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3 years ago

What is the coefficient of the term 13a, in the expression 13a+ 6b, a, plus, 6b?

Mathematics

1 answer:

tamaranim1 [39]3 years ago

4 0

The coefficient of the term 13a is 13

The coefficient of 6b in expression is a + 6b is 6

Solution:

Given expression is:

13a + 6b

We have to find the coefficient of 13a

The coefficients are the numbers that multiply the variables or letters

Here in 13a , a is the variable that is multiplied with number 13

Thus the coefficient of the term 13a is 13

Second expression:

a + 6b

Here, in 6b , b is the variable that is multiplied with number 6

so 6 is the coefficient

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3 years ago

Read 2 more answers

Two lighthouses are located 75 miles from one another on a north-south line. If a boat is spotted S 40o E from the northern ligh

yuradex [85]

Answer:

The northern lighthouse is approximately $24.4\; \rm mi$ closer to the boat than the southern lighthouse.

Step-by-step explanation:

Refer to the diagram attached. Denote the northern lighthouse as $\rm N$ , the southern lighthouse as $\rm S$ , and the boat as $\rm B$ . These three points would form a triangle.

It is given that two of the angles of this triangle measure $40^{\circ}$ (northern lighthouse, $\angle {\rm N}$ ) and $21^{\circ}$ (southern lighthouse $\angle {\rm S}$ ), respectively. The three angles of any triangle add up to $180^{\circ}$ . Therefore, the third angle of this triangle would measure $180^{\circ} - (40^{\circ} + 21^{\circ}) = 119^{\circ}$ (boat $\angle {\rm B}$ .)

It is also given that the length between the two lighthouses (length of $\rm NS$ ) is $75\; \rm mi$ .

By the law of sine, the length of a side in a given triangle would be proportional to the angle opposite to that side. For example, in the triangle in this question, $\angle {\rm B}$ is opposite to side $\rm NS$ , whereas $\angle {\rm S}$ is opposite to side ${\rm NB}$ . Therefore:

$\begin{aligned} \frac{\text{length of NS}}{\sin(\angle {\rm B})} = \frac{\text{length of NB}}{\sin(\angle {\rm S})} \end{aligned}$ .

Substitute in the known measurements:

$\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of NB}}{\sin(21^{\circ})} \end{aligned}$ .

Rearrange and solve for the length of $\rm NB$ :

$\begin{aligned} & \text{length of NB} \\ =\; & (75\; \rm mi) \times \frac{\sin(21^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 30.73\; \rm mi\end{aligned}$ .

(Round to at least one more decimal places than the values in the choices.)

Likewise, with $\angle {\rm N}$ is opposite to side ${\rm SB}$ , the following would also hold:

$\begin{aligned} \frac{\text{length of NS}}{\sin(\angle {\rm B})} = \frac{\text{length of SB}}{\sin(\angle {\rm N})} \end{aligned}$ .

$\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of SB}}{\sin(40^{\circ})} \end{aligned}$ .

$\begin{aligned} & \text{length of SB} \\ =\; & (75\; \rm mi) \times \frac{\sin(40^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 55.12\; \rm mi\end{aligned}$ .

In other words, the distance between the northern lighthouse and the boat is approximately $30.73\; \rm mi$ , whereas the distance between the southern lighthouse and the boat is approximately $55.12\; \rm mi$ . Hence the conclusion.