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kicyunya [14]
2 years ago
14

Jabari drives 20.4 miles in 17 minutes. He passes a sign which gives the speed limit as 55 mph. By how much, in mph, did Jabari'

s average speed exceed the speed limit?
Mathematics
1 answer:
MAXImum [283]2 years ago
8 0

Jabari's average speed exceed the speed limit by 17 miles per hour

Given:

Total distance = 20.4 miles

Time taken = 17 minutes

convert minutes to hour

60 minutes = 1 hour

17 minutes = 0.283333 hour

Average speed = Total distance ÷ Time taken

= 20.4 ÷ 0.283333

= 72.00008470598200

Approximately, 72 miles per hour

Speed limit = 55 miles per hour

Exceeded speed limit = 72 - 55

= 17 miles per hour

Therefore, Jabari's average speed exceed the speed limit by 17 miles per hour

Learn more about average speed:

brainly.com/question/4931057

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seropon [69]

Answer:

D) 8

Step-by-step explanation:

3x - 7 - 2x + 5 = 6

3x - 2x  = 6 + 7 - 5

  x   = 8

Note: 6 + 7 -5 =  13 -5 = 8

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Oliver sewed a quilt 3 2/3 feet long and 2 1/4 feet wide. what is the area of her quilt?
Hunter-Best [27]

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8.25 feet

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3 2/3 times 2 1/4 equals 8.25

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Expand and simplify 2(5x - 1) – (2x - 5) ​
horrorfan [7]

Answer:

2(5x - 1) – (2x - 5)

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Step-by-step explanation:

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Read 2 more answers
Find the least common multiple (LCM) of 8y^6+ 144y^5+ 640y^4 and 2y^4 + 40y^3 + 200y^2.
Mice21 [21]

Answer:

<em>LCM</em> = 8y^{4}(y+ 10)^{2}(y + 8)

Step-by-step explanation:

Making factors of 8y^{6}+ 144y^{5}+ 640y^{4}

Taking 8y^{4} common:

\Rightarrow 8y^{4} (y^{2}+ 18y+ 80)

Using <em>factorization</em> method:

\Rightarrow 8y^{4} (y^{2}+ 10y + 8y + 80)\\\Rightarrow 8y^{4} (y (y+ 10) + 8(y + 10))\\\Rightarrow 8y^{4} (y+ 10)(y + 8))\\\Rightarrow \underline{2y^{2}} \times  4y^{2} \underline{(y+ 10)}(y + 8)) ..... (1)

Now, Making factors of 2y^{4} + 40y^{3} + 200y^{2}

Taking 2y^{2} common:

\Rightarrow 2y^{2} (y^{2}+ 20y+ 100)

Using <em>factorization</em> method:

\Rightarrow 2y^{2} (y^{2}+ 10y+ 10y+ 100)\\\Rightarrow 2y^{2} (y (y+ 10) + 10(y + 10))\\\Rightarrow \underline {2y^{2} (y+ 10)}(y + 10)        ............ (2)

The underlined parts show the Highest Common Factor(HCF).

i.e. <em>HCF</em> is 2y^{2} (y+ 10).

We know the relation between <em>LCM, HCF</em> of the two numbers <em>'p' , 'q'</em> and the <em>numbers</em> themselves as:

HCF \times LCM = p \times q

Using equations <em>(1)</em> and <em>(2)</em>: \Rightarrow 2y^{2} (y+ 10) \times LCM = 2y^{2} \times  4y^{2}(y+ 10)(y + 8) \times 2y^{2} (y+ 10)(y + 10)\\\Rightarrow LCM = 2y^{2} \times  4y^{2}(y+ 10)(y + 8) \times (y + 10)\\\Rightarrow LCM = 8y^{4}(y+ 10)^{2}(y + 8)

Hence, <em>LCM</em> = 8y^{4}(y+ 10)^{2}(y + 8)

5 0
2 years ago
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