# Posts by Steve

Total # Posts: 52,081

**Algebra**

each element must be identical in the two matrices (you already have 1=1 and 3=3), so you just have to solve 5x-8 = 2 4m-1 = 7m+6 then use those values to fill in the unknown elements.

**Calculus**

you have a point and a slope, so just use the point-slope form, which you learned in Algebra I: y+4 = 5(x-2)

**algebra**

?121 = ?

**math**

0.7x = 3.43 x = 3.43/0.7 = 4.9

**math**

4/100 x = 0.8 x = 0.8 * 100/4

**maths**

10/2 (2a+9d) = 220 5/2 (2a+4*2d) = 100

**Maths**

1,2,4,7 cannot be made I think you can easily show that all the rest are multiples of 3 or 5, or can be made using them and makeable remainders.

**KSU**

7:3

**Math**

take logs. 100 log 99 = 199.56 99 log 100 = 99*2 = 198

**algebra**

Add up the wages: 8x + 11(23-x) = 208

**One-to-one function**

think about it. If f(1) = 2 and f(2) = 2 how can you pick an inverse g(2)?

**Calculus**

now you have given even less information than in your last post! There are many many functions passing through (4,-11) tangent to that line. The simplest is, of course, f(x) = (7x-61)/3 Even the next simplest one could be a parabola opening either up or down, and there are ...

**math**

man, I can't even figure out the question...

**Quadratic Equation**

well, you have the function. Using it, you can find the initial speed (64 ft/s), maximum height (at the vertex), and how long it takes to hit the ground (when h=0). Incidentally, it is odd that the initial height is given in meters, but the -16t^2 is in ft/s^2.

**Calculus**

No way to tell, unless you know you are working with something simple, like a quadratic. If y = ax^2+bx+c and you have two points, then you also have 2ax+b = slope of the tangent line Then you have three equations to solve for a,b,c.

**Pre-Calc**

there are lots of online graphing sites, like desmos, wolframalpha, etc. Find one you like and produce your graphs; see where they intersect.

**Math**

(2x 3 y2 3z)(3xyz^4)^3 = (18xy^2z)(27x^3y^3z^12) = 486x^4y^5z^13 (2x^3y^2 3z)(3xyz^4)^3 = 162x^6y^5z^13 Or something else, depending on how you have mangled the expression

**Math**

(-2y4^z3)^2(-xy)^4 = (4y^8z^6)(x^4y^4) = 4x^4y^12z^6

**Physics**

(0,-6)+(10,0)+(0,5) = (10,-1) I assume you can find vector lengths. The distance, naturally, is just 6+10+5

**Chemistry**

assuming that is the number of molecules of CO2, then just divide by Avogadro's number: 7.7x10^24 / 6.02x10^23 = 1.28x10^1 = 12.8 moles

**physics**

1.2x10^-6 / 0.25x10^-12 = 4.8x10^6

**Math**

I will try to unravel your garbled language, and get x^2 = 100/2 + 12+1 If that's not right, maybe you can fix it up. Not quite sure what "dozen one" means.

**Easy Math**

f(x) = 1/x g(x) = x^2-5 f?g = f(g) = 1/g ...

**Math trigonometry**

Assuming you meant 8sin^2?+10 sin? cos?+3cos^2? = 0 (2sin?+cos?)(4sin?+3cos?) = 0 see what you can do with that.

**Calculus**

we have ?/3 h^2(3r-h) = 5?/3 3rh^2-h^3 = 5 3h^2 dr/dh + 6rh - 3h^2 = 0 h dr/dh + 2r - h = 0 dr/dh = (h-2r)/h = 1 - 2r/h ...

**MATHS**

13601 = 87*156 + 29

**science**

If there are g grams of gold, then add up the volumes (mass/density): g/19.3 + (9.35-g)/10.5 = 0.690 g = 4.6 So, the % mass of gold is 4.62/9.35 = 0.49 = 49%

**calc**

draw a diagram. Using similar triangles, if her shadow has height h when she is x cm from the light, then h/500 = 120/x or, hx = 60000 so, h dx/dt + x dh/dt = 0 Now, find h when x = 150, and you know that dx/dt = -60 cm/s; you can then find dh/dt

**Math**

well, there are 4 sides, right? for the cost, just multiply area by cost/m^2

**Math**

factor out the ?27 and you have ?27^2 (6 - 1/4)^2 = 27 (23/4)^2 = ...

**finite math**

well, it is certainly not (a) 30. Note that you can choose any of 6 appetizers. For each of those 6, you can pick any of 8 salads. and so on. So, what do you think?

**Math - typo**

I hope you caught my top. The total area is (c+d)^3

**Math Division polynomials**

Note that (c^2+2cd+d^2) = (c+d)^2 That means the area is (c+d)^2 * (c+d) = (c^2)^3 Now, divide the total area by the area covered by one can to get the number of cans needed.

**geometry**

relative to C=(-1,4), P = (-1,-1) Q = (5,-3) So, scaling those coordinates by 2, P -> (-2,-2) Q -> (10,-6) making P' = (-3,2) Q' = (9,-2) That is, P' = C+2(P-C) Q' = C+2(Q-C)

**Computer programming**

sum=0 for i = 1..4 {sum += cat[i]} for i = 1..4 {pct[i] = cat[i]/sum * 100}

**Math**

Too bad you didn't show your work... -16(5^2)+430 = -16 * 25 + 430 = -400 + 430 = 30 How ever did you get 270? 5^2 is not 10

**Math**

just plug in 5 for t and evaluate it.

**Maths**

30% = 30/100 = 3/10 = 0.3

**Maths**

just plug and chug: (4/7+5/21)÷(4/7-5/21) = (12/21+5/21)÷(12/21-5/21) = (12+5)/(12-5) ...

**Optional mathematics**

(3-4sin^2x) (1-3tan^2x) = (3-4sin^2x) * (cos^2x-3sin^2x)/cos^2x = (3-4sin^2x)/cos^2x * (cos^2x-3sin^2x) = (3sec^2x-4tan^2x)(cos^2x-3sin^2x) = (3+3tan^2x-4tan^2x)(cos^2x-3sin^2x) = (3-tan^2x)(cos^2x-3sin^2x) see if you can finish up from here

**math**

the product of weight and distance must be equal on both sides, so 60*8 = 80x

**geometry**

of which lines? I don't mind helping to find answers, but I hate having to provide the questions as well ...

**geometry**

so, where is P?

**Maths**

1.20 * 0.90 * 0.95 = 1.026 so he gets 2.6% profit

**math**

4526.5/4050 = 1.1176 or 11.76% in two years. That means it gets 5.88% each year so, 1 lakh grows by 0.0588 each year ...

**math**

If all 30 heads are chickens, then there are 60 legs. But we have 86 legs, so there are 26 extra legs. If we replace a chicken with a cow, we get 2 more legs. We need 26 more legs, requiring 13 cows. So, there are 13 cows and only 17 chickens.

**geometry**

you only show one line.

**Math**

3*33 + 2*35 + x = 6*34 or, if you notice that 34 is midway between 33 and 35, the extra day had to be a 35, to balance them out.

**math**

234 = 2m m = 117 so, 117 mangoes, 234 apples ...

**Calculus**

(a) Apparently you are supposed to find the slopes over the various intervals [a,b]. As always, that is [f(b)-f(a)]/(b-a) Just plug in the numbers. (b) By looking at the way the results approach a single value, that is the estimate of the speed. It should be 50.28

**calculus 3**

see related questions below

**calculus 3**

Draw a diagram of the velocities. If the plane's heading is ?, we need the final x- and y-components to be N45W. So, 180sin? - 32cos60 = v sin135 180cos? - 32sin60 = v cos135 ? = 20° v^2 = 180^2+32^2-2*180*32 cos40° v = 157 mph

**College Physics**

Assuming it was thrown horizontally, the height h = 4.9*2.33^2 = 26.60 meters The vertical speed vy = 9.8*2.33 = 22.83 m/s The horizontal speed vx is just 17.3/2.33 = 7.42 m/s The final speed is of course ?(vx^2+vy^2) = 24.00 m/s

**Math**

you want dy/dt = 2 dx/dt x^2+4 = y 2x dx/dt = dy/dt So, you want 2x dx/dt = 2 dx/dt 2x = 2 x = 1 So, at (1,5) dy/dt = 2 dx/dt or, you want dy/dx = 2 y = x^2+4 dy/dx = 2x = 2 x = 1

**math**

I assume that by "five times less" you mean "1/5 as much" -- if so, then we have, using the scarf as the basic measure, x + x+5 + 5(x+5) = 555 7x = 525 ...

**Math**

there are 4 numbers, so add 'em up and divide the sum by 4.

**Calculus - 7?**

am I missing something here?

**Math**

r/d = 3/4 (r/2)/(d+r/2) = r/(2d+r) = 1/((2d+r)/r) = 1/(2d/r + 1) = 1/(2(4/3)+1) = 1/(11/3) = 3/11

**maths**

see related questions below.

**Geometry**

you're kidding, right? Just solve the equation I showed you. Don't forget your Algebra I now that you're taking geometry! subtract x from both sides, subtract 2 from both sides.

**Geometry**

since P and Q are congruent, their measures are equal. So, x + 7 = 2x + 2 find x, then you have something to work with.

**Algebra**

you haven't eliminated anything. Adding the equations does no good unless x or y disappears. Try writing them as x + 2y = 1 8x + 2y = 22 Now if you subtract them, y is eliminated...

**Math**

ok. now what? I know. Why not write them as 22x ? y = -23 22x + 24y = 2 guess what comes next?

**CAlculus**

the line 4x ? 2y = 3 has slope = 2 so, you want a line with slope = -1/2. Now you have a point (2,9) and a slope (-1/2), so your line is g-9 = -1/2 (x-2)

**Math**

.88p = 704

**Linear equations**

2k+50 = 500 Looks like a lot of unsupervised kids!

**math**

That ?(mn) is a problem. Try setting n = secx-tanx and see where that takes you.

**Maths**

there is a common difference, so log_y(x) - 1 = log_z(y)-log_y(x) log_z(y)-log_y(x) = -15log_x(z) - log_z(y) Not sure just where you want to go with this, since you only have two equations in three unknowns, but if you play around a bit, you should arrive at the fact that both...

**trigonometry**

sec(A) = r/x, so r = 4 x = 3 y = -?7 so, cot(A) = x/y = -3/?7

**maths**

(26*8)cm^2 / 2cm = 104 cm or 4*26 = 13*8 = 104

**Number theory**

take a look at the powers of 2 and their remainders: 2^1: 2 2^2: 4 2^3: 1 The remainders repeat like this, with a period of 3. Since 50 (mod 3) = 2, 2^50 (mod 7) = 4

**physics**

g = -9.8 m/s^2 So, if it loses 9.8 m/s every second, it had to start out at 3*9.8 m/s to drop to zero in 3 seconds. algebraically, v(t) = vi - 9.8t so, vi-9.8*3 = 0 vi = 3*9.8

**math**

BC has slope -2/5 AC has slope 6 Those two lines are not perpendicular, so there is a misprint somewhere. Now, if you just want a parallelogram, you want AD to have slope -2/5, and BD to have slope 6. That means you want the intersection of the lines y+4 = -2/5 (x-3) y-4 = 6(x...

**physics**

(100/s)÷(6/s^2) = 100/6 s It starts with a speed of 100, and loses 6 every second...

**surface area**

I'll assume that the surfaces lying on the ground are not counted. They are all 1x1 anyway, for an additional 3 u^2 area. The 1/6 block has its top and 3 faces exposed: 1+3*(1/6) = 3/2 The 1/3 block has its top, two sides, and half of another side exposed: 1+2*(1/3)+(1/6...

**Science**

That is Avogadro's number -- the number of atoms or molecules in a mole of matter. There are that many electrons in a mole of hydrogen atoms.

**trigonometry - maths**

180rev/1min * 1min/60s * 2? rad/rev = 6? rad/s

**trigonometry - maths**

well, the two angles add to pi/2, so if x is the smaller angle, x + (x+2pi/5) = pi/2 2x = pi/10 x = pi/20 So, the two angles are pi/20 and 9pi/20

**College linear word equation**

42000 + 0.10 * 800000 However, if you are looking for a linear equation, pay = 42000 + 0.10 * sales

**Maths**

or, how about this? (a^2-b^2)sin? + 2abcos? = (a^2+b^2) sin? + 2ab/(a^2-b^2)cos? = (a^2+b^2)/(a^2-b^2) Multiply the the fractions top and bottom by 1/a^2, and we have sin? + 2(b/a)/(1-(b/a)^2)cos? = (1+(b/a)^2)/(1-(b/a)^2) Now let tanx = b/a sin? + tan(2x)cos? = (1+tan^2x)/(1-...

**Maths**

(a^2-b^2)sin? + 2abcos? = (a^2+b^2) (a^2-b^2)tan? + 2ab = (a^2+b^2)sec? (a^2-b^2)^2 tan^2? = ((a^2+b^2)sec?-2ab)^2 (a^2-b^2)^2(sec^2?-1) = (a^2+b^2)^2sec^2? - 4ab(a^2+b^2)sec? + 4a^2b^2 ((a^2-b^2)^2-(a^2+b^2)^2)sec^2? + 4ab(a^2+b^2)sec? + 4a^2b^2 = 0 -4a^2b^2sec^2? + 4ab(a^2+b...

**Math**

suppose they were all chickens. That would require 35*2 = 70 legs You have 8 extra legs. Each chicken you replace with a cow requires 2 extra legs. So, it looks like 4 of the 35 animals are cows.

**Geometry**

at least in the case of an equilateral triangle, a little diagramming should convince you that the small circles have 1/2 the radius of the larger circle. So, the sum of their areas is the same as the area of the large circle. Not sure about other triangles, though ...

**math trigonometry**

the shortest distance from a point to a line is along a direction perpendicular to that line. So, you want a right triangle, and the desired distance x can be found by x/1000 = sin 75°

**ftcc**

well, it has 8/3 that many atoms of hydrogen ...

**Math 50**

(x-6)^2 + x^2 = (x+6)^2 or, consider that a 3-4-5 triangle has sides which differ by 1. You have a triangle whose sides differ by 6.

**math**

surely you can see that the probability of three consonants is 21/26 * 20/25 * 19/24

**math**

google is your friend. You might start here: https://www.geogebra.org/material/show/id/130909

**Geometry**

since the area of ABE is 100, and the altitude is 20, the length of AB is 10. Since the area of ABCD is 3/4 the area of ABE, CDE has 1/4 the area. So, CDE~ABE, with a scale factor of 1/2. So, CD = AB/2 = 5 check: the altitude of ABCD is 1/2 * 20 = 10 the area of ABCD is thus (...

**Math**

note that the 2nd difference is a constant, so we expect a quadratic rule. Using the method of your choice, you should be able to arrive at y = (x^2+x+6)/2

**Calc**

not quite. The 1st derivative is (-2x^2+4)/?(4-x^2) The 2nd derivative is (2x^3-12x)/(4-x^2)3/2

**Mensuration maths**

why are B and C the same? v = ?r^2h = 6468? cm^3 a = ?r^2 + 2?rh = 1120? cm^2 As for the distance, the most obvious answer is to go right across the bottom and straight up, for a distance of 28+33=61 cm. To figure minimum distance, you have to find the length of 1/2 of a turn ...

**Math**

20-8 = 12 so, how many small dogs will it take to make up the other 12 lbs?

**physics**

I'll assume he's lying on the ground... The arrow's height is y = tan? x - 4.9/(v cos?)^2 x^2 so, we need 400?3 - 19.6*400^2/v^2 = 40 v = 69.309 m/s y = ?3 x - 4.9/(69.309/2)^2 x^2 see the graph at http://www.wolframalpha.com/input/?i=plot+y%3D%E2%88%9A3+x+-+4.9%2F...

**Math**

12...x...17 so each row has 30 seats Now do the same idea to get the number of rows.

**calc**

#2 Let the rectangle extend from -x to +x, with height y. Then the area a = 2xy = 2x?(25-x^2) now find x such that da/dx = 0 #3 If the square cross-section has side x, and the package has length y, then the girth is 4x, and the volume is v = x^2y = x^2(108-4x)

**math**

surely you know that "at least" means "greater than or equal" at least 8 means 8 or more.

**I am in class7**

I don't see a formula, but to get the average, just add them up and divide by 7.

**Maths**

120(20-x) - 10x = 1880

**Maths (!!)**

surely you are joking! sin/cos + sin/1 is certainly NOT 2sin/(cos+1) any more than 4/3 + 4/1 = 2*4/(3+1) = 2 It just happens that your mistakes eliminated each other. sin = tan*cos, so tan+sin = tan(1+cos) tan-sin = tan(1-cos) divide and you get (1+cos)/(1-cos) divide top and ...