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Steve 2.0
Answers (58)
Yo I think that answer was wrong :skull:
AcTuAlLy if you BoThEreD to do WERK (sis) then you might know der!v@t!ve$
Find the lowest common denominator. The just sort the numerators. For example, 5/8,11/12,7/4,9/16 = 30/48, 44/48, 84/48, 27/48 Now you just have to sort 30,44,84,27 do the others in like wise.
f◦g = √(g-1) = √(x^2-1) You know that √n has domain n>=0. SO, you need (x^2-1) >= 0. g◦f = f^2 = (√(x-1))^2 = x-1 I'm pretty sure you can handle that one, right?
you can find the radius: it's the distance between the points.
well, the distance between those points is the radius of the circle.
Good catch.
since LMO+NMO = LMN, x+34 + NMO = 6x-28 But, MO bisects LMN, so LMO=NMO x+34 + x+34 = 6x-28 2x+68 = 6x-28 4x = 96 x = 24 Now you can figure the actual degree measures.
linear pairs add to 180, so (assuming the usual typos) x-30 + x+76 = 180 x = 77 Now you can figure the two angles.
tan 29.3° = 9/16, so width = diagonal * sin 29.3° length = diagonal * cos 29.3°
well, sin x = 12/37 cos x = 35/37 so, x is close to zero either (b) or (c) push the buttons and find out
pick one of the acute angles. Call it t. Label the opposite side x, and the hypotenuse h. Then sin(t) = x/h Do the same for the other angle and side, or use the Pythagorean Theorem to find the other side.
right on
max distance is when θ=45° so, use that to figure x- and y-components of initial velocity,
42
me too.
since the period is proportional to r^(3/2), replacing r with (9r) produces a period 9^(3/2) = 27 times longer.
nope. There are exactly as many roots as the degree of the polynomial. In this case, 2, since the highest power is x^2. There are three terms, but only two solutions. Recall your quadratic formula. It only produces two values.
right on
you are correct
6d^4+9d^3-12d^2 first, factor out the d^2 d^2(6d^2+9d-12) Now, how about a 3? 3d^2(2d^2+3d-6) Now, note that the discriminant is 57, so the roots will contain √57. No rational roots exist. Can't factor it any further using integers.
since time = distance/speed, Mrs Healthy took 800m/(2400 m/hr) = 1/3 hr = 20 min So, Mrs Wise's speed = 800m/16min = 50m/min = 3000m/hr
well, what is (20x)^2 - pi(4x)^2 ?
and it is correct or, as we usually write it, 1.25x10^-4 or 1.25 * 10^-4
oops #2 is (4x^3 + 2e^2x)/(x^4+e^2x)
recall the chain rule fy/fx = dy/du du/dx #1: u=3x^3 dy/dx = -sin(u) (9x^2) = -9x^2 sin(3x^3) #2: u = x^4 + e^2x) dy/dx = 1/u du/dx = (4x^3 + 2e^2x)/ln(x^4+e^2x)
well, now you know that d = p-3 - (p-5) = 2 2p-15 - (p-3) = 2, so p=14 so, a = p-5 = 9 n/2 (2*9 + (n-1)*2) = 180 n = 10
the radius is 60 135-60 = ?
if all you have is angles, there is no way to find the sides. All similar triangles have the same angles, but different sides. However, you do know that X/sin40 = Y/sin80 what's the stuff about a plane on top? What does that have to do with anything?
#1 any rational function with vertical asymptotes will do this: 1/(x-1)(x+1) #2 f(x) = x+1 for x 1 just any function where there is a finite break at some x value
∫te^at dt needs to be done with integration by parts, with u = t, du = dt dv = e^at dt, v = 1/a e^at Do all that and you have ∫0.02te^(-0.05t) dt = -(0.4t+8)e^(-0.05t) now just plug in the numbers
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x/2 + 1 < 5/2 x/2 < 3/2 x < 3
(-2)^-3 (-24) = 1/(-2)^3 * (-24) = -24/-8 = 3
-(13x)^0 y^-7 z anything^0 = 1, so -z/y^7
figure the length of the diameter d, using your distance formula. Then the area of the circle is A = πd^2/4
v = 2pi*r*w just watch the units: w = /min r = ft, so c=2pi*r = ft
For x people, we have A: 50+25x B: 130+20x So, you want 50+25x = 130+20x
4s+6h = 52 2h = s+1 Now you can find both costs.
any line y = -1/12 x + c will be parallel to that line, for any value of c that you like.
If you mean |x|-15 then as usual, the graph is shifted down 15 units. It's a lot easier to figure out what you mean if you just use proper notation. If you meant |x-15| then none of the choices is correct.
64 = 4^3 so log_4(64) = 3log_4(4) = 3*1 = 3
The population by 4000 for 15 years, or 60,000 So, it must have been 450,000-60,000 = 390,000 in 2010 p(x) = 390 + 4x
First, it's "heading" or "course," not "bearing." Just figure the x- and y-displacements for each leg of the trip. Add them up for a final location. Then figure the distance and (yes) bearing of the port from there. What do you get? I'll check your answer,
Or, you can reason that since y=kx, y/x is constant. So, 14/-4 = y/-6 y=21 And you don't even have to figure out what k is.
correct
well, just plug in n=9 A(9) = -14 + (9-1)(2) = 2
Hmmm. z - 1/5 > 1/4 z > 1/4 + 1/5 z > 9/20 You don't swap direction unless multiplying or dividing by a negative value. Addition and subtraction do not affect the sense of the inequality.
yes, if you lose the space. More common would be 3A.
looks good to me
