An antenna is connected to a car battery. Will the antenna emit electromagnetic radiation? Why or why not? Explain
the last sentence says 21 <= n <= 99 one digit prime and n=3k+1 means we have something in 25,34,43,52,58,61,67,70,76,79,85,91,94 reverse digits is prime means we have 34,67,91 sum of digits is prime means 34,67 four factors leaves 34, which has factors 1,2,17,34
∫ -dt/√(1+2t-t^2) = ∫ -dt/√(2-(1+t)^2) Now recall that ∫ du/√(a^2-u^2) = arcsin(u/a) So, that gives us -arcsin (1+t)/√2 = -arcsin (x+2)/((x+1)√2) I don't see where the √3 comes from, nor the arcsinh. I suspect a typo in...
since these events are independent of each other, you just multiply their probabilities. Since there are 3 hearts out of 15 cards, P(heart) = 3/15 = 1/5 P(heart,heart) = 1/5 * 1/5 = 1/25
since sqrt4 is just 2, you have (-8±2*2)/2 = (-8±4)/2 = -4±2
1/3 m + 15
x+y = 12 x-y = 24 2x = 36 x = 18 y = -6 I figure you can pick the larger one.
4x^2-3 = 11 4x^2 = 14 x^2 = 7/2 x = ±√(7/2)
distance flown is d/11000 = csc 15° so, time needed is d/265 seconds
well, you have the formula: t = 2π√(L/40) just plug in your t and find L. by the way, the word is pendulum, not either of the garbled attempts you wrote.
line with slope m through (h,k): (y-k) = m(x-h) line with slope m and y-intercept at (0,b): y = mx+b Neither is always better. Which to use depends on the data given or required.
well, since 1 rotation is 2π, can you not figure out three rotations? By the way, it holds for any circle, not just the unit circle.
I'd say the first step is to list the factors of 48: 1 2 3 4 6 8 12 16 24 48 Take it from there.
go with Damon.
Should have said 19+(21-14)(19-3)/(14-6) = 33
19+(21-19)(19-3)/(14-6) = 23
s = 176 + 8*1.25 x = s - 1.25w y = 1.25w I'm sure you can interpret these, and feel free to modify them as needed to fit your needs.
a = 5 r = 3 Sn = a(1-r^n)/(1-r) = 5(3^n - 1)/2 So, we want 5(3^n-1)/2 > 10^8 3^n-1 > 4*10^7 3^n > 4*10^7 - 1 n > log(4*10^7 - 1)/log3 Now, using base 10 logs, and ignoring the useless -1, n > ( n > 7+log4)/log3 n > 15.9 So the first 16 terms will sum to mo...
well, what is 1/4 of 12? 12/4 = ?
I'd say 50 g of chocolate are in the chocolate bar. Now, you probably want to know how much cocoa is there: .30*50 = 15g
(√2^√2)^√2 = √2^(√2*√2) = √2^2 = 2 √2^√2 = (2^1/2)^√2 = 2^(1/√2) That should get you started
cube: v = Bh pyramid: 1/3 B(h/2) = 1/6 Bh So, (A)
y = √(3x)/(x^2-4) y' = -√3(3x^2+4) ------------------ 2√x (x^2-4)^2 At x=3 that is -√3(31)/(2√3 (25)) = -31/50
if a quartic has one complex zero, it will have two. They come in conjugate pairs. An easy way to make some up is to have two quadratic factors with negative discriminants: (x^2+4x+10)(3x^2-8x+7) = 3x^4+4x^3+5x^2-52x+70 you can easily make up others.
(D) in any atom the protons always equal the electrons, which is the atomic number. The only thing left that can vary is the number of neutrons, which gives a different mass number (protons+neutrons)
2x(√3x^2)*2√x (2)(√3)(2) * (x)(√x^2)(√x) 4x^2 √(3x) assuming you meant √(3x^2) and not (√3)x^2
a = 2pi r (r+h) If r and h are shrunk by a factor of 3 each, than we have 2pi (r/3)(r/3 + h/3) = 2/9 pi r (r+h) = 1/9 a as with all geometric figures, when the linear dimensions are scaled by a factor of f, the area is scaled by f^2 and the volume is scaled by f^3.
h(t) = 80 - 4.9t^2 now plug in t=3
Math 5th grade
Is there some kind of grid in the rectangle and circle, or are the numbers to be placed anywhere on a large blank canvas?
Last math problem help
this is just the same as several others. Only the numbers are different. If he has two accounts, with $x at 8% and $y at 9%, then the amounts in the accounts must add up to the total invested: x+y = 10437 Now, add up the interest earned. The amounts must equal the total intere...
the cost is the same for m miles if 15+.11m = 18+.08m Now just solve for m.
well, since c+p=60, p = 60-c 4.00c + 2.50(60-c) = 3.00*60 That is, add up the values of the separate nuts, and it must equal the value of the final mixture. 4.00c + 150 - 2.5c = 180 1.5c = 30 c = 20 So, 20 lbs cashews and 40 lbs peanuts Note that since the final cost of $3.00 ...
c+p = 60 4.00c + 2.50p = 3.00*60 Now just solve for c and p
.72 + .15(6-.1)/.1 = $9.57 Or, you can think of it as .15 * 60 plus an extra .57 for the 1st 1/10 mile.
If the base is square of area x^2, then the height is 11/x^2 area = x^2 + 4(11/x^2) cost = 1.1 x^2 + (1.6)(4)(11/x^2) = 1.1x^2 + 70.4/x^2 dc/dx = 2.2x - 140.8/x^3 = 2.2x(x^4-64)/x^3 dc/dx=0 when x=2√2
plain old power rule here. f'(x) = -6x^2 + 72x - 162 = -6(x^2 - 12x + 27) f"(x) = -12x + 72 = -12(x-6) So, f' = 0 when x = 3 or 9 f"(3) > 0 so f is concave up (min) f"(9) < 0 so f is concave down (max) Note that you can also just rely on what you k...
c+a = 52 c = 3a 3a+a = 52 a = 13 so, c=39
someone who is disgruntled likely feels 1- optimistic 2- generous 3- depressed 4- irritated my answer is D
draw a diagram. Label these points: T = top of pole B = base of pole on road S = tip of shadow on road Draw a horizontal line from S to where it intersects the extension of TB below the road. Call that point C. Now, we want the height h = BT Let x = SC y = CB Now, we have y/60...
sorry - I mean (1/6)(2x-y)^3 + C
since x is the variable of integration, y acts just like any other constant, so it's just (1/8)(2x-y)^4 + C
log(n) is undefined for n <= 0, so we need 1a 2x-5 > 0 x > 5/2 1b x^2+3x-4 > 0 x < -4 or x>1 #2 4^x + 12 = 4^2x (4^x)^2 - 4^x - 12 = 0 (4^x - 4)(4^x + 3) = 0 4^x = 4 or -3 x = 1
Q: What do you call two crows on a tree limb? A: Attempted murder!
plus ça change, plus cest la même chose.
if you know that the slope is zero, then it is clear that the graph is a horizontal line through (0,-1).
3 ln 2 ln 4 + 5 ln x = ln 3 + 2 ln 5 ln (2^3 / 4 * x^5) = ln(3 * 5^2) ln 2x^5 = ln 75
Foundations Math 12
P(1+.039/52)^(48) = 225.50 P = 217.59 225.50-217.59 = 7.91 interest
7.03 x 10^9
14^2/15^2 = .871 r^2/10^2 = .871 r = 9.3 so, the crust edge will be 0.7 inch
if you meant 3 1/2 years, then you will need 25000/(1+.12/2)^(2*3.5) = $16,626.43
since P/T remains constant 111/(23+271) = P/(475+271) P = 281.65 kPa
duct tape or bubble gum and baling wire
since tanθ = cot(90-θ), tan(4x-4) = tan(90-(3x+10)) 4x-4 = 80-3x 7x = 84 x = 12
we know that y = a(x+1)-1 4 = a(1)-1 so a = 5 y = 5(x+1)-1 y = 5x^2 + 10x + 4
since the lines are not the same or parallel, then they have a single solution. I'm sure you can come up with it. So, the equations are independent and consistent. dependent means they are the same line inconsistent means they are parallel lines.
2x+2y = 102 x^2+y^2 = 39^2 A 5-12-13 right triangle looks promising. Scale that up to a 15-36-39 size, and we see that 15+36=51, so our rectangle is 15 by 36
Quick math help
for elimination, you want all the equations to look the same, so it's easy to add/subtract. The usual way is to give them the form ax+by = c So, for #1, we have 4y=13+2x 6x+4y=9 On the first one, subtract 2x from each side and you have -2x+4y = 13 the 2nd is ok, so you now...
Quick math help
since I just showed you how to do three others just like these, do you have any ideas on these? elimination or substitution will work fine.
#1 we want to find p such that (80)(0.90)p = 80(3.50)(1.60) #2 assuming the retail price p above, we want to find q such that q = 1/p * 100
well, just plug in your data. A = 36730 * .82^(7.5)
assuming you meant log4√(a^2/b) then that would be log4a - 1/2 log4b
science - Damon?
Yeah, you may be right. It could be that the hydraulics involve prove that the weight of air in the beaker remains constant regardless of how much is displaced, but I'd have to research it. I guess you can do that on your own, or hope Damon or someone will stop by with gre...
since you displace some of the air in the beaker, its side of the lever should rise.
Quick math help
to do them by elimination, make the following steps: #2 4x-y = 9 3x+y = 12 add them and eliminate y. Then solve for x. #3 x+3y = 13 x+y = 5 subtract them and eliminate x, then solve for y #4 9a+3b = 18 a+3b = 10 subtract and solve for a
The problem is that Katy cannot accurately transcribe the exercise or proofread her submission. On the off chance that she wants to know the radius of a circle with the same perimeter as the square, then since c = 2πr, 32 = 2πr r = 16/π On the off-off chance tha...
If you mean |4x-1| = 1 then you need to recall that |n| = n if n>=0 |n| = -n if n<0 So, we have two cases, which must be solved separately 4x-1 >= 0 4x-1=1 4x=2 x = 1/2 Since x=1/2 satisfies the condition that 4x-1 >= 0, it is a solution 4x-1 < 0 -(4x-1) = 1 -4x...
20*.72 = 14.4 (144+(.76-.56))/20 = .73 so (C)
the slope of the line from (-1,-12) to (5,12) = 4, so think of the Mean Value Theorem.
Quick English help
#1. dark and light are opposites, so (B) #2. ok #3. in a dinner you eat the food, but in a meal you don't eat the recipe. An agenda is a plan for a meeting, so (A)
8.37g/3.1cm^2 = 2.7 g/cm^3 Looks like aluminum to me.
area of field = 3km^2 area of lake = (.9)(.3) = .27 km^2 so, assuming the pilot just flies around randomly till she crashes, P(lake) = .27/3 = .09
you know the vertex of y=a(x-h)^2+k is at (h,k). The minimum value of (x-h) is zero, when x=h. For any other value of x, (x-h)^2 gets larger. So, you want to get your function in that form and then you can just read off the coordinates of the vertex. f(x) = 2x^2+8x+9 = 2(x^2+4...
w(w+8) = 172 w = 2(√47 - 2) Now you can get the other values. Q1 since the larger is 9 times the area, it has 3 times the perimeter. So, the two squares have perimeters 50 and 150. Q2 x^2 + y^2 =425 4x+4y = 100 The squares have sides 5 and 20 Q3 If there are x students p...
80*(6380/7980)^2 = 51 kg is mass, which does not change. The weight would become the weight of a 51kg man.
using common notation, we have 2(x+h)^2 + 5(x+h) - 6 - (2x^2 + 5x - 6) 2h^2 + h(4x+5)
(5+√3)/(4+√3) * (4-√3)/(4-√3) (5+√3)(4-√3)/(16-3) (17-√3)/13
well, they told us that f(x) = 10/(x^2+1) Looks to me like f(0) = 10/(0+1) = 10
(a) well, we have a point and a slope (5), so y-8 = 5(x+1) (b) dy = 10/(x^2+1) dx, so if dx=1, dy = 10/2 (1) = 5 So, the estimate is 8+5 = 13 (c) F(0)-F(-1) = 7.854 F(0) = 7.854 + F(-1) = 7.854+8 = 15.854 I hope that (c) had a typo and didn't want f(0), since f(0) = 10. (f...
If 4^x = 4+√19, then take logs, recalling that log(x^n) = n*log(x) x*log(4) = log(4+√19) and so on log(x^2) = (log x)^2 2 logx = (logx)^2 logx(2-logx) = 0 logx = 2 or 0 x = b^2 or 1 to whatever base b you are using for the logs.
I assume you meant (1/2)^(3-2x). If so, make everything powers of 2 and you have 2^(2x) * 2^(2x-3) = 2^3 * 2^(2x) so, adding powers, 2x + 2x-3 = 3 + 2x x=3
assuming base 5 throughout, 2log(x+2) = 2 + log(x-2) log[(x+2)^2 / (x-2)] = 2 (x+2)^2 = 25(x-2) x = 3 or 18
do a little multiplying, and we have 25*5^x + 5*5^x = 5 30*5^x = 5 5^x = 1/6 x = -log(6)/log(5)
well, that would be cos^2(pi/15) + sin^2(pi/15) = 1 In fact, (f+g)(x) = 1 for any x.
we need 25-1.25t-4.9t^2 = 0 t=2.135 so, we have a final v of v = -1.25 - 9.8(2.135) = ?
T10+T11+T12 = (a+9d)+(a+10d)+(a+11d), so 3a+30d = 129 Similarly, for the last three terms, 3a+57d = 237 a=3 d=4 and the sequence is 3,7,11,...,83
Calculus 3 - Damon is correct
and you would be correct. I knew there was something not right.
the cylinder has parametric equations x = t y = t^2 Intersect that with the ellipsoid and you get x = t y = t^2 z = 1/5 √(25-t^2-5t^4)
it takes 8/300 hours to get to cruising height. That's 1.6 minutes in the remaining 3.4 minutes it travels 3.4/60*450 = 25.5 km during the climb, it traveled 8cot26° = 16.4 km along the ground. After flying the additional 25.5 km, it is 41.9 km away along the ground. S...
convenience stores like 7-11 big box stores like Sam's club
since the square has side x, the surrounding area has side 6+x+6 = x+12 So, the green area is (x+12)^2 - x^2 = 24x+144
Rational Exponents - Please Help!!!!
#1 (0.00252d^(9/4))/e = (0.00252*90^(9/4))/1.4 90^(1/4) = ∜90 = 3.0801 90^(9/4) = 3.0801^9 = 24948.5693 So, we now have 0.00252 * 24948.5693 / 1.4 = 44.9074 If you visit wolframalpha.com you can play around with arbitrarily complex expressions, just to see how things com...
at time t, we must have s^2 + (vt)^2 = (ut)^2 (u^2 - v^2)t^2 = s^2 t = s/√(u^2-v^2)
h/140 = sin 36° gives the height in yards
s0 = 28.7 @ 95.0° = -2.501,28.591 v0 = 4.85 @ 40.0° = 3.715,3.119 a = 1.94 @ 200° = -1.823,-0.664 Now, we just plug in our usual equations: v = v0 + at s = s0 + v0*t + 1/2 at^2 and then convert back to polar form as needed.
Caculus f'(x) - typo
typo near the end y = 2x + 8/3
y = ax^3 y' = 3ax^2 at x=-2, y' = 12a The line y=2x+b has slope 2, so 12a=2 a = 1/6 at x = -2, y = -8a = -4/3 so the line must go through (-2,-4/3) So, the tangent line has slope 2 and goes through (-2,-4/3) y = 2x+b -4/3 = -4 + b b = 8/3 So, now we have y = 2x - 8/3 i...
p = 4w just take the number of words and multiply by the score for each word.
Simplify - correction
oops, the answer is 9, not ∛9 9 is in fact ∛729