R=73.3m^{\frac{3}{4}}=73.3\cdot(72.64)^{\frac{3}{4}}=1823.84[/tex]I will show you the question
Given a coordinate plane
For each point, we will make a rotation about the origin
A: ( 9, 3) Clockwise by 90
The rule of rotation will be:
[tex](x,y)\rightarrow(y,-x)[/tex]So, the image of A will be A' = ( 3, -9)
B: ( -4, 7 ) counterclockwise 90
the rule of rotation will be:
[tex](x,y)\rightarrow(-y,x)[/tex]So, the image of B will be B' = ( -7, -4)
C: ( 6, -5) by 180
The rule of rotation will be:
[tex](x,y)\rightarrow(-x,-y)[/tex]so, the image of C will be C' = (-6, 5)
D: ( -8, -2) clockwise by 90
So, the image of D will be D' = ( -2, 8)
