Volume of a cylinder:
[tex]V=h*r^2*\pi[/tex]For cylinder A:
[tex]10\pi cm^3=h*r^2*\pi[/tex]For cylinder B:
[tex]V_B=2h*(2r)^2*\pi[/tex]Simplify the equation for volumen of cylinder B:
[tex]\begin{gathered} V_B=2h*4r^2*\pi \\ V_B=8*(h*r^2*\pi) \end{gathered}[/tex]in the equation for the volume of cylinder A you have the value of h*r^2*π:
[tex]\begin{gathered} V_B=8*(10\pi cm^3) \\ V_B=80\pi cm^3 \end{gathered}[/tex]Then, the volume of cylinder B is 80π cubic centimeters.Estimate the difference between 7,472 and 3,827 by rounding each number to the nearest hundred.
Answer:
The difference is aproximately 3700.
Step-by-step explanation:
First, we'll round each number to the nearest hundred:
[tex]\begin{gathered} 7472\rightarrow7500 \\ 3827\rightarrow3800 \end{gathered}[/tex]Now, we can estimate the difference:
[tex]7500-3800=3700[/tex]This way, we can conlcude that the difference is aproximately 3700.
can you please help me on e. f. and g.
His temperature was 100.1 degree farad initially which is around 6 pm. At 7 pm it became 101 degree farad.
[tex]\begin{gathered} \text{slope = }\frac{y_2-y_1}{x_2-x_1}=\frac{101-100.1}{7-6}=\frac{0.9}{1}=0.9 \\ m=0.9 \end{gathered}[/tex]y = mx + b
where
m = slope
b = y - intercept
let find the y intercept
[tex]\begin{gathered} 101=0.9(7)+b \\ 101-6.3=b \\ b=94.7 \end{gathered}[/tex]Therefore, the equation is
[tex]y=0.9x+94.7[/tex]e. let us draw a graph
His temperature will be critical above 22 minutes past 9 pm.
f . He should go to emergency room.
g.
[tex]\begin{gathered} y=0.9x+94.7 \\ 98.6=0.9x+94.7 \\ 98.6-94.7=0.9x \\ 3.9=0.9x \\ x=\frac{3.9}{0.9} \\ x=4.33333333333 \end{gathered}[/tex]His temperature will be normal around past 4 pm which is 98.6 degree farad.
Ms. Morgan is the cafeteria manager. She keeps track of how many students select each type of drink. Today during breakfast, 32 children picked milk while 44 children picked juice. What is the ratio of the numbe of children who picked juice to those who picked milk?
Answer:
ratio of those who picked juice to milk
it refers to division
Identify the rate of change and Intial Value in this equationy = 3x +6
The rate of change is 3.
The initial value is 6.
Step - by - Step Explanation
What to find?
• Rate of change.
,• Initial value.
Given:
y = 3x + 6
The rate of change is also the same as the slope.
To find the slope of the gien equation, compare the equation with y=mx + b.
Where m is the slope (rate of change).
Comparing the two equations, m = 3
Hence, the rate of change is 3.
The initial value also known as the y-intercept, is the value of y at x=0.
y = 3(0) + 6
y = 6
Hence, the initial value is 6.
Hi I have a meeting at my house in about
The derivative is the instantaneous rate of change of a function with respect to one of its variables. This is equivalent to finding the slope of the tangent line to the function at a point.
The function is given to be:
[tex]T(t)=Ate^{-kt}[/tex]where A and k are positive constants.
We can find the derivative of the function as follows:
[tex]T^{\prime}(t)=\frac{d}{dt}(Ate^{-kt})[/tex]Step 1: Pull out the constant factor
[tex]T^{\prime}(t)=A\cdot\frac{d}{dt}(te^{-kt})[/tex]Step 2: Apply the product rule
[tex]\frac{d(uv)}{dx}=u \frac{dv}{dx}+v \frac{du}{dx}[/tex]Let
[tex]\begin{gathered} u=t \\ v=e^{-kt} \\ \therefore \\ \frac{du}{dt}=1 \\ \frac{dv}{dt}=-ke^{-kt} \end{gathered}[/tex]Therefore, we have:
[tex]T^{\prime}(t)=A(t\cdot(-ke^{-kt})+e^{-kt}\cdot1)[/tex]Step 3: Simplify
[tex]T^{\prime}(t)=A(-kte^{-kt}+e^{-kt})[/tex]QUESTION A
At t = 0, the instantaneous rate of change is calculated to be:
[tex]\begin{gathered} t=0 \\ \therefore \\ T^{\prime}(0)=A(-k(0)e^{-k(0)}+e^{-k(0)}) \\ T^{\prime}(0)=A(0+e^0) \\ Recall \\ e^0=1 \\ \therefore \\ T^{\prime}(0)=A \end{gathered}[/tex]The rate of change is:
[tex]rate\text{ }of\text{ }change=A[/tex]QUESTION B
At t = 30, the instantaneous rate of change is calculated to be:
[tex]\begin{gathered} t=30 \\ \therefore \\ T(30)=A(-k(30)e^{-k(30)}+e^{-k(30)}) \\ T(30)=A(-30ke^{-30k}+e^{-30k}) \\ Collecting\text{ }common\text{ }factors \\ T(30)=Ae^{-30k}(-30k+1) \end{gathered}[/tex]The rate of change is:
[tex]rate\text{ }of\text{ }change=Ae^{-30k}(-30k+1)[/tex]QUESTION C
When the rate of change is equal to 0, we have:
[tex]0=A(-kte^{-kt}+e^{-kt})[/tex]We can make t the subject of the formula using the following steps:
Step 1: Apply the Zero Factor principle
[tex]\begin{gathered} If \\ ab=0 \\ a=0,b=0 \\ \therefore \\ -kte^{-kt}+e^{-kt}=0 \end{gathered}[/tex]Step 2: Collect common terms
[tex]e^{-kt}(-kt+1)=0[/tex]Step 3: Apply the Zero Factor Principle:
[tex]\begin{gathered} e^{-kt}=0 \\ \ln e^{-kt}=\ln0 \\ -kt=\infty \\ t=\infty \end{gathered}[/tex]or
[tex]\begin{gathered} -kt+1=0 \\ -kt=-1 \\ t=\frac{-1}{-k} \\ t=\frac{1}{k} \end{gathered}[/tex]The time will be:
[tex]t=\frac{1}{k}[/tex](G.lla, 1 point) Use the circle shown to answer the question. ♡ If MAC = 64. and m 2 ABC 16) find the value of x. A. 12 B 36 C. 25 D. 24
12
1) In this case, we have two chords within that circle. And since the arc = 64º and the m ∠ABC = 4x -16
2) Applying one Theorem that states that
3) So we can write:
[tex]\begin{gathered} (4x-16)\text{ =}\frac{64}{2} \\ 4x-16\text{ =32} \\ 4x\text{ =32+16} \\ 4x\text{ = 48} \\ x=12 \end{gathered}[/tex]So the value of x = 12
Mr. Santos cycled a total of 16 kilometers by making 4 trips to work. After 5 trips to work, how many kilometers will Mr. Santos have cycled in total? 5 Kilometers
According to the information given in the exercise, you know that he cycled a total of of 16 kilometers by making 4 trips to work.
Let be "d" the total amount of kilometers Mr. Santos will have cycled after 5 trips to work.
Based on the above, you can set up the following proportion:
[tex]\frac{16}{4}=\frac{d}{5}[/tex]Finally, you must solve for the variable "d" in order to find its value. This is:
[tex]\begin{gathered} 4=\frac{d}{5} \\ \\ (4)(5)=d \\ d=20 \end{gathered}[/tex]Therefore, the answer is:
[tex]20\operatorname{km}[/tex]Find the point that partitions segment AB in a 1:3 ratio (_,_)Find the point that partitions segment AD in 1.1 ratio (_,_)
AB in 1:3 ratio, Find a pointwhere on one side there is 1/4 of AB and in the other side 3/4 of AB:
Explain why the product of 20 x 30 is equal to 600.
BIU
Answer:
600
Step-by-step explanation:
2 X 3 = 6
20 has one 0
30 has one 0
one 0 and one 0 is two 0s
6 plus two 0s = 600
Let f(x) = 8x^3 - 3x^2Then f(x) has a relative minimum atx=
1) To find the relative maxima of a function, we need to perform the first derivative test. It tells us whether the function has a local maximum, minimum r neither.
[tex]\begin{gathered} f^{\prime}(x)=\frac{d}{dx}\mleft(8x^3-3x^2\mright) \\ f^{\prime}(x)=\frac{d}{dx}\mleft(8x^3\mright)-\frac{d}{dx}\mleft(3x^2\mright) \\ f^{\prime}(x)=24x^2-6x \end{gathered}[/tex]2) Let's find the points equating the first derivative to zero and solving it for x:
[tex]\begin{gathered} 24x^2-6x=0 \\ x_{}=\frac{-\left(-6\right)\pm\:6}{2\cdot\:24},\Rightarrow x_1=\frac{1}{4},x_2=0 \\ f^{\prime}(x)>0 \\ 24x^2-6x>0 \\ \frac{24x^2}{6}-\frac{6x}{6}>\frac{0}{6} \\ 4x^2-x>0 \\ x\mleft(4x-1\mright)>0 \\ x<0\quad \mathrm{or}\quad \: x>\frac{1}{4} \\ f^{\prime}(x)<0 \\ 24x^2-6x<0 \\ 4x^2-x<0 \\ x\mleft(4x-1\mright)<0 \\ 0Now, we can write out the intervals, and combine them with the domain of this function since it is a polynomial one that has no discontinuities:[tex]\mathrm{Increasing}\colon-\infty\: 3) Finally, we need to plug the x-values we've just found into the original function to get their corresponding y-values:[tex]\begin{gathered} f(x)=8x^3-3x^2 \\ f(0)=8(0)^3-3(0)^2 \\ f(0)=0 \\ \mathrm{Maximum}\mleft(0,0\mright) \\ x=\frac{1}{4} \\ f(\frac{1}{4})=8\mleft(\frac{1}{4}\mright)^3-3\mleft(\frac{1}{4}\mright)^2 \\ \mathrm{Minimum}\mleft(\frac{1}{4},-\frac{1}{16}\mright) \end{gathered}[/tex]4) Finally, for the inflection points. We need to perform the 2nd derivative test:
[tex]\begin{gathered} f^{\doubleprime}(x)=\frac{d^2}{dx^2}\mleft(8x^3-3x^2\mright) \\ f\: ^{\prime\prime}\mleft(x\mright)=\frac{d}{dx}\mleft(24x^2-6x\mright) \\ f\: ^{\prime\prime}(x)=48x-6 \\ 48x-6=0 \\ 48x=6 \\ x=\frac{6}{48}=\frac{1}{8} \end{gathered}[/tex]Now, let's plug this x value into the original function to get the y-corresponding value:
[tex]\begin{gathered} f(x)=8x^3-3x^2 \\ f(\frac{1}{8})=8(\frac{1}{8})^3-3(\frac{1}{8})^2 \\ f(\frac{1}{8})=-\frac{1}{32} \\ Inflection\: Point\colon(\frac{1}{8},-\frac{1}{32}) \end{gathered}[/tex]Which expression represents the area of the rectangle below in square units
Area of rectangle is given by:-
[tex]\begin{gathered} l\times b \\ =(3x+2)\times2x \\ =6x^2+4x \end{gathered}[/tex]So the correct answer is
[tex]6x^2+4x[/tex]What is the vertical shift for the absolute value function below?f(x) = 9|x + 11 + 2
Since the function is shifted 2 units up, the vertical shift is 2
Determine the reasonableness of a solution to a logarithmic equation
SOLUTION
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Write the given equation
[tex]\log_3x=7[/tex]STEP 2: State the law of logarithm
[tex]\begin{gathered} If\text{ }\log_ab=c \\ \Rightarrow b=a^c \\ By\text{ substitution,} \\ \therefore\log_aa^c=c \end{gathered}[/tex]STEP 3: Substitute the given values in the question to get the correct answer
[tex]\begin{gathered} \log_3x=7 \\ x=3^7 \\ By\text{ substitution,} \\ \log_3(3^7)=7 \end{gathered}[/tex]Hence, Answer is:
[tex]\log_3(3^7)=7[/tex]OPTION A
Bobby says the dilation can be represented by (1\3X, 1,\3Y)Betty says the dilation can be represented by (3X, 3Y)who is correct and why?
Bobby is right because the measurements were made smaller so the dilation factor must be a number less than 1, and 1/3 is less than 1
A cash register contains only five dollar and ten dollar bills. It contains twice as many fives as tens and the total amount of money in the cash register is 740 dollars. How many tens are in the cash register?
ANSWER
There are 37 tens in the cash register
EXPLANATION
Given that;
The total amount in the cash register is $740
The cash register contain five dollar and ten dollar
Follow the steps below to find the number of ten dollar in the cash register.
Let x represents the number of $5 and $10 in the cash register.
Recall, that the register contain twice as many $5 as ten dollars and this can be expressed mathematically as
[tex]\text{ 5\lparen2x\rparen+ 10\lparen x\rparen= 740}[/tex]Evaluate x in the above expression
[tex]\begin{gathered} \text{ 10x + 10x = 740} \\ \text{ 20x = 740} \\ \text{ Divide both sides by 20} \\ \text{ }\frac{\text{ 20x}}{\text{ 20 }}\text{ = }\frac{\text{ 740}}{\text{ 20}} \\ \text{ x = 37} \end{gathered}[/tex]Therefore, we have 37 tens in the cash register
Prove that every differentiable function is continuous
To prove :
every differentiable function is continuous.
thus, every differentiable function is continuous.
Janelle is conducting an experiment to determine whether a new medication is effective in reducing sneezing. She finds 1,000 volunteers with sneezing issues and divides them into two groups. The control group does not receive any medication; the treatment group receives the medication. The patients in the treatment group show reduced signs of sneezing. What can Janelle conclude from this experiment?
Answer:
Step-by-step explanation:
In circle G with m_FGH = 150 and FG = 12 units find area of sector FGH.Round to the nearest hundredth.Fa.
The formula for the area of sector is,
[tex]A=\frac{\theta}{360}\pi(r)^2[/tex]Substitute the values in the formula to obtain the area of sector FGH.
[tex]\begin{gathered} A=\frac{150}{360}\cdot\pi(12)^2 \\ =188.4955 \\ \approx188.50 \end{gathered}[/tex]So area of sector FGH is 188.50.
Find the volume of the cone.9 cmr= 6 cmV = [?] cm3
The radius of cone is r = 6 cm.
The height of cone is h = 9 cm.
The formula for the volume of cone is,
[tex]V=\frac{1}{3}\pi\cdot r^2\cdot h[/tex]Substitute the values in the formula to determine the volume of cone.
[tex]\begin{gathered} V=\frac{1}{3}\pi\cdot(6)^2\cdot9 \\ =339.29 \\ \approx339.3 \end{gathered}[/tex]Thus, volume of cone is 339.3 cm^3.
Simplify by combining like terms,8t3 + 8y + 7t3 + 6y + 9t2
The simplification of the expression will be; 15t³ + 9t² + 14y
What are equivalent expressions?Those expressions that might look different but their simplified forms are the same expressions are called equivalent expressions. To derive equivalent expressions of some expressions, we can either make them look more complex or simple.
Given that the expression as 8t³ + 8y + 7t³ + 6y + 9t²
Now combining like terms;
8t³ + 7t³ + 9t² + 8y + 6y
Simplify;
15t³ + 9t² + 14y
It cannot be solved further because of unlike terms in the expression.
Therefore, the simplification of the expression will be; 15t³ + 9t² + 14y
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Using the conjugate zeros theorem to find all zeros of a polynomial
We know that 1+i is a root of the polynimial. This also implies that 1-i is also a root of the polynomial. In other words, the term
[tex](x-1+i)(x-1-i)[/tex]is a factor of our polynomial. This last expression can be written as
[tex](x-1+i)(x-1-i)=x^2-2x+2[/tex]so, in order to find the remaining zero, we can compute the following division of polynomials,
which gives
Therefore, the remaining root is x=1.
In summary, the answer is:
[tex]1+i,1-i,1[/tex]In how many ways can 3 students from a class of 23 be chosen for a field trip?aYour answer is:
SOLUTION:
This is a combination problem.
The number of ways 3 students from a class of 23 be chosen for a field trip is;
[tex]23C_3=\frac{23!}{(23-3)!3!}=1771\text{ }ways[/tex]When the polynomial mx^3 - 3x^2 +nx +2 is divided by x+3, the remainder is -4. When it is divided by x-2, the remainder is -4. Determine the value of m and n.
Answer:
[tex]\begin{gathered} m\text{ =-2} \\ n\text{ =11} \end{gathered}[/tex]Explanation:
Here, we want to find the value of m and n
If we substituted a supposed root into the parent polynomial, the value after evaluation is the remainder. If the remainder is zero, then the value substituted is a root.
for x+ 3
x + 3 = 0
x = -3
Substitute this into the first equation as follows:
[tex]\begin{gathered} m(-3)^3-3(-3)^2-3(n)+\text{ 2 = -4} \\ -27m\text{ -27-3n+ 2 = -4} \\ -27m\text{ -3n = -4}+27-2 \\ -27m-3n\text{ = 21} \\ -9m\text{ - n = 7} \end{gathered}[/tex]We do this for the second value as follows:
x-2 = 0
x = 2
Substitute this value into the polynomial:
[tex]\begin{gathered} m(2)^3-3(2)^2+2(n)\text{ + 2 = -4} \\ 8m\text{ - 12 +2n + 2 = -4} \\ 8m\text{ + 2n = -4-2+12} \\ 8m\text{ + 2n = 6} \\ 4m\text{ + n = 3} \end{gathered}[/tex]Now, we have two equations so solve simultaneously:
[tex]\begin{gathered} -9m-n\text{ = 7} \\ 4m\text{ + n = 3} \end{gathered}[/tex]Add both equations:
[tex]\begin{gathered} -5m\text{ = 10} \\ m\text{ =-}\frac{10}{5} \\ m\text{ = -2} \end{gathered}[/tex]To get the value of n, we simply susbstitute the value of m into any of the two equations. Let us use the second one:
[tex]\begin{gathered} 4m\text{ +n = 3} \\ 4(-2)\text{ + n = 3} \\ -8\text{ + n = 3} \\ n\text{ = 8 + 3} \\ n\text{ = 11} \end{gathered}[/tex]Write the equation of the circle given the following graph.
Given:
Equation of a circle on a graph with center(3, -2).
To find:
Equation of a circle.
Explanation:
General eqution of a circle is
[tex](x-h)^2+(y-k)^2=r^2[/tex]Solution:
From the graph, we can see that center is (3, -2) and radius equal 3.
So, equation of a circle is
[tex](x-3)^2+(y+2)^2=3^2[/tex]Hence, this is the equation of a circle.
The graph used Is below ill attach a picture of the question and options after
Using the triangle sum theorem:
[tex]\begin{gathered} m\angle L+m\angle K+20=180 \\ 2m\angle L=180-20 \\ 2m\angle L=160 \\ m\angle L=\frac{160}{2} \\ m\angle L=80 \end{gathered}[/tex]Using the exterior angle theorem:
[tex]\begin{gathered} m\angle E=m\angle L+m\angle J \\ m\angle E=80+20 \\ m\angle E=100 \end{gathered}[/tex]Answer:
100
solve the quadratic equation below.3x^2-9=0
Solve the System of Equations8x + 15y = -1174x + 9y=-75Write your answer as an ordered pair: (x,y)
We have to solve the system of linear equations:
[tex]\begin{gathered} 8x+15y=-117 \\ 4x+9y=-75 \end{gathered}[/tex]We can substract 2 times the second equation for the first equation and solve for y:
[tex]\begin{gathered} (8x+15y)-2(4x+9y)=-117-2(-75) \\ 8x+15y-8x-18y=-117+150 \\ 0x-3y=33 \\ y=\frac{33}{-3} \\ y=-11 \end{gathered}[/tex]Now, we can solve for x:
[tex]\begin{gathered} 4x+9y=-75 \\ 4x+9(-11)=-75 \\ 4x-99=-75 \\ 4x=-75+99 \\ 4x=24 \\ x=\frac{24}{4} \\ x=6 \end{gathered}[/tex]Answer: (x,y)=(6,-11)
Triangle RST has the coordinates R(0 , 2), S(2 , 9), and T(4 , 2). Which of the following sets of points represents a dilation from the origin of triangle RST? A. R'(0 , 2), S'(8 , 9), T'(16 , 2) B. R'(0 , 2), S'(2 , 36), T'(16 , 2) C. R'(4 , 6), S'(6 , 13), T'(8 , 6) D. R'(0 , 8), S'(8 , 36), T'(16, 8)
The set of points that represents a dilation from the origin of triangle RST are: D. R'(0 , 8), S'(8 , 36), T'(16, 8).
What is dilation?In Mathematics, dilation is a type of transformation which changes the size of a geometric object, but not its shape. This ultimately implies that, the size of the geometric object would be increased or decreased based on the scale factor used.
For the given coordinates of triangle RST, the dilation with a scale factor of 4 from the origin (0, 0) or center of dilation should be calculated as follows:
Point R (0, 2) → Point R' (0 × 4, 2 × 4) = Point R' (0, 8).
Point S (2, 9) → Point S' (2 × 4, 9 × 4) = Point S' (8, 36).
Point T (4, 2) → Point T' (4 × 4, 2 × 4) = Point T' (16, 8).
In conclusion, the other sets of points do not represents a dilation from the origin of triangle RST.
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As of the given condition ordered pair in the option D R'(0 , 8), S'(8 , 36), T'(16, 8), represents the dilated coordinates of the former triangle.
Given that,
Coordinates of the triangle, R(0 , 2), S(2 , 9), and T(4 , 2).
The scale factor for the dilation = 4
The scale factor is defined as the ratio of the modified change in length to the original length.
Here,
According to the question,
The dilated coordinate is given as,
R' = (0×4 , 2×4) = (0, 8)
S' = (2×4, 9×4) = (8, 36)
T' = (4×4, 2×2) = (16, 8)
Thus, As of the given condition ordered pair in the option D R'(0 , 8), S'(8 , 36), T'(16, 8), represents the dilated coordinates of the former triangle.
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How do the coordinates of the blue point relate to the solution of the equation 3x = x + 4
we have the following:
They are related in the way taht if we replace, in both equations it gives the same result:
[tex]\begin{gathered} 3x=2\cdot3=6 \\ x+4=2+4=6 \end{gathered}[/tex]=Given f(x) = -0.4x – 10, what is f(-12)? If it does not exist,enter DNE.
We have the function:
[tex]f\mleft(x\mright)=-0.4x-10[/tex]And we need to find its value when x = -12. So, replacing x with -12, we obtain:
[tex]f(-12)=-0.4(-12)-10=4.8-10=-5.2[/tex]Notice that the product of two negative numbers is a positive number.
Therefore, the answer is -5.2.
