Can anyone help me with an Algebra 1 question?

I am having problems with these questions. Thanks if you can help!(:DIRECTIONS: SOLVE THE FORMULA FOR THE INDICATED VARIABLE.1) Volume of a rectangular prism: V=lwh. Solve for w2) Surface area of a prism: S=2B+Ph.

Solve for h3)Length of movie projected at 24 frames persecond: l=24f. Solve for f(These are all seperate questions) Asked by emy12851 38 months ago Similar questions: help Algebra question Science > Math.

You can divide both sides of an equality relationship by a number or variable, as long as that number is never zero. All of these questions come down to the same thing. When you have an equality relationship like A = B then you can divide both sides of the relationship by some number or variable without problem, IF you can satisfy one requirement.

The number or variable that you divide by must never be zero. Let me look at the third problem first. If you know the total length of a movie in seconds, projected at 24 frames per second, and you know that L = 24f then if you solve for f, this tells you the total number of frames.

You can solve for f by dividing both sides of this equality relationship by 24. Does this operation satisfy the requirement I mentioned above? 24 is not zero, so there is no problem.

L/24 = 24*f/24 But, we know that 24*f/24 can be simplified to just f. The 24 in both the numerator and the denominator will "cancel" each other out, since 24/24 = 1. And when you multiply any number or value by 1, you get the number you started with.So the total number of frames in the movie, f, is given by L/24.

Now let us return to problem 1. Given that the Volume of a prism is V = Lwh (I’ve used an upper case L again here, because l is too easy to confuse with the number 1.) We must solve for w. You can use the same approach we took to solve problem 3.

You divide both sides of an equality relationship by a variable or number, as long as that number is never zero. V/h = Lwh/h Again, as long as the height of the prism is never exactly zero, this is a valid operation. Since h/h will "cancel" to leave just the product Lw, we now have V/h = Lw Do this one more time, since we wish to solve for w on its own.

Divide by L. Again, L must never be zero, so we cannot solve for w on prisms with zero lengths or zero heights. V/(Lh) = Lw/L Here again, we see that Lw/L will simplify to just w, since L/L is just 1 as long as L is not zero.

So we are left with the desired result V/(Lh) = w The second problem is no harder to solve, if we apply simple rules. We are given the surface area of a prism as S = 2B + Ph Here B is the area of the base of the prism, h is the height, and P is the length of the perimeter of the base. We are asked to solve for h, the height of the prism.

We first subtract the 2B term from both sides of the equality. We do this to get the Ph term isolated. S - 2B = 2B + Ph - 2B Since addition is commutative, we can swap terms on the right hand side, to yield S - 2B = pH + 2B - 2B What do we have if we subtract any value from itself?

Yes.Zero. The terms 2B - 2B simplify to zero. (Don’t ask me if 2B or not 2B.

Thankfully, that is not the question.) So we now have S - 2B = Ph + 0 But we know that this must reduce to just Ph on the right hand side, since adding zero to anything will not change its value. S - 2B = Ph Finally, we use the same rule we learned above one more time. If we wish to solve for h, divide both sides of the equality by P.

This is acceptable as long as P is never zero. Can the perimeter of the base of our prism ever be zero? Not so for any non-degenerate prism, so we are on firm ground.(S - 2B)/P = Ph/P The final step is to recognize that Ph/P simplifies to just h.(S - 2B)/P = h Remember to reduce your algebra problems to a series of these simple steps, keeping an eye on your goal.

And, never divide by zero..

(S - 2B)/P = h Remember to reduce your algebra problems to a series of these simple steps, keeping an eye on your goal. And, never divide by zero.

Remember to reduce your algebra problems to a series of these simple steps, keeping an eye on your goal. And, never divide by zero.

1 To solve for a variable in algebra means to rewrite the equation you are given so that the variable required is by itself on one side of the equation with all of the other variables and numbers (coefficients) on the other side of the equation. For example the formula for the perimeter of a rectangle is given by adding together the lengths of the four sides (two pf which are L for length and the other two of which are W for width)P = 2L + 2WIf you are asked to solve the equation for L you need to end up with the letter L all by itself on one side. Since the L shows up on one side only let us start there.

On the side with the 2L is also a 2W. You don't want the Ws there so look to see how they are "attached" to the 2L. Since they are added, you need to UNadd them (subtract them).

To keep the equation true or "equivalent" you must subtract the 2W from BOTH sidesP - 2w = 2L + 2W -2WP - 2W = 2L Now the letter L is on one side of the equation but there is still a 2 there as well. YOu want the L by itself so since the 2 is multipled (the meaning of 2L) you need to undo the mutiplication and again must divide BOTH sides by the 2 (And divide EVERYTHING by 2)(P-2W) / 2 = 2L/2orP/2 - 2w/2 = Lfinally P/2 - W = LDid you follow that? Try your first problem now.

What do you want to detach from the w and how is it attached? .

To solve for a variable in algebra means to rewrite the equation you are given so that the variable required is by itself on one side of the equation with all of the other variables and numbers (coefficients) on the other side of the equation. For example the formula for the perimeter of a rectangle is given by adding together the lengths of the four sides (two pf which are L for length and the other two of which are W for width)P = 2L + 2WIf you are asked to solve the equation for L you need to end up with the letter L all by itself on one side. Since the L shows up on one side only let us start there.

On the side with the 2L is also a 2W. You don't want the Ws there so look to see how they are "attached" to the 2L. Since they are added, you need to UNadd them (subtract them).

To keep the equation true or "equivalent" you must subtract the 2W from BOTH sidesP - 2w = 2L + 2W -2WP - 2W = 2L Now the letter L is on one side of the equation but there is still a 2 there as well. YOu want the L by itself so since the 2 is multipled (the meaning of 2L) you need to undo the mutiplication and again must divide BOTH sides by the 2 (And divide EVERYTHING by 2)(P-2W) / 2 = 2L/2orP/2 - 2w/2 = Lfinally P/2 - W = LDid you follow that? Try your first problem now.

What do you want to detach from the w and how is it attached?

Emy12851 replied to post #1: 2 Okay so for #1. Am I doing this right? V=lwhSo I need w on one side and l,h,and be on the otherSo it's w*l so I am dividing by lwhich would be V/1=whthen I need to move the hso divide by h.

Which would beV/1/h=w? Is this right? .

Okay so for #1. Am I doing this right? V=lwhSo I need w on one side and l,h,and be on the otherSo it's w*l so I am dividing by lwhich would be V/1=whthen I need to move the hso divide by h.

Which would beV/1/h=w? Is this right?

Unixcorn replied to post #2: 3 You can see it clearer if you do a little more work at the beginning...V = lwh You want the w by itself so look at the equation again but like thisV = (lh)w In other words, you want that w but you don't want the other letters BOTH of which multiply the w (I rearranged the factors which is legal becauase the commutative property for multliplication says the order is not important if you have ONLY multiplication going on)Since the (lh) multiplies the w , divide by both the lh at one time rather than one after another. What you did was LEGAL (you made no mistake) BUT you left the answer in what is called a complex fraction (more than one division sign) and it is harder for me to type how to correct that. Let me try though.... DIVIDING by 3 (for example) is the same and MULIPLYING by 1/3 M = 3P solve for P We usually say let us divide by 3 but let me this time say let me mutiply by 1/3 on both sides(1/3) M = (1/3) 3/ P(1/3)M = PM/3 = PWhy say multiply by reciprocal instead of divide?

Just for the kind of problem you tried to do. Take your step V/l = wh (Now you want to remove that h which is multiplied on so let us MULTIPLY by the RECIPROCAL 1/h(1/h) (V/l) = (1/h)wh Write those fraction on the left so they LOOK like fractions to youYou end up withV/(hl) = (wh)/(h) (the 1 times V is V)V/(hl) = w (the h divided into h is one) .

You can see it clearer if you do a little more work at the beginning...V = lwh You want the w by itself so look at the equation again but like thisV = (lh)w In other words, you want that w but you don't want the other letters BOTH of which multiply the w (I rearranged the factors which is legal becauase the commutative property for multliplication says the order is not important if you have ONLY multiplication going on)Since the (lh) multiplies the w , divide by both the lh at one time rather than one after another. What you did was LEGAL (you made no mistake) BUT you left the answer in what is called a complex fraction (more than one division sign) and it is harder for me to type how to correct that. Let me try though.... DIVIDING by 3 (for example) is the same and MULIPLYING by 1/3 M = 3P solve for P We usually say let us divide by 3 but let me this time say let me mutiply by 1/3 on both sides(1/3) M = (1/3) 3/ P(1/3)M = PM/3 = PWhy say multiply by reciprocal instead of divide?

Just for the kind of problem you tried to do. Take your step V/l = wh (Now you want to remove that h which is multiplied on so let us MULTIPLY by the RECIPROCAL 1/h(1/h) (V/l) = (1/h)wh Write those fraction on the left so they LOOK like fractions to youYou end up withV/(hl) = (wh)/(h) (the 1 times V is V)V/(hl) = w (the h divided into h is one).

DIVIDING by 3 (for example) is the same AS multiplying by 1/3When you write fractions so that the numerator is clearly on top of the fraction and the denominator is clearly on the botton, you don't need to parentheses. When you type or write things in a long line as we are doing here, you need the parentheses to show very clearly which terms and factors are where.

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