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Mkey [24]
3 years ago
14

Asafa runs 0.75 kilometers in 6 minutes. Jemaine runs 3,000 meters in half an hour. Who ran at a faster rate?

Mathematics
1 answer:
Lorico [155]3 years ago
5 0

Answer:

  • Asafa ran at a faster rate

Step-by-step explanation:

<u>Asafa's speed:</u>

  • 0.75 km/ 6 min = 750 m/ 6 min = 125 m/min

<u>Jemaine's speed:</u>

  • 3000 m / 30 min = 100 m/min

Asafa ran at faster rate as 125 m/min > 100 m/min

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THE VALUE OF 9 IN 495,123 IS 10000 <span>TIMES THE VALUE OF 9 IN 63,129</span>
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2 years ago
line k has a slop of -5. Line J is perpendicular to line K and passes through the point (5,9). Create the equation for line j.
sveticcg [70]

Answer:

The equation of the line would be y = 1/5x + 8

Step-by-step explanation:

Since the slope of k is -5, the slope of j has to be 1/5. This is because perpendicular lines have opposite and reciprocal slopes.

Now we can use the slope and the point in point-slope form to get the equation.

y - y1 = m(x - x1)

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y - 9 = 1/5x - 1

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6 0
2 years ago
Subtract 4 from 1/6 of 42
ivanzaharov [21]
\frac{1}{6} \times 42 - 4= \frac{42}{6}-4 = 7-4=3
5 0
3 years ago
Read 2 more answers
Given: △ABC, E∈ AB m∠ABC=m∠ACE AB=34, AC=20 Find: AE
LenaWriter [7]

Answer:

AE = 11.76 units

Step-by-step explanation:

For better understanding of the solution, see the attached figure :

Given : E ∈ AB, m∠ABC = m∠ACE, AB = 34 and AC = 20

To find AE :

In ΔABC and ΔACE,

m∠ABC = m∠ACE ( Given )

∠A = ∠A    ( Common angle for both the triangles )

By AA postulate of similarity of triangles, ΔABC ~ ΔACE

So, proportion of the corresponding will be equal.

\implies \frac{AC}{AB}=\frac{CE}{BC}=\frac{AE}{AC}\\\\\implies\frac{AC}{AB}=\frac{AE}{AC}\\\\\implies\frac{20}{34}=\frac{AE}{20}\\\\\implies AE=\frac{20\times 20}{34}\approx 11.76

Hence, AE = 11.76 units

4 0
3 years ago
I need ill post question 20 after i really need help
ivanzaharov [21]
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3. -17
4. -13
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7. -45
8. 36
9. -75
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17. 25
18. -282

4 0
2 years ago
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