Friday

January 20, 2017
Total # Posts: 47,946

**Algebra**

X=A + B= 168618472 if b = 17472947 than what is A +Z = X what is x

*December 19, 2016*

**Math**

if you want the square root, then that would be 4/7

*December 19, 2016*

**HELP ME PLZ MATH**

21/28 = 7*3 / 7*4 = ?

*December 19, 2016*

**MATH**

(152-133)/10

*December 19, 2016*

**Algebra 1a**

the slope is 7/10 Now just use the point-slope form of the line.

*December 19, 2016*

**Trigonometry**

tan(?-30) = -5 tan 78.7 = 5 since tan? is negative in QII and QIV, ?-30 = (180-78.7) or (360-78.7) so, now you can figure ?

*December 19, 2016*

**Financial mathematics**

P(1+r)^20 = P(1+r)^10 * (1+r)^10 So, the amount after 10 years is 800000/(1+.053/12)^(12*10) = 471433.81

*December 19, 2016*

**maths**

the base has area (40-26)/2 = 7 divide by 2 since the faces come in pairs.

*December 19, 2016*

**chemistry**

first step: what is your balanced equation? then start using moles.

*December 19, 2016*

**Algebra**

#10 wrong -- see #11 #11 right #14 has a typo, but I'm pretty sure you're wrong. #24 what does a tiny correlation coefficient mean? #25 use your favorite best-fit line tool. Many online.

*December 19, 2016*

**Math**

better try using (16,0) and (0,-4)

*December 19, 2016*

**connections math**

x/5 = -2 x = -2*5 x = -10 Ernesto lost 14.5 pounds ... x - 14.5 = 156.5 assuming that he weighed 156.5 after the weight loss The ones you answered you got correct.

*December 19, 2016*

**math(geometric series)**

512/729 = 2^9/3^6 = 2^8/3^8 * 2*3^2

*December 19, 2016*

**tangent help me maths**

you want to find where 2x+5y+11=0 intersects x²+y²+2x-8y-12=0 just doing a straight substitution, we get 4x^2+4y^2+8x-32y-48 = 0 2x = -(5y+11) so, (2x)^2+4y^2+4(2x)-32y-48 = 0 (5y+11)^2 + 4y^2 - 4(5y+11) - 32y - 48 = 0 29y^2+58y+29 = 0 29(y+1)^2 = 0 y = -1 Since ...

*December 19, 2016*

**Algebra**

two exterior angles are also 90° The remaining exterior angles (all the same), are 180-120 = 60° The exterior angles add to 360°, so there are 3 more of them, each 60°. The figure is a pentagon similar to a home plate in baseball. Consider that the interior ...

*December 19, 2016*

**Math**

just mark off any distance x on QR and measure it off 10 times. Let the last point marked be R. Draw line PR. Then construct parallel lines at each marked point on QR, and their intersections with PQ will divide it into 10 equal segments.

*December 19, 2016*

**Math**

I think you can show with little effort that CDEF is a parallelogram. Then find angle ?=DEF of CDEF and its area is 5*5*sin?

*December 19, 2016*

**math**

their sum is ?/2, so if the smaller angle is x, then x + x+2?/5 = ?/2

*December 19, 2016*

**math**

draw an altitude. It forms a 30-60-90 right triangle, where the sides are in the ratio 1:?3:2

*December 19, 2016*

**math**

back it up an hour, and it the same motion as moving from 12:00 to 3:00. Is that easier?

*December 18, 2016*

**geography**

since there are 24 hours in a day (360°), each hour covers about 15°.

*December 18, 2016*

**Geometry**

exterior + interior = 180 The interior angle could be either the vertex angle or one of the two base angles

*December 18, 2016*

**maths**

huh? really? They gave you the numbers! 96:50.40 I guess you could simplify that a bit to 9600/5040 = 40/21

*December 18, 2016*

**Calculus**

depends on the man's height, h. If the shadow's height is s and the light is x m from the wall, then using similar triangles, we have x/s = (x-4)/h x/(x-4) = s/h -4/(x-4)^2 dx/dt = 1/h ds/dt So, plugging your numbers, -4/36 * -1/2 = 1/h ds/dt ds/dt = h/18

*December 18, 2016*

**physics**

draw the diagram. You have a nice 30-60-90 right triangle with a diagonal of 100, so calculating the sides is easy. right?

*December 18, 2016*

**math**

If m=1, you have (x+1)/(x-1) = 1 x+1 = x-1 1 = -1 So, there is no solution, as you already noted. This is made clear when you solve for x: (x+m)/(x?1)=1 x+m = x-1 m = -1 That is the only value of m which works.

*December 18, 2016*

**Math**

the lines intersect at (2,-7). Now just use the 2-point form of the line.

*December 18, 2016*

**unitar**

sounds feasible, all right.

*December 18, 2016*

**Math**

that's what happens when you guess instead of calculate. You are decreasing a larger number, so the % is less: 1.25 * r = 1 r = 1/1.25 = 0.80 = 80% So, you just have to give a 20% discount to get back to the original price.

*December 18, 2016*

**Math**

if y=cost and x=units, then y = 4000 + (30-22)/(6-4) x

*December 18, 2016*

**Math**

y = 1200 + (1350-1200)/2 x Now just plug in x=3 (3 years since 1985)

*December 18, 2016*

**physical science**

PV=kT if T is constant, PV is constant. So, you want 208V = 106*3

*December 18, 2016*

**Math**

The same way I already showed you ...

*December 18, 2016*

**Mensuration**

Just subtract the bare floor from the total area: 10*8 - (10-2*2)(8-2*2)

*December 18, 2016*

**Algebra II Honors**

(25+x)(10+x) = 1000 Or, you could just double the dimensions. Then the area grows by a factor of 4, to 1000 ft^2.

*December 18, 2016*

**Math**

sure. at t=0, we are at the maximum value. But sin(0) = 0 and sin(t) is a maximum 1/4 period later. Go to the website and play around with different shift values. If you replace 3 with 0, you will see that the graph starts at y=10, but we want to shift it so that it starts at ...

*December 18, 2016*

**Math**

I suspect they are expecting you to say the cosine model is simpler, since then there is no shift needed: we start at the max or min. Then all you need is y = ±4cos(?/6 x) + 10

*December 18, 2016*

**Math**

high tide = 14m low tide = 6m so, the center line is (14+6)/2 = 10m, and the amplitude is (14-6)/2 = 4m The 1/2 period (max to min) is 6 hours, so the period is 12 hours. SO, we can start with y = 4sin(?/6 x) + 10 However, sin(x)=0 at x=0 (high tide), and we want y to be at ...

*December 18, 2016*

**Math**

next steps: 2x^2+x-21 = 0 (2x+7)(x-3) = 0 x = -7/2 or 3 since real-world figures do not have negative dimensions, the rectangle is thus 3x7. Which you probably already guessed, since the only factors of 21 are 3 and 7!

*December 18, 2016*

**calculus**

it sure is (x-x^3)-(x^2-x) = -x^3-x^2+2x You can't combine x^2 and x^3 to make x^5!! They're added, not multiplied.

*December 17, 2016*

**calculus**

since f > g on (0,1), the area is just ?[0,1] (x-x^3)-(x^2-x) dx = 5/12

*December 17, 2016*

**math**

Since the diameter is symmetric about (0,0), B=(-3,5)

*December 17, 2016*

**math**

note that y grows by 19 when x grows by 1. So, start with y = 19x Now adjust that so it fits the table.

*December 17, 2016*

**Calculus Finals Review sheet!! Explanation needed**

inflection points are where y"=0. That is, where y' has a max or min. [[x-2]]/x -> [[3.5-2]]/3.5 = 1/3.5 upward acceleration is where the velocity is increasing. That is, where the graph is concave up.

*December 17, 2016*

**algebra**

try typing your expressions in at wolframalpha.com

*December 17, 2016*

**algebra**

without the typo, we have (3x-2)(2x-3) = 3x(2x-3) - 2(2x-3) = 6x^2-9x - (4x-6) = 6x^2-9x - 4x+6 <------- = 6x^2-13x+6

*December 17, 2016*

**Dilation**

just multiply all the coordinates by 2.

*December 17, 2016*

**Go with @Joe**

Sorry about my typo. e^.07t is correct

*December 17, 2016*

**Math**

e^(1.07t) = 2

*December 17, 2016*

**precalculus**

the odd powers of (-y)^k have a minus sign.

*December 17, 2016*

**precalculus**

1 1 1 Now form new rows by adding 1 at each end and each element is the sum, of the two above it. google is your friend. It will provide many discussions of this topic.

*December 17, 2016*

**a.p.**

or, we know that Sn = n/2 (2a + (n-1)d) = 3n^2+2n an + n(n-1)/2 d = 3n^2+2n an + d/2 n^2 - d/2 n = 3n^2+2n d/2 n^2 + (a - d/2)n = 3n^2+2n d/2 = 3 so d=6 (a-3)=2 so a=5 Tn = 5 + 6(n-1) = 6n-1

*December 17, 2016*

**algebra**

-3(4y-9)+2(2y+6) = (-3)(4y) + (-3)(-9) + (2)(2t) + (2)(6) = ...

*December 17, 2016*

**Math - Coordinate Geometry**

huh? You can't apply the distance formula? AB = ?((5+2)^2+(4-3)^2) = ?(49+1) = ?50 CD = ?((-3-4)^2+(-4+3)^2) = ?(49+1) = ?50 So, AB=CD Now do the same steps for AD and BC.

*December 17, 2016*

**maths**

the fixed setup cost must be 75-45=30 so, 1000 copies cost 30 + 10*15

*December 17, 2016*

**maths**

100 per 10,000 so, use that to find for another 23,000 people: 675+230=905

*December 17, 2016*

**Math**

path is total area minus the pool: (8+1)(5+1) - 8*5 = ?

*December 17, 2016*

**Math**

the dimensions are 8x and 5x. So, 8x * 5x = 169 40x^2 = 169 x^2 = 169/40 x = 13/?40 Now you can determine 8x and 5x

*December 16, 2016*

**Cardinality of Intersection**

15+60-x = 63

*December 16, 2016*

**math**

This should help out: http://www.jiskha.com/display.cgi?id=1317372273

*December 16, 2016*

**math**

well, 8 = 2^3 a^27 = (a^9)^3 ...

*December 16, 2016*

**precalculus**

#1 ok #2 will be the x^4 term #3 will be the x^10 y^2 term #4 Nope. It is 8C6 (3x)^2 (-2y)^6 = 28*3^2*2^6 x^2y^6 = 16128 x^2y^6

*December 16, 2016*

**precalculus**

correct

*December 16, 2016*

**Algebra**

A+69-13 = 107

*December 16, 2016*

**math**

It is (somewhat) clearly a motion problem. v(t) = t^2-2t+1 s(t) = 1/3 t^3 - t^2 + t + C Now, knowing that v(1) = 4 does not help to find C. Moreover, given the equation, it is clear that v(1)=0, not 4. So, let's assume that s(1) = 4. Then we have 1/3 - 1 + 1 + C = 4 C = 3 ...

*December 16, 2016*

**Algebra**

x^2 + 10x = 18 x^2 + 10x + 25 = 18+25 (x+5)^2 = 43 x+5 = ±?43 x = -5±?43 work the other in like wise.

*December 16, 2016*

**math**

x(t) = 9t^2/2 - t^3/2 v(t) = 9t - 3t^2/2 a(t) = -3t Now it should be easy to answer the questions.

*December 16, 2016*

**algebra**

300+5x = 700-6x love the duplication. :-(

*December 16, 2016*

**algebra**

t^0 = 1 negative exponents cause switch between top and bottom (4t)^2t^0/t^6v-8 = 16t^2 * v^8/t^6 = 16v^8/t^4 or 16 t^-4 v^8 p^-3/p^-2r^3 = 1/p^3 / (r^3/p^2) = p^2/(p^3r^3) = 1/(pr^3) or p^-1 r^-3

*December 16, 2016*

**geometry**

the one that's about 78.5/3.14

*December 16, 2016*

**Math (please help!!)**

the last one the clue is that a right angle is perpendicular hence, it belongs with perpendicular bisectors

*December 16, 2016*

**Algebra 2**

3.stretch: 3x^3 right: 3(x-4)^3 down: 3(x-4)^3 - 3

*December 16, 2016*

**math**

x^2 + 15^2 = 19^2

*December 16, 2016*

**math**

surely you can at least do part 1. ?

*December 16, 2016*

**Algebra**

x >= 2 1.75 + 0.55x <= 10.00

*December 16, 2016*

**harmonic progression**

again with this stuff? n ?k(k+1)(2k+1) k=1 = ?(2k^3+3k^2+k) = 2?k^2 + 3?k^2 + ?k Now just consult your reference for special sums of powers to get a 4th-degree polynomial for n.

*December 16, 2016*

**Math - Coordinate Geometry**

The slope of 2x-3y+3=0 is 2/3. The parallel line you want has a slope of 2/3 and passes through (6,-4) So, using the point-slope form of the line, y+4 = 2/3 (x-6)

*December 16, 2016*

**calculus**

y = arctan(x) tan(y) = x sec^2(y) y' = 1 y' = 1/sec^2(y) = 1/(1+tan^2(y)) = 1/(1+x^2)

*December 16, 2016*

**algebra**

(-x)^2 (-x)^5 = (-x)^7 = -x^7 (-3m^3)^2 = (-3)^2 (m^3)^2 = 9m^6 (x/y)^-3 = (y/x)^3 = y^3/x^3 or, (x/y)^-3 = x^-3 / y^-3 = x^-3 y^3

*December 15, 2016*

**Algebra**

x @ 2pts y @ 1pt x+y = 50 2x+1y = 87

*December 15, 2016*

**Algebra**

3x-500 = 840

*December 15, 2016*

**Pre geometry**

check your theorems on transversals of parallel lines. alternate/corresponding interior/exterior angles are equal adjacent angles are supplementary I suspect your answer is 132 degrees, but I have no diagram to consult.

*December 15, 2016*

**uchs**

so, I guess the width is 6+7t

*December 15, 2016*

**Math**

(5u)+(5u - 8/3) ----------------- 2 though I'm not sure how 8/3 fewer is possible.

*December 15, 2016*

**MATH 2**

147 is 1.4 std above the mean, so ...

*December 15, 2016*

**Math**

a+c = 163 11a+6c = 1578

*December 15, 2016*

**ST KYLINS**

My digits in the ones period are zeros. xxx,xxx,000 now what else do you know?

*December 15, 2016*

**calculus**

this should help http://www.jiskha.com/display.cgi?id=1481848543

*December 15, 2016*

**Maths**

I agree. Since the distance between the foci is only 2, there is no way that the condition can be met.

*December 15, 2016*

**Maths**

google is your friend. Try this to get going: https://people.richland.edu/james/lecture/m116/conics/hypdef.html

*December 15, 2016*

**Algebra II**

V = k/P, so VP = k, a constant so, you want V so that 2.2V = 20*16 By the way, L/cm^2 is not pressure. 1L = 1000cm^3, so L/cm^2 = cm Pressure is force/area, as in N/m^2 I've never heard of this L/cm^2 unit. But then, I'm not a physicist...

*December 15, 2016*

**Algebra**

h^2+27^2 = 38^2

*December 15, 2016*

**algebra**

9.6/3.2 * 10^(2-5) = 3.0 * 10^-3

*December 15, 2016*

**algebra**

2.5^4 * (10^-2)^4 = 39.0625 * 10^-8 = 3.90625 * 10^-7

*December 15, 2016*

**Math**

so the square must be 4x4 what do you mean when you say "all sides"? Only the sides, not the rows or columns, or diagonals? For 4 numbers to add to 25, either one or three of them must be odd. See what you can do with that.

*December 15, 2016*

**calc**

e^(x^2-x+a) if x < 1 as x->1-, f(x) -> e^a (4^x-x^2)) if 1<x<2 as x->1+, f(x) -> 4^1-1 = 3 as x->2-, f(x) -> 4^2-2^2 = 12 6In(x-b) if x>2 as x->2-, f(x) -> 6ln(2-b) so, we need e^a = 3 a = ln3 6ln(2-b) = 12 ln(2-b)=2 2-b = e^2 b = 2-e^2 But ...

*December 15, 2016*

**Calc**

?(5-x) is defined for x<5 so, using x<1 works there lnx is defined for x>0 so, x>1 works there except where x=25 So, f(x) is defined and continuous on (-?,1)U(0)U(1,25)U(25,?)

*December 15, 2016*

**science**

none, if it's limestone

*December 15, 2016*

**algebra**

just move the decimal point. to the left for negative powers to the right for positive powers

*December 15, 2016*