Answer:
In mathematics, a real number is a value of a continuous quantity that can represent a distance along a line (or alternatively, a quantity that can be represented as an infinite decimal expansion).
Suppose you make napkin rings by drilling holes with different diameters through two wooden balls (which also have different diameters). You discover that both napkin rings have the same height 5h. Use cylindrical shells to compute the volume V of a napkin ring of height 5 h created by drilling a hole with radius r through the center of a sphere of radius R and express the answer in terms of h .
Answer:
V = 1/6 π ( 5h)^3
Step-by-step explanation:
Height of napkin rings = 5h
Compute the volume V of a napkin ring
let a = 5
radius = r
express answer in terms of h
attached below is the detailed solution
find and sketch the domain of the function. f(x,y)=√(4-x^2-y^2) +√(1-x^2)
Answer:
Hello
Step-by-step explanation:
The domain is limited with 2 lines parallel: -1 ≤ x ≤ 1
and the disk ? (inside of a circle) of center (0,0) and radius 2
[tex]dom\ f(x,y)=\{(x,y) \in \mathbb{R} ^2 | \ -1\leq x \leq -1\ and \ ( -\sqrt{4-x^2} \leq \ y \leq \sqrt{4-x^2}\ ) \ \}\\[/tex]
What is the complete factorization of the polynomial below?
x3 + 8x2 + 17x + 10
A. (x + 1)(x + 2)(x + 5)
B. (x + 1)(x-2)(x-5)
C. (x-1)(x+2)(x-5)
O D. (x-1)(x-2)(x + 5)
Answer: A (x+1)(x+2)(x+5)
Step-by-step explanation:
Bob's truck averages 23 miles per gallon. If Bob is driving to his mother's house, 72 miles away, how many gallons of gas are needed? Round to the nearest tenth.
Answer:
3.1 gallons
Step-by-step explanation:
To solve this, we need to figure out how many gallons of gas go into 72 miles. We know 23 miles is equal to one gallon of gas, and given that the ratio of miles to gas stays the same, we can say that
miles of gas / gallons = miles of gas / gallons
23 miles / 1 gallon = 72 miles / gallons needed to go to Bob's mother's house
If we write the gallons needed to go to Bob's mother's house as g, we can say
23 miles / 1 gallon = 72 miles/g
multiply both sides by 1 gallon to remove a denominator
23 miles = 72 miles * 1 gallon /g
multiply both sides by g to remove the other denominator
23 miles * g = 72 miles * 1 gallon
divide both sides by 23 miles to isolate the g
g = 72 miles * 1 gallon/23 miles
= 72 / 23 gallons
≈ 3.1 gallons
An isosceles right triangle has a hypotenuse that measures 4√2 cm. What is the area of the triangle?
PLEASE HELP
Answer:
8
Step-by-step explanation:
As it's an isosceles right triangle, it's sides are equal, say x. x^2+x^2=(4*sqrt(2))^2. x=4, Area is (4*4)/2=8
(4-1) + (6 + 5) = help plz
An isosceles trapezoid has a consecutive-sides of length: 10,6,10 and 14. Find the measure of each angle if the trapezoid.
Answer:
Angle A = Angle D = 69° 30'
Angle B = Angle C = 110° 30'
Step-by-step explanation:
B ___ C
/ \
/ \
A ________ D
AB and CD are 10
BC is 6
AD is 14
If we divide the trapezoid, we can imagine a line.
B_ F_C
/ | \
/ | \
A ___E____ D
AE = ED = 7 (14/2)
BF = FC = 3
So now, we draw another line from B or C to AE or ED
B_ F_ C
/ | | \
/ | | \
A ___E_ G_ D
EG = GD = 3.5 (7/2)
There is a right triangle now, GCD
GD is 3.5 and CD is 10. To determine angle D, we can apply trigonometric function:
CD is H, and GD is A
cos D = A/H
cos D = 3.5/10 → 0.35
angle D = 69° 30'
By theory, we know that angle D and angle A, are the same so:
Angle D = Angle A = 69° 30'
Angle B = Angle C
We also make a cuadrilateral, which is EFCD.
Angle D is 69° 30', Angle E is 90°, Angle F is also 90°
Sum of angles in cuadrilateral is 360°
360° - 69° 30' - 90° - 90° = Angle C = Angle B
Angle C = Angle B = 110° 30'
Let's confirm the angles in the trapezoid:
69° 30' + 110° 30' + 69° 30' + 110° 30' = 360°
A + B + C + D
A rope is 56 in length and must be cut into two pieces. If one piece must be six times as long as the other, find the length of each piece. Round your answers to the nearest inch, if necessary.
Answer:
48, 6
Step-by-step explanation:
The ratio of the pieces is 6 to 1
Add them together to get the total
6+1 = 7
Divide the total length by 7
56/7 = 8
Multiply the ratios by 8
6*8 = 48
1*8 = 6
The peices are 48 and 6
what’s the missing side of the polygons
Answer:
the missing side is 21!!!!!!!!
a triangle has sides of 6 m 8 m and 11 m is it a right-angled triangle?
Answer:
No
Step-by-step explanation:
If we use the Pythagorean theorem, we can find if it is a right triangle. To do that, set up an equation.
[tex]6^{2}+8^{2}=c^2[/tex]
If the triangle is a right triangle, c would equal 11
Solve.
[tex]36+64=100[/tex]
Then find the square root of 100.
The square root of 100 is 10, not 11.
So this is not a right triangle.
I hope this helps!
1. Consider a lottery with three possible outcomes:-$125 will be received with probability 0.2-$100 will be received with probability 0.3-$50 will be received with probability 0.5a. What is the expected value of the lottery
Answer:
The expected value of the lottery is $80
Step-by-step explanation:
To get the expected value, we have to multiply each outcome by its probability
Then we proceed to add up all of these to get the expected value of the lottery
we have this as ;;
125(0.2) + 100(0.3) + 50(0.5)
= 25 + 30 + 25 = $80
Which one goes where?
"RS tangent to circle a..." is first statement Reason: Given
Second Reason: "Radius perpendicular to tangent"
Second Statement: "AR is parrallel to BS" Reason: "2 lines perpendicular..."
Which of the following exponential equations is equivalent to the logarithmic
equation below?
log 970 = x
A.x^10-970
B. 10^x- 970
C. 970^x- 10
D. 970^10- X
Given:
The logarithmic equation is:
[tex]\log 970=x[/tex]
To find:
The exponential equations that is equivalent to the given logarithmic equation.
Solution:
Property of logarithm:
If [tex]\log_b a=x[/tex], then [tex]a=b^x[/tex]
We know that the base log is always 10 if it is not mentioned.
If [tex]\log a=x[/tex], then [tex]a=10^x[/tex]
We have,
[tex]\log 970=x[/tex]
Here, base is 10 and the value of a is 970. By using the properties of exponents, we get
[tex]970=10^x[/tex]
Interchange the sides, we get
[tex]10^x=970[/tex]
Therefore, the correct option is B, i.e., [tex]10^x=970[/tex].
Note: It should be "=" instead of "-" in option B.
khai niem hinh cat don gian ?
Answer:
khai niem hinh cat don gian?
Prove the following identities : i) tan a + cot a = cosec a sec a
Step-by-step explanation:
[tex]\tan \alpha + \cot\alpha = \dfrac{\sin \alpha}{\cos \alpha} +\dfrac{\cos \alpha}{\sin \alpha}[/tex]
[tex]=\dfrac{\sin^2\alpha + \cos^2\alpha}{\sin\alpha\cos\alpha}=\dfrac{1}{\sin\alpha\cos\alpha}[/tex]
[tex]=\left(\dfrac{1}{\sin\alpha}\right)\!\left(\dfrac{1}{\cos\alpha}\right)=\csc \alpha \sec\alpha[/tex]
Question :
tan alpha + cot Alpha = cosec alpha. sec alphaRequired solution :
Here we would be considering L.H.S. and solving.
Identities as we know that,
[tex] \red{\boxed{\sf{tan \: \alpha \: = \: \dfrac{sin \: \alpha }{cos \: \alpha} }}}[/tex][tex] \red{\boxed{\sf{cot \: \alpha \: = \: \dfrac{cos \: \alpha }{sin \: \alpha} }}}[/tex]By using the identities we gets,
[tex] : \: \implies \: \sf{ \dfrac{sin \: \alpha }{cos \: \alpha} \: + \: \dfrac{cos \: \alpha }{sin \: \alpha} }[/tex]
[tex]: \: \implies \: \sf{ \dfrac{sin \: \alpha \times sin \: \alpha }{cos \: \alpha \times sin \: \alpha} \: + \: \dfrac{cos \: \alpha \times cos \: \alpha }{sin \: \alpha \times \: cos \: \alpha } } [/tex]
[tex] : \: \implies \: \sf{ \dfrac{sin {}^{2} \: \alpha }{cos \: \alpha \times sin \alpha} \: + \: \dfrac{cos {}^{2} \: \alpha }{sin \: \alpha \times \: cos \: \alpha } } [/tex]
[tex]: \: \implies \: \sf{ \dfrac{sin {}^{2} \: \alpha }{cos \: \alpha \: sin \alpha} \: + \: \dfrac{cos {}^{2} \: \alpha }{sin \: \alpha \: cos \: \alpha } } [/tex]
[tex]: \: \implies \: \sf{ \dfrac{sin {}^{2} \: \alpha \: + \: cos {}^{2} \alpha}{cos \: \alpha \: sin \alpha} } [/tex]
Now, here we would be using the identity of square relations.
[tex]\red{\boxed{ \sf{sin {}^{2} \alpha \: + \: cos {}^{2} \alpha \: = \: 1}}}[/tex]By using the identity we gets,
[tex] : \: \implies \: \sf{ \dfrac{1}{cos \: \alpha \: sin \alpha} }[/tex]
[tex]: \: \implies \: \sf{ \dfrac{1}{cos \: \alpha } \: + \: \dfrac{1}{sin\: \alpha} }[/tex]
[tex]: \: \implies \: \bf{sec \alpha \: cosec \: \alpha}[/tex]
Hence proved..!!Which of the following is a solution to 2sin2x+sinx-1=0?
Answer:
270 degrees
Step-by-step explanation:
If you plug in 270 in place of the x's, the function is true!
This is correct for Plate/Edmentum users!! Hope I could help :)
Tara created a 1 inch cube out of paper.
1 in
If she doubles the volume of her cube, which statement could be true?
A Tara added two inches to the height, length and width of the cube.
B Tara added two inches to the height of the cube.
C Tara doubled the measurements of the cube's height, length and width.
D Tara doubled the measurement of the cube's height.
Answer:
answer D
Step-by-step explanation:
V=L*W*H=1 ==> L=1,W=1,H=1
A:
L-> L+2=1+2=3
W -> W+2 = 1+2=3
H -> H+2=1+2=3
V=3*3*3=27 not the doubled of the volume's cube
A is false
B:
H -> H+2=1+2=3
V=1*1*3=3 not the doubled of the volume's cube
B is false
C:
H -> 2*H=2*1=2
L -> 2*L=2*1=2
W -> 2*W = 2*1=2
V=2*2*2=8 not the doubled of the volume's cube
C is false
D:
H-> H*2=1*2=2
L=1
W=1
V=1*1*2=2 is the doubled of the volume's cube
D is true
Fraces bonitas para decirle a tu nv?
minimo 6
Answer:
it's. is now the MA plz I miss you
si pudiera escoger entre vivir eternamente y vivir dos veces
yo escogeria vivir dos veces porque vivir una vida eterna sin ti a mi lado seria el mayor sufrimiento, ahora vivir dos veces me dejaria tranquilo porque despues del final de mi vida podria volver a encontrarme contigo y vivir todos los momentos bellos una vez mas y eso seria un sueño volviendose realidad
Help please guys thanks
Answer:
5
Step-by-step explanation:
(625 ^2)^(1/8)
Rewriting 625 as 5^4
(5^4 ^2)^(1/8)
We know that a^b^c = a^(b*c)
5^(4*2)^1/8
5^8 ^1/8
5^(8*1/8)
5^1
5
Answer:
[tex]5[/tex]
Step-by-step explanation:
[tex] { {(625}^{2} )}^{ \frac{1}{8} } \\ { ({25}^{2 \times 2} )}^{ \frac{1}{8} } \\ {25}^{4 \times \frac{1}{8} } \\ {5}^{2 \times 4 \times \frac{1}{8} } \\ {5}^{ \frac{8}{8} } \\ {5}^{1} \\ = 5[/tex]
Pleaseee Help. What is the value of x in this simplified expression?
(-1) =
(-j)*
1
X
What is the value of y in this simplified expression?
1 1
ky
y =
-10
K+m
+
.10
m т
9514 1404 393
Answer:
x = 7
y = 5
Step-by-step explanation:
The applicable rule of exponents is ...
a^-b = 1/a^b
__
For a=-j and b=7,
(-j)^-7 = 1/(-j)^7 ⇒ x = 7
For a=k and b=-5,
k^-5 = 1/k^5 ⇒ y = 5
The administration conducted a survey to determine the proportion of students who ride a bike to campus. Of the 123 students surveyed 5 ride a bike to campus. Which of the following is a reason the administration should not calculate a confidence interval to estimate the proportion of all students who ride a bike to campus. Which of the following is a reason the administration should not calculate a confidence interval to estimate the proportion of all students who ride a bike to campus? Check all that apply.
a. The sample needs to be random but we don’t know if it is.
b. The actual count of bike riders is too small.
c. The actual count of those who do not ride a bike to campus is too small.
d. n*^p is not greater than 10.
e. n*(1−^p)is not greater than 10.
Answer:
b. The actual count of bike riders is too small.
d. n*p is not greater than 10.
Step-by-step explanation:
Confidence interval for a proportion:
To be possible to build a confidence interval for a proportion, the sample needs to have at least 10 successes, that is, [tex]np \geq 10[/tex] and at least 10 failures, that is, [tex]n(1-p) \geq 10[/tex]
Of the 123 students surveyed 5 ride a bike to campus.
Less than 10 successes, that is:
The actual count of bike riders is too small, or [tex]np < 10[/tex], and thus, options b and d are correct.
Question 4 of 10
If A = (-1,-3) and B = (11,-8), what is the length of AB?
A. 12 units
B. 11 units
C. 14 units
D. 13 units
SUBMIT
Step-by-step explanation:
AB = square root of [(xA-xB)^2+(yA-yB)^2]
AB=Squarerootof(-1-11)^2 +(-3-(-8))^2=Squarerootof(-12)^2+(5)^2)
AB=Squarerootof((144)+25)= Squarerootof(169)=13 the answer is 13 units
The choice D is the right one
Chang has 2 shirts: a white one and a black one. He also has 2 pairs of pants, one blue and one tan. What is the probability, if Chang gets dressed in the dark, that
he winds up wearing the white shirt and tan pants? Show your work.
Answer:
1/4
Step-by-step explanation:
White = w
Black = B
Blue = b1
Tan = t
Wb1
Wt
Bbi
Bt
The answer will be 1/4, because there are 4 ways it can work and only 1 way it can be white shirt and tan pants.
Answer:
1/4
Step-by-step explanation:
it would be 1/4 because there are 4 different clothing pieces in total and there is only one way it would work the way the problem says.
What is the explicit formula for the sequence ? -1,0,1,2,3
Answer:
B
Step-by-step explanation:
substitute the values in the eq. Ot is also arithmetic progression.
If $6^x = 5,$ find $6^{3x+2}$.
If 6ˣ = 5, then
(6ˣ)³ = 6³ˣ = 5³ = 125,
and
6³ˣ⁺² = 6³ˣ × 6² = 125 × 6² = 125 × 36 = 4500
Which of the following displays cannot be used to compare data from two different sets?
Answer:
Scatter plot charts are good for relationships and distributions, but pie charts should be used only for simple compositions — never for comparisons or distributions.
The cost of producing a custom-made clock includes an initial set-up fee of $1,200 plus an additional $20 per unit made. Each clock sells for $60. Find the number of clocks that must be produced and sold for the costs to equal the revenue generated. (Enter a numerical value.)
Answer:
30 clocks
Step-by-step explanation:
Set up an equation:
Variable x = number of clocks
1200 + 20x = 60x
Isolate variable x:
1200 = 60x - 20x
1200 = 40x
Divide both sides by 40:
30 = x
Check your work:
1200 + 20(30) = 60(30)
1200 + 600 = 1800
1800 = 1800
Correct!
Determine the domain and range of the graph
Answer:
5 ≤ x ≤ 10 5 ≤ y ≥ -1
Step-by-step explanation:
Which point is a solution to y equal greater than or less too
4x + 5?
Answer:
4x+ 4
Step-by-step explanation:
Answer theas question
(1) Both equations in (a) and (b) are separable.
(a)
[tex]\dfrac xy y' = \dfrac{2y^2+1}{x+1} \implies \dfrac{\mathrm dy}{y(2y^2+1)} = \dfrac{\mathrm dx}{x(x+1)}[/tex]
Expand both sides into partial fractions.
[tex]\left(\dfrac1y - \dfrac{2y}{2y^2+1}\right)\,\mathrm dy = \left(\dfrac1x - \dfrac1{x+1}\right)\,\mathrm dx[/tex]
Integrate both sides:
[tex]\ln|y| - \dfrac12 \ln\left(2y^2+1\right) = \ln|x| - \ln|x+1| + C[/tex]
[tex]\ln\left|\dfrac y{\sqrt{2y^2+1}}\right| = \ln\left|\dfrac x{x+1}\right| + C[/tex]
[tex]\dfrac y{\sqrt{2y^2+1}} = \dfrac{Cx}{x+1}[/tex]
[tex]\boxed{\dfrac{y^2}{2y^2+1} = \dfrac{Cx^2}{(x+1)^2}}[/tex]
(You could solve for y explicitly, but that's just more work.)
(b)
[tex]e^{x+y}y' = 3x \implies e^y\,\mathrm dy = 3xe^{-x}\,\mathrm dx[/tex]
Integrate both sides:
[tex]e^y = -3e^{-x}(x+1) + C[/tex]
[tex]\ln(e^y) = \ln\left(C - 3e^{-x}(x+1)\right)[/tex]
[tex]\boxed{y = \ln\left(C - 3e^{-x}(x+1)\right)}[/tex]
(2)
(a)
[tex]y' + \sec(x)y = \cos(x)[/tex]
Multiply both sides by an integrating factor, sec(x) + tan(x) :
[tex](\sec(x)+\tan(x))y' + \sec(x) (\sec(x) + \tan(x)) y = \cos(x) (\sec(x) + \tan(x))[/tex]
[tex](\sec(x)+\tan(x))y' + (\sec^2(x) + \sec(x)\tan(x)) y = 1 + \sin(x)[/tex]
[tex]\bigg((\sec(x)+\tan(x))y\bigg)' = 1 + \sin(x)[/tex]
Integrate both sides and solve for y :
[tex](\sec(x)+\tan(x))y = x - \cos(x) + C[/tex]
[tex]y=\dfrac{x-\cos(x) + C}{\sec(x) + \tan(x)}[/tex]
[tex]\boxed{y=\dfrac{(x+C)\cos(x) - \cos^2(x)}{1+\sin(x)}}[/tex]
(b)
[tex]y' + y = \dfrac{e^x-e^{-x}}2[/tex]
(Note that the right side is also written as sinh(x).)
Multiply both sides by e ˣ :
[tex]e^x y' + e^x y = \dfrac{e^{2x}-1}2[/tex]
[tex]\left(e^xy\right)' = \dfrac{e^{2x}-1}2[/tex]
Integrate both sides and solve for y :
[tex]e^xy = \dfrac{e^{2x}-2x}4 + C[/tex]
[tex]\boxed{y=\dfrac{e^x-2xe^{-x}}4 + Ce^{-x}}[/tex]
(c) I've covered this in an earlier question of yours.
(d)
[tex]y'=\dfrac y{x+y}[/tex]
Multiply through the right side by x/x :
[tex]y' = \dfrac{\dfrac yx}{1+\dfrac yx}[/tex]
Substitute y(x) = x v(x), so that y' = xv' + v, and the DE becomes separable:
[tex]xv' + v = \dfrac{v}{1+v}[/tex]
[tex]xv' = -\dfrac{v^2}{1+v}[/tex]
[tex]\dfrac{1+v}{v^2}\,\mathrm dv = -\dfrac{\mathrm dx}x[/tex]
[tex]-\dfrac1v + \ln|v| = -\ln|x| + C[/tex]
[tex]\ln\left|\dfrac yx\right| -\dfrac xy = C - \ln|x|[/tex]
[tex]\ln|y| - \ln|x| -\dfrac xy = C - \ln|x|[/tex]
[tex]\boxed{\ln|y| -\dfrac xy = C}[/tex]
