Solve each equation:

1 answer:

6 0

You sure seem to be asking a lot of questions lately. I'd like to see that you've been trying with these problems at least because if you can't get that first one it's almost like you missed the whole lesson.

1. 20 = 4b + 7 + 5
Add the 7 and 5.
20 = 4b + 12
Subtract 12 from each side.
8 = 4b
Divide each side by 4.
2 = b

2. 7 = 6k - 7k
6k - 7k = -1k. (the k acts as a sort of unit)
7 = -1k
Divide each side by -1.
-7 = k

3. 3.23 - 2m = 3 - 2(5m - 2)
Distribute the ×-2 to each term inside the parentheses.
3.23 - 2m = 3 - 10m + 4
Add the 3 and 4.
3.23 - 2m = 7 - 10m
Add 10m to each side.
3.23 + 8m = 7
Subtract 3.23 from each side.
8m = 3.77
Divide by 8.
m = 0.47125

4. -88/45=1/3r+2/5r
To get rid of the fractions, let's multiply everything by 45.
-88 = 15 + 18r
Subtract 15 from each side.
-103 = 18r
Divide by 18.
-103/18 = r
As a mixed number, r = -5 and 13/18
As a decimal, r = -5.7222...

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Please help with this problem

Andre45 [30]

Answer:

The length of the short side is 14.5 units, the length of the other short side is 18.5 units, and the length of the longest side is 23.5 units.

Step-by-step explanation:

The Pythagorean Theorem

If a and b are the lengths of the legs of a right triangle and c is the length of the hypotenuse, then the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.

This relationship is represented by the formula:

$a^2+b^2=c^2$

Applying the Pythagorean Theorem to find the lengths of the three sides we get:

$(x)^2+(x+4)^2=(x+9)^2\\\\2x^2+8x+16=x^2+18x+81\\\\2x^2+8x-65=x^2+18x\\\\2x^2-10x-65=x^2\\\\x^2-10x-65=0$

Solve with the quadratic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=-10,\:c=-65:\quad x_{1,\:2}=\frac{-\left(-10\right)\pm \sqrt{\left(-10\right)^2-4\cdot \:1\left(-65\right)}}{2\cdot \:1}\\\\x_{1}=\frac{-\left(-10\right)+ \sqrt{\left(-10\right)^2-4\cdot \:1\left(-65\right)}}{2\cdot \:1}=5+3\sqrt{10}\\\\x_{2}=\frac{-\left(-10\right)- \sqrt{\left(-10\right)^2-4\cdot \:1\left(-65\right)}}{2\cdot \:1}=5-3\sqrt{10}$