Thursday

January 19, 2017
Total # Posts: 512

**math**

Put the equations into slope-intercept form: y = mx + b m = slope b = y intercept For your first problem... subtract x from both sides x + 8y = 16 8y = -x + 16 divide both sides by 8 y = -(1/8)x + 2 So, the slope is -(1/8) I am not sure why you set y to 0 in your solution. ...

*August 23, 2010*

**Math**

Put the equation into the form ax^2 + bx + c = 0 then use the quadratic formula to find the solutions.

*August 22, 2010*

**math**

You can put the equations into the form ax^2 + bx + c = 0 then use the quadratic formula to find the solutions.

*August 22, 2010*

**math**

The x intercept occurs where the line intercepts the x axis. The value of y at that point is 0. If you substitute 0 for y in your equation, what does x then equal?

*August 17, 2010*

**math**

The y intercept is where the line intercepts the y axis. So, it occurs when x = 0. Plug the value of 0 into your equation. What do you get?

*August 17, 2010*

**Math**

If you multiply the top row number by 5 and then add 1, you get the bottom row number.

*August 10, 2010*

**Math**

I am not sure why your teacher circled it. When the problem is solved, 8 fits into the sequence top and bottom.

*August 10, 2010*

**Math**

Now for the second part, how does the second row relate to the first. 2, 3, 5, 8, 12 11, 16, 26, 41, ? First I would check what would be added to the top row to get the bottom row. That would be 9, 13, 21, 33. Those don't seem very related. What would be multiply the top ...

*August 10, 2010*

**Math**

Yes! Excellent!

*August 10, 2010*

**Math**

For determining sequences a good start is to look at the difference between adjacent numbers. For the sequence: 2, 3, 5, ?, 12 write the differences. This sequence is not a hard one.

*August 10, 2010*

**Algebra**

t = 3t^3 + 22t^2 + 8t subtract t from both sides 3t^3 + 22t^2 + 7t = 0 t is common to each term t(3t^2 + 22t + 7) = 0 So, t = 0 is one solution. Factor 3t^2 + 22t + 7 to get the other two solutions.

*August 7, 2010*

**science**

You had the right idea when you said "the weight of the book is cancelled out by the support of the table". The force due to gravity is opposed by a force exerted on the book by the table. The NET force on the book is zero. So, the answer is b.

*August 7, 2010*

**Math**

That looks good!

*August 3, 2010*

**Math**

Yes, there is a solution. Start by getting a common denominator.

*August 3, 2010*

**pre-algebra**

It looks like some parenthesis would help clear this up. 2/(x-5) = 1/(x+2) cross multiply... 2*(x+2) = x-5 = 2x + 4 = x - 5 subtract 4 from both sides 2x = x - 9 subtract x from both sides x = -9 check by substituting -9 back into the original equation. 2/(-9 - 5) = 1/(-9 + 2 ...

*August 2, 2010*

**Math-please check my answer**

That's right!

*August 1, 2010*

**math-please check my answer**

No, the slope is not 3. (y2 - y1)/(x2 - x1) point 1: x1 = 3, y1 = 5 point 2: x2 = -5, y2 = -2 So... (-2 - 5)/(-5 - 3) =(-7)/(-8) =7/8

*July 31, 2010*

**Math**

You're welcome.

*July 30, 2010*

**Math**

You are doing it correctly. 5(59.4) = 4(61) + x multiply it out... 297 = 244 + x subtract 244 from each side 53 = x

*July 30, 2010*

**math**

First write the problem. 4 < x^(1/2) < 9 Square each term. There are no negative signs, so the inequality signs will not change. Squaring... 16 < x < 81

*July 26, 2010*

**algebra**

Your very welcome!

*July 19, 2010*

**algebra**

Well, let's try your answer: (x-5)(x+5) = x^2 - 25. So, that doesn't work. Doing the factoring I mentioned previously gives: 7x^2(x - 5) - 6(x - 5) Now there are two terms. The terms have a common factor of (x-5). Using our friendly distributive law we can write: (7x^2...

*July 19, 2010*

**algebra**

7x^3 - 35x^2 - 6x + 30 Factor 7x^2 from the first two terms and -6 from the last two terms.

*July 19, 2010*

**math**

For this type of problem, try to get just the variable on one side and just the numbers on the other side. (1/2)a - 5 = (1/3)a - 1 Add 5 to both sides to get rid of the number on the left. (1/2)a - 5 + 5 = (1/3)a - 1 + 5 combining... (1/2)a = (1/3)a + 4 Now subtract (1/3)a ...

*July 19, 2010*

**math**

The numerator can be factored: (x^4 - 8x^2 + 15) = (x^2 - 3)(x^2 - 5) Use that and cancel the common (x^2 - 3) in the numerator and denominator.

*July 17, 2010*

**math**

1. The solution is an inequality. Also when you multiply or divide by a negative number, the inequality is reversed. from you step: -3x/-3 < 6/-3 Since you are dividing both sides by a negative number the inequality is flipped: x > -2 2. OK

*July 15, 2010*

**math**

4x3x2 are common in both the numerator and denominator. Canceling those leaves just 5 in the numerator.

*July 14, 2010*

**math**

Turn the factorial expressions into product series, cross out common factors in the numerator and denominator.

*July 14, 2010*

**math**

The distance is the hypotenuse of a right triangle where one leg is the x distance and the other leg is the y distance. In this case, the y values are the same so the distance between the points is simply the x distance.

*July 14, 2010*

**Physics Check**

A more complete explanation... The car is traveling horizontally at a constant speed. When it drives off the cliff, it keeps going at the same speed (disregarding friction) until it is stopped by hitting the ground. A forumla for (d) distance traveled with constant ...

*July 12, 2010*

**Physics Check**

#1 correct #2 correct #3 wrong #4 correct #5 wrong ---- #6 find t in seconds: 122.5m = (0.5)(g)(t^2) 122.5m = (0.5)(9.8m/s^2)t^2 reducing... 25s^2 = t^2 t = +5 seconds and t = -5 seconds Disregard the negative time. That is how long it took to fall 122.5m. So, if the car ...

*July 12, 2010*

**Math (Algebra)**

Be careful. You posted (72)/(-68) reduces to (-18)/(-17). You had a sign change in the numerator. Probably just a typo, but you might want to check.

*July 10, 2010*

**math**

The slope is 2.

*July 10, 2010*

**Calculus/Physics**

In the equation s(t) = (s0) + (v0)t + (1/2)(a)(t^2) s(t) gives the height for some time t. s0 is the initial height, which is the height of the cliff. v0 is the initial velocity, for this problem 20m/s. The last term is the distance due to gravity. As t^2 increases, this term ...

*July 8, 2010*

**Calculus/Physics**

You have the right formula. Let S0 be the height of the cliff in meters. When the stone stops it is at 0m height. So, 0m = S0 + (v0)t + (1/2)gt^2 t =7 s g=-9.8m/(s^2) 0 = S0 + (20m/s)(7s) + (1/2)(-9.8m/(s^2))(49s^2) 0 = S0 + 140m - 240.1m S0 = 100.1m

*July 8, 2010*

**Algebra**

You have an error in your first step when you multiplied: -2(5 + 4a) It is not -10 + 8a .

*July 6, 2010*

**Math**

Do you see a numeric factor that is common in each term? Factor that number from each term. What is left?

*July 4, 2010*

**college**

v1 = velocity one second later vi = initial velocity of rock = 4.9 m/s t = time in seconds g = acceleration due to gravity = -9.8 m/s^2 v1 = vi + gt Plug the numbers in and solve for v1.

*July 1, 2010*

**Math**

Check how much each number differs from the previous one. Is there any pattern?

*June 26, 2010*

**algebra**

This is similar to your 7:58pm post. In that case we interpreted x^2y as x^(2y) rather than (x^2)y. In that case y was 1, and the value was the same. In this case, I don't see any answers if x is raised to the 2y power. If it is interpreted as 5(x^2)y for the numerator and...

*June 26, 2010*

**algebra**

putting a few parenthesis in for clarification... 2x^(2y)/(4xy) does = 1 for x=2 and y=1 So, you are correct!

*June 26, 2010*

**alg**

I think that the 4 is probably inside the "crook" of the radical sign to indicate the fourth root of (81/16).

*June 25, 2010*

**alg**

I got a, also.

*June 25, 2010*

**algebra**

If the problem is asking for the fourth root of 625, then c is correct. That would be the case if the 4 is in the crook of the radical sign.

*June 25, 2010*

**algebra**

a is NOT correct.

*June 25, 2010*

**algebra**

Glad to help.

*June 25, 2010*

**algebra**

Yes, I got b, also.

*June 25, 2010*

**algebra**

Your welcome.

*June 25, 2010*

**algebra**

Yes, it is correct.

*June 25, 2010*

**college physics**

Yes, thanks!

*June 24, 2010*

**college physics**

OK, the ramp makes a 40 degree angle with the horizontal. A sketch might help visualizing this. If you move 15m down the ramp what is the change in the y direction (the height)? It is 15m * sin(40degrees). So, the potential energy change is: =(80kg) * 9.8m/sec^2 * (- 15m * sin...

*June 24, 2010*

**college physics**

gravitational potential energy = mgh where: m = mass g = gravitational acceleration h = height So, the change in gravitational potential energy is: delta p = 80kg * 9.8m/sec^2 * (h1 - h2)m

*June 24, 2010*

**Calculus**

This same question was answered Wednesday for your 10:58pm post. Check back a few pages to see it.

*June 17, 2010*

**Calculus**

It probably helps to sketch the ships. The distance between the ships is a right triangle with one leg (the east/west leg) being a constant 50km and the other leg being the TOTAL distance traveled by both ships. Express the distance as a function of time then take the ...

*June 16, 2010*

**math**

secant of an angle = hypotenuse/adjacent cosine of an angle = adjacent/hypotenuse What happens when you multiply (hypotenuse/adjacent) x (adjacent/hypotenuse) ?

*January 23, 2010*

**Math115**

1. 5x - 15y = 2 2. 5x - 15y = 3 Subtract equation 2 from equation 1. or 0 = -1 So, you are correct, there is no solution.

*January 23, 2010*

**Physics**

1. Work is the dot product of force and distance. What force is neede to lift the box? 2. The force of gravity here is F= (mass of the ball) times (acceleration of gravity). The work is that force through 2.5 meters.

*January 23, 2010*

**Trig.**

Left off the trailing parenthesis... Area of sector of a circle with radius r for sector given in degrees (theta): A = (1/2)(r^2)(theta)((pi)/(180))

*January 23, 2010*

**Trig.**

Area of sector of a circle with radius r for sector given in degrees (theta): A = (1/2)(r^2)(theta)((pi)/(180)

*January 23, 2010*

**PSAT - Math**

Mike's time equals Paul's time. Let D = Paul's distance in miles Mikes time = (D+4miles)/(7mph) Paul's time = D/5mph (D+4miles)/(7mph) = D/(5mph) Ssolve for D, then find D + 4.

*December 26, 2009*

**Algebra 1**

Yes, you are right!

*December 21, 2009*

**MATH**

log2(8)=3 log9(3)=0.5 So, (3)^x = (0.5)^(x+1) ln of both sides (x)(ln(3) = (x+1)(ln(0.5) (ln(3) - ln(0.5))(x)= ln(0.5) calculate ln(3) and ln(0.5) solve for x

*December 13, 2009*

**algebra**

Let N be the number. Remember that 1 percent of a number is 0.01 times the number. 25 percent of a number is 0.25 times that number. So.. (0.25)(N) = 20 Divide both sides by 0.25 to find N.

*November 28, 2009*

**algebra**

X + y = 56 y = 3x Substituting that into the first equation... x + 3x = 56 combining... 4x = 56 Divide both sides by 4 to find the value for x. Once you have the value for x, use the second equation of y = 3x to find the value for y.

*November 16, 2009*

**algebra**

a. is correct b. is not correct You have the right idea. y = 3x. Now substitute 3x in for y in the equation and solve it.

*November 16, 2009*

**4th grade**

What happens when the sound waves hit the hard walls?

*November 16, 2009*

**Math**

let D = diameter Radius is half the diameter. So, r = (1/2) D Substitute that in the given equation and then you will have an equation using the diameter.

*November 9, 2009*

**Maths**

No. How many oranges can you get for 3 peaches? If oranges are half the price of peaches, then 3 peaches are worth 6 oranges. 3 bananas cost the same as 2 oranges, So, you can trade 2 oranges for 3 bananas. How many times can you trade 2 oranges for 3 bananas if you start with...

*November 4, 2009*

**5th grade-math**

Let d = the number of dimes. Let q = the number of quarters. What is known... d * $0.10 + q * $0.25 = $1.95 and... q = d - 2 Substitute the above equation for q into the first equation. That gives an equation only in d. Solve for d. Then plug that value into the second ...

*November 3, 2009*

**math**

In the explanation, the angle x and the x length are different items.

*November 1, 2009*

**math**

a) let x = the x length let y = the y length let r = the length from the origin to the x,y point. r^2 = x2 + y^2 sine of x = y/r so y and r are given in the problem. solve for x. cosine of x = x/r

*November 1, 2009*

**college**

For the pseudocode you need to know if variables passed to the module are passed by reference or passed by value. It is likely that they are passed by value. Look at all your display statements. Think what each will show. The "Display" function probably adds end of ...

*November 1, 2009*

**Math**

First calculate the volume of the carton. Then calculate the volume of the tank on the truck. Be careful. The dimensions of the tank on the truck are in feet while the dimensions of the cartons are in inches.

*October 25, 2009*

**math**

You're very welcome.

*October 25, 2009*

**math**

a_n = 21 + (n-1)*(-8) Expand that using the distributive law. The first term, 21, is not multiplied by -8. Only the terms in parenthesis are multiplied by -8.

*October 25, 2009*

**math**

You are close. You have an error: Your step: a_n = 21 + (n - 1) - 8 should be: a_n = 21 + (n - 1) * (-8)

*October 25, 2009*

**Math**

7.53 x 10^3 =7.53 x 1000 = ?

*October 24, 2009*

**physics**

One method to solve this is conservation of energy. initial kinetic energy = final potential energy let m = mass of stone let g = acceleration of gravity let v = initial velocity let h = maximum height (1/2)mv^2 = mgh So, v^2 = 2gh h = (v^2)/g

*September 25, 2009*

**Physics but algebra problem**

Go back and look at the two acceleration equations posted above. Don't forget to include gravity. for 89N: a = (89N - mg)/m m and g are known for 99N; a = (99 -mg)/m The acceleration at 99N should be twice the 89N a.

*September 23, 2009*

**Physics but algebra problem**

My mistake! stupied = stupid See, we all make errors.

*September 23, 2009*

**Physics but algebra problem**

No problem. That is why we are here. You are not being difficult or stupied at all. No one is born knowing this stuff. Some more hints: 2(89N - mg)/m = (99N -mg)/m Let's get rid of the denominator. Multiply both sides by m to get: 2(89N - mg) = (99N - mg) Multiplying out ...

*September 23, 2009*

**Physics but algebra problem**

Mia's post accidently went here. Ignore it.

*September 23, 2009*

**Physics but algebra problem**

Let a = the acceleration for 89N Let m = the bowling balls mass. Let g = acceleration of gravity. So, 2a is the acceleration for 99N. Set up the equation that shows the 99N acceleration is 2 times the 89N acceleration: 2(89N - mg)/m = (99N - mg)/m Solve for m.

*September 23, 2009*

**Physics but algebra problem**

Remember that the acceleration of the bowling ball is the NET force of your lifting and gravity. So, the acceleration of your 89N lifting is: a = (89N - mg)/m where g is gravity. Likewise for the 99N lift: a = (99N - mg)/m

*September 23, 2009*

**Physics but algebra problem**

It doesn't look like there is a solution. First equation: 89N = ma Second equation: 99N = m2a For the second equation, if you divide both sides by 2 you get: 99N/2 = m2a/2 49.5N = ma The first equation says ma = 89N. The second equation says ma = 49.5N Assuming a is ...

*September 23, 2009*

**math**

Expand all the terms by multiplying them out: 8(2x-3) = -16 16x - 24 = -16 Put all the variable terms on one side and the "just numbers" terms on the other side. 16x = -16 + 24 Now, solve for x.

*September 23, 2009*

**Geometry**

No, the problem requests to the nearest tenth which would be: 13.2 ft. Also, don't forget the units.

*September 10, 2009*

**Geometry**

Your welcome! Glad to help.

*September 10, 2009*

**Geometry**

This is a little tricky. Let R = the radius Let Pi = 3.14 as stated. The circumference is (Pi)(2)(R) The problem already calculates (2)(R) for you as 4.2 ft. So, circumference = (Pi)(4.2) or =(3.14)(4.2) ft. Multiply and round as requested by the problem.

*September 10, 2009*

**math 115**

No. |-2| = 2 So, |10| + |-2| =|10| + 2 = ?

*September 6, 2009*

**algebra**

Let a side have ratio value 5. Let b side have ratio value 12. Let c side have ratio value 13. add the ratios of the sides: 5 + 12 + 13 = T It is given that the perimeter (all sides added) is 15 inches. The length of each side has the same relationship to its ratio as the ...

*August 18, 2009*

**geometry**

Answered below for the jj post.

*August 18, 2009*

**math**

For a proper fraction, the absolute value of the numerator is less than the absolute value of the denominator. So, 1/3 is a proper fraction 2/3 is a proper fraction Is (2/3)/(1/3) a proper fraction?

*August 18, 2009*

**pre- algebra**

Check your answer... If x = 5 3(5) = (96-75) 15 = 21 So, 5 is not a correct answer.

*July 19, 2009*

**math am i correct**

Yes, you are correct.

*July 19, 2009*

**trig**

assuming that 2cos^2x means 2(cos(x))^2 substitute using (cos(x))^2 = 1 - (sin(x))^2 2(1 - (sin(x))^2) + 3sin(x) = 3 2 - 2(sin(x))^2 + 3sin(x) = 3 for clarity, let u = sin(x) 2 - 2u^2 + 3u = 3 rearranging... 2u^2 - 3 u + 1 = 0 Solve this quadratic to find the values of sin x ...

*July 18, 2009*

**math**

x + 2/x + 3 = x - 1/x + 1 It's usually helpful to get rid of denominators, so multiply both sides by x x^2 + 2 + 3x = x^2 - 1 + x subtract x^2 from both sides 2 + 3x = -1 + x collect like terms 2x = -3 So, x = ?

*July 3, 2009*

**Math**

No, -|-8| - |-11-8| = -8 - |-19| = -8 - 19 = -27

*July 3, 2009*

**math**

The distance is the hypotenuse of the right triangle with a vertical leg distance of 4.6 - 0.6 = 4.0 and a horizontal leg distance of (-4.5) - (-6.2) = 1.7 The hypotenuse = ( (4.0)^2 + (1.7)^2 )^0.5 = (16.0 + 2.89)^0.5 = 4.35

*June 23, 2009*

**Algebra**

For a: root(-21) = root( (21)*(-1) ) = root(21) * root(-1) = root(21) * i That should get you started

*June 23, 2009*