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3 years ago

Suppose AP = 2 cm and BQ = 7 cm. What is the length of PC?

Mathematics

1 answer:

viktelen [127]3 years ago

3 0

Step-by-step explanation:

is this a quadrilateral? or polygon or triangle?

what are the angles (I believe there should be at least an angle given) else PC can be any value that is positive

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NEED HELP PLEASE BEGGING PLEASE!!!!

Sliva [168]

Answer:

-54

Step-by-step explanation:

$8(-9) - 6(-3)\\$

Multiply 8 by -9 and 6 by -3

$-72 - (-18)$

Subtract $Value = -54$

Hope this helps

plz mark brainliest

6 0

3 years ago

Consider multiplying 26.2 by 16.43 what would a mathematician say the answer is

IgorC [24]

In mathematics multiplying decimals is just like multiplying a whole number and ignoring the decimals., then just put the decimals in the answer – it is how many decimals does the two numbers combined.

First step:

Line the numbers on the right and do not a line the decimal points

26.2

× 16.43

Second Step:

Multiply the numbers starting on the right, just like multiplying a whole number and ignore the decimals.

26.2

× 16.43

786

1048

1572

+ 262

430466

Third step: add the products

26.2

× 16.43

786

1048

1572

+ 262

430.466 is the mathematician’s answer.

8 0

3 years ago

A ship leaves port at noon and has a bearing of S29oW. The ship sails at 20 knots. How many nautical miles south and how many na

ira [324]

Answer:

Approximately $58.2\; \text{nautical miles}$ (assuming that the bearing is ${\rm S$29^{\circ}$W}$ .)

Step-by-step explanation:

Let $v$ denote the speed of the ship, and let $t$ denote the duration of the trip. The magnitude of the displacement of this ship would be $v\, t$ .

Refer to the diagram attached. The direction ${\rm S$29^{\circ}$W}$ means $29^{\circ}$ west of south. Thus, start with the south direction and turn towards west (clockwise) by $29^{\circ}$ to find the direction of the displacement of the ship.

The hypothenuse of the right triangle in this diagram represents the displacement of the ship, with a length of $v\, t$ . The dashed horizontal line segment represents the distance that the ship has travelled to the west (which this question is asking for.) The angle opposite to that line segment is exactly $29^{\circ}$ .

Since the hypotenuse is of length $v\, t$ , the dashed line segment opposite to the $\theta = 29^{\circ}$ vertex would have a length of:

$\begin{aligned}& \text{opposite (to $\theta$)} \\ =\; & \text{hypotenuse} \times \frac{\text{opposite (to $\theta$)}}{\text{hypotenuse}} \\ =\; & \text{hypotenuse} \times \sin (\theta) \\ =\; & v\, t \, \sin(\theta) \\ =\; & v\, t\, \sin(29^{\circ})\end{aligned}$ .

Substitute in $\begin{aligned} v &= 20\; \frac{\text{nautical mile}}{\text{hour}}\end{aligned}$ and $t = 6\; \text{hour}$ :

$\begin{aligned} & v\, t\, \sin(29^{\circ}) \\ =\; & 20\; \frac{\text{nautical mile}}{\text{hour}} \times 6\; \text{hour} \times \sin(29^{\circ}) \\ \approx\; & 58.2\; \text{nautical mile}\end{aligned}$ .