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1 year ago

Yo i need sum help question in picutre

Mathematics

1 answer:

balandron [24]1 year ago

3 0

Answer:

$2875

Step-by-step explanation:

$50,000 > $17001, so the progressive tax rate will be 5.75%

5.75% of 50,000 = 0.0575 * 50000 = 2875

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In a recent snail race, a snail traveled a distance of 1 1/6 inches in 2 minutes. Find the number if inches traveled per minute

Alecsey [184]

Answer:

For inch/min =0.9167 inch/min

For min/inch = 1.0908 min/inch

Step-by-step explanation:

The snail in the race traveled 11/6 inches in 2 minutes.

The number of inches per minute is equal to = (11/6)/(2)

= 11/12

=0.9167 inch/min

The number of minutes per inch is equal to = 1/(inch per min)

= 1/0.9167

= 1.0908 min/inch

Minutes per inch is gotten by taking the inverse of inch/min or dividing the number of minutes by number of inches

6 0

3 years ago

What is the probability that a point chosen at random in the rectangle is also in the blue triangle?

V125BC [204]

Answer:

1/2

Step-by-step explanation:

I did the quiz on Edge 2021 and got it right with 100%.

Please give me brainliest!

3 0

3 years ago

How would you show a rate of change of 3.5 meters per minute on a graph name some of the points

olga2289 [7]

Y=3.5x
Pts are (1,3.5),(2,7),(3,10.5)

8 0

3 years ago

Pls help if pls do most important 15 number 20 number 26 number 28 number 30 number 17 number 25 number 22 number 24 number if u

Tamiku [17]

Answer:

Step-by-step explanation:

15. $\frac{cotx+1}{cotx-1} = \frac{cotx+cotx.tanx}{cotx-cotx.tanx} = \frac{cotx(1+tanx)}{cotx(1-tanx)} = \frac{1+tanx}{1-tanx} \\$

16. $\frac{1+cos\alpha }{sin\alpha } = \frac{sin\alpha }{1-cos\alpha }$

$=> (1+cos\alpha )(1-cos\alpha ) = sin^{2} \alpha \\=> 1-cos^{2}\alpha =sin^{2} \alpha \\=> sin^{2}\alpha +cos^{2}\alpha =1\\$

=> The clause is correct

17. Do the same (16)

18. $\frac{sinx}{1-cosx}+\frac{sinx}{1+cosx} = \frac{sinx(1+cosx) + sinx(1-cosx)}{(1+cosx)(1-cosx)} = \frac{sinx + sinx.cosx + sinx - sinx.cosx}{1-cos^{2}x} = \frac{2sinx}{sin^{2}x }= \frac{2}{sinx} = 2cosecx$

19.

$\frac{sinA}{1+cosA} + \frac{1+cosA}{sinA}=\frac{2}{sinA} \\=> \frac{sinA}{1+cosA} + \frac{1+cosA}{sinA} - \frac{2}{sinA} \\ = 0\\=> \frac{sinA}{1+cosA} + (\frac{1+cosA}{sinA} - \frac{2}{sinA} \\) = 0\\=> \frac{sinA}{1+cosA} + \frac{1+cosA-2}{sinA} = 0\\=> \frac{sinA}{1+cosA} +\frac{cosA-1}{sinA} = 0\\=> \frac{sin^{2} A+(cosA-1)(1+cosA)}{(1+cosA)sinA}=0\\=> \frac{sin^{2} A+cos^{2} A-1}{(1+cosA)sinA}=0\\=> \frac{0}{(1+cosA)sinA} =0\\$