(5+3i)(3+5i) PLEASEEE HELP

Mathematics

1 answer:

Alex17521 [72]2 years ago

7 0

Answer:

34i....................

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The function shown below was created to track the different intervals of speed that an automobile travels over a period of 28 se

Ksenya-84 [330]

Answer:

1. The car is slowing down at a rate of 2.5mph/s

2. The greatest acceleration is 10 mph/s.

3. In the interval 4s to 16s the speed remains constant and has magnitude 25 mph.

Step-by-step explanation:

1. The deceleration of the car is from 16 seconds to 24 seconds is the slope $m$ of the graph from 16 to 24:

$m=\dfrac{\Delta speed }{\Delta time } = \dfrac{5-25}{24-16} =-2.5mph/s$

the negative sign indicates that it is deceleration.

2. The automobile experiences the greatest change in speed when the slope is greatest because that is when acceleration/deceleration is greatest.

From the graph we see that the greatest slope of the graph is between 28 and 24 seconds. The acceleration the interval is the slope $m$ :

$m= \dfrac{45-5}{28-24}= 10mph/s$

3. The automobile experiences no acceleration in the interval 4 s to 16 s—that's the graph is flat.

The speed of the automobile in that interval, as we see from the graph, is 25 mph.

7 0

2 years ago

Read 2 more answers

(1) a) 5270 cl= ? dal

Ivanshal [37]

Answer:

Check below

Step-by-step explanation:

1) These metric volume units can be easily converted by dividing or multiplying by 10 and its multiple. Like this: each step up on the ladder multiply by 10. Each step down divide by 10

$a)5270 \:cl=527\:dl\\b)kl=5.3 \times 10=530 hl\\c)58000ml=5.8 dal =(\frac{58000}{10000})\\d)0.39 dal=3.9 l\\e)7.8 dal=780 dl\\f)57.2 ml=5.72 cl$ .

2) When it comes to area, the "ladder scheme" remains valid but now we'll multiply or divide by

$10^3$

Bear in mind these useful relations:

$1litre =\frac{1}{10^3}m^{3}$

$1 cm^{3} =1 ml$

$a) 3.7 l=3700 cm^{3}\\b)29.6 dm^{3}\\29.6 dm^{3} \rightarrow 29.6\times10^{3} cm^3=29.6\times10^{3} ml = 29.6\times10^{2} cl \:or\:2960 cl\\c)8.1l=8100 cm^{3}\\d)5l=\frac{5}{10^{-3}} m^{3}\\e)5.2 cl=52ml=52 cm^{3}=52\times10^{3}=52000 \:mm^{3}\\f)375 ml=375 cm^{3}\\g) 1300 cm^{3}=1300 ml=13 dl\\h) 8.6 ml= 8.6 cm^{3}\times\: 10^{3}=8600 mm^{3}\\i) 51.8\times 10^{11} l \rightarrow 51.8\times 10^{-11}km^{3}$