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mash [69]
3 years ago
12

a. 0.98 – 0.053 b. 0.67 – 0.4 c. 0.3 – 0.002 d. 3.2 – .789 e. 6.53 – 4.298 f. 6 – 4.32 g. 7 – 3.574 h. 4.83 – 1.8 i. 3.7 – 1.8 j

. 16.17 – 11.632
Mathematics
1 answer:
Pavlova-9 [17]3 years ago
3 0

a. 0.927

We have:

0.98 – 0.053

We can re-write it as:

0. 9 8 0 -

0. 0 5 3

Moving digits to the right:

0. 9 7 10 -

0. 0 5 3

Digit-per-digit subtraction:

0. 9 2 7


b. 0.27

We have:

0.67 – 0.4

We can re-write it as:

0. 6 7 -

0. 4 0

Digit-per-digit subtraction:

0. 2 7


c. 0.298

We have:

0.3 – 0.002

We can re-write it as:

0. 3 0 0 -

0. 0 0 2

Moving digits to the right:

0. 2 10 0 -

0. 0 0 2

Again:

0. 2 9 10 -

0. 0 0 2

Digit-per-digit subtraction:

0. 2 9 8


d. 2.411

We have:

3.2 – .789

We can re-write it as:

3. 2 0 0 -

0. 7 8 9

We need to rewrite the first term by moving digits to the right several times:

3. 2 0 0 = 2. 12 0 0 = 2. 11 10 0 = 2. 11 9 10

So now we have:

2. 11 9 10 -

0. 7 8 9

Digit-per-digit subtraction:

2. 4 1 1


e. 2.232

We have:

6.53 – 4.298

We can re-write it as:

6. 5 3 0 -

4. 2 9 8

We need to rewrite the first term by moving digits to the right several times:

6. 5 3 0 = 6. 5 2 10 = 6. 4 12 10

So now we have:

6. 4 12 10 -

4. 2 9 8

Digit-per-digit subtraction:

2. 2 3 2


f. 1.68

We have:

6 – 4.32

We can re-write it as:

6. 0 0 -

4. 3 2

We need to rewrite the first term by moving digits to the right several times:

6. 0 0 = 5. 10 0 = 5. 9 10

So now we have:

5. 9 10 -

4. 3 2

Digit-per-digit subtraction:

1. 6 8


g. 4.426

We have:

7 – 3.574

We can re-write it as:

7. 0 0 0 -

3. 5 7 4

We need to rewrite the first term by moving digits to the right several times:

7. 0 0 0 = 6. 10 0 0 = 6. 9 10 0 = 6. 9 9 10

So now we have:

6. 9 9 10 -

3. 5 7 4 =

Digit-per-digit subtraction:

3. 4 2 6


h. 3.03

We have:

4.83 – 1.8

We can re-write it as:

4. 8 3 -

1. 8 0

We can immediately do the digit-per-digit subtraction:

3. 0 3


i. 2.9

We have:

3.7 – 1.8

We can re-write it as:

3. 7 -

1. 8

We need to rewrite the first term by moving digits to the right:

3. 7 = 2. 17

So now we have:

2. 17 -

1. 8 =

Digit-per-digit subtraction:

2. 9


j. 4.538

We have:

16.17 – 11.632

We can re-write it as:

1 6 . 1 7 0 -

1 1 . 6 3 2

We need to rewrite the first term by moving digits to the right:

1 6. 1 7 0 = 1 6. 1 6 10 = 1 5. 11 6 10  

So now we have:

1 5. 11 6 10 -

1 1. 6 3 2 =

Digit-per-digit subtraction:

0 4. 5 3 8

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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.

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