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3 years ago

7, 10, 13, 16, find the 43rd term A)75 B)133 C)121 D)101

Mathematics

2 answers:

uranmaximum [27]3 years ago

7 0

I think it’s 121 i could be wrong tho://

Eddi Din [679]3 years ago

5 0

Answer:

Step-by-step explanation:

a,n = 3n+4

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3 years ago

Chris made $136 for 8 hours of work.

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2 years ago

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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