Thursday

September 29, 2016
Total # Posts: 7

**Algebra**

Not sure on a few of my answers, can you help? 1. Find the slope and y intercept : f(x)=-5x-7 slope = -5 intercept = (0,-7). Is this right? -1/2x=-5/6 my answer was 5/3 was this right? If I have these two sets of coordinate (-4,0) and (0,2) will my slope be -1/2? Same question...

*January 8, 2010*

**amendment to #2**

5x+4y=2, 4x-5y=4....I said neither on whether the line was parallel or perpendicular or neither.

*January 8, 2010*

**MATH/ALGEBRA**

I am going over some math homework and need to know if I am getting the right answers. 1. Solving by elimination: 2x+3y=3, and 4x+6y=6 I say the determinant is o, there is no solution 2. y-11>2y-2 Is 1,-15,-17,-2 solutions: My answer = 1 is not, -15 is, -17 is, and -2 is ...

*December 19, 2009*

**Please take a look**

Here is a question, but I am having a problem finding the right systems of equation. Are the ones I have written correct? Question: The mortgage department of the company is selling two model homes that are located on the same block. The square footage, as well as the type of ...

*December 17, 2009*

**Algebra**

Here is a question, but I am having a problem finding the right systems of equation. Are the ones I have written correct? Question: The mortgage department of the company is selling two model homes that are located on the same block. The square footage, as well as the type of ...

*December 17, 2009*

**Math-stuck**

I know that to find the height of a woman that has a tibia bone measuring 43.4 would be computed like this H(t)=0.984(t)+28.6, and I would arrive at this :69.5696, but I how would I prepare a function to figure out a womeans tibia in inches when she is 5 foot, 5 inches. I am ...

*December 3, 2009*

**MATH/Algebra**

I know that to find the height of a woman that has a tibia bone measuring 43.4 would be computed like this H(t)=0.984(t)+28.6, and I would arrive at this :69.5696, but I how would I prepare a function to figure out a womeans tibia in inches when she is 5 foot, 5 inches. I am ...

*December 3, 2009*

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