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3 years ago

(X+2)(x+4)what’s the product using FOIL method?

Mathematics

2 answers:

rodikova [14]3 years ago

7 0

$(x*x)+(x*4)+(2*x)+(2*4)$

$x^{2} +4x+2x+8\\x^2+6x+8$

the answer is therefore: $x^2+6x+8$

Lesechka [4]3 years ago

5 0

Answer: x^2+6x+8

Step-by-step explanation: Using FOIL, you multiply the First, Outer, Inner, and Last parts of the problem. :)

x^2+4x+2x+8

x^2+6x+8

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3 years ago

A post is supported by two wires (one on each side going in oppositedirections) creating an angle of 80° between the wires. The

Vladimir [108]

Using the Sine rule,

$\frac{\sin A}{A}=\frac{\sin B}{B}=\frac{\sin C}{C}$ $\begin{gathered} \text{Let A = 14m,} \\ Substituting the variables into the formula,Where the length of the wires are, AP = xm and BP = ym[tex]\begin{gathered} \frac{\sin80^0}{14}=\frac{\sin40^0}{x} \\ \text{Crossmultiply,} \\ x\times\sin 80^0=14\times\sin 40^0 \\ Divide\text{ both sides by }\sin 80^0 \\ x=\frac{14\sin40^0}{\sin80^0} \\ x=9.14m \end{gathered}$

Hence, the length of wire AP (x) is 9.14m.

For wire BP (y)m,

Sum of angles in a triangle is 180 degrees,

$A^0+P^0+B^0=180^0$ $\begin{gathered} \text{Where A}^0=\text{ unknown,} \\ P^0=80^0\text{ and,} \\ B^0=40^0 \\ A^0+80^0+40^0=180^0 \\ A^0+120^0=180^0 \\ A^0=180^0-120^0 \\ A^0=60^0 \end{gathered}$

Using the side rule to find the length of wire BP,

$\begin{gathered} \frac{\sin 60^0}{y}=\frac{\sin 80^0}{14} \\ \text{Crossmultiply,} \\ y\times\sin 80^0=14\times\sin 60^0 \\ \text{Didive both sides by }\sin 80^0 \\ y=\frac{14\times\sin 60^0}{\sin 80^0} \\ y=12.31m \end{gathered}$

Hence, the length of wire BP (y) is 12.31m

Therefore, the length of the wires are (9.14m and 12.31m).

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1 year ago

<img src="https://tex.z-dn.net/?f=%20%7B%28x%20-%20y%29%7D%5E%7B2%7D%20" id="TexFormula1" title=" {(x - y)}^{2} " alt=" {(x - y)

Lubov Fominskaja [6]

Step-by-step explanation:

$= {(x - y)}^{2} \\ = {x}^{2} - 2xy + {y}^{2}$

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e-lub [12.9K]

Yes 25 to 16 is proportional to 5 to 4.
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It results in the same proportion which is 5 to 4.

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