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hammer [34]
3 years ago
7

Kelli swam upstream for some distance in one hour. She then swam downstream the same river for the same distance in only 6 minut

es. If the river flows at 5 km/hr, how fast can Kelli swim in still water?
Mathematics
1 answer:
solmaris [256]3 years ago
4 0

Answer:

6.11km/hr

Step-by-step explanation:

Let the speed that Kelli swims be represented by Y

Speed of the river = 5km/hr

Distance = Speed × Time

Kelli swam upstream for some distance in one hour

Swimming upstream takes a negative sign, hence:

1 hour ×( Y - 5) = Distance

Distance = Y - 5

She then swam downstream the same river for the same distance in only 6 minutes

Downstream takes a positive sign

Converting 6 minutes to hour =

60 minutes = 1 hour

6 minutes =

Cross Multiply

6/60 = 1/10 hour

Hence, Distance =

1/10 × (Y + 5)

= Y/10 + 1/2

Equating both equations we have:

Y - 5 = Y/10 + 1/2

Collect like terms

Y - Y/10 = 5 + 1/2

9Y/10 = 5 1/2

9Y/ 10 = 11/2

Cross Multiply

9Y × 2 = 10 × 11

18Y = 110

Y = 110/18

Y = 6.1111111111 km/hr

Therefore, Kelli's can swim as fast as 6.11km/hr still in the water.

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Plz hurry!!!! thank you!!!!
Nikitich [7]

Answer:

Area of trapezium = 4.4132 R²

Step-by-step explanation:

Given, MNPK is a trapezoid

MN = PK and ∠NMK = 65°

OT = R.

⇒ ∠PKM = 65° and also ∠MNP = ∠KPN = x (say).

Now, sum of interior angles in a quadrilateral of 4 sides = 360°.

⇒ x + x + 65° + 65° = 360°

⇒ x = 115°.

Here, NS is a tangent to the circle and ∠NSO = 90°

consider triangle NOS;

line joining O and N bisects the angle ∠MNP

⇒ ∠ONS = \frac{115}{2} = 57.5°

Now, tan(57.5°) = \frac{OS}{SN}

⇒ 1.5697 = \frac{R}{SN}

⇒ SN = 0.637 R

⇒ NP = 2×SN = 2× 0.637 R = 1.274 R

Now, draw a line parallel to ST from N to line MK

let the intersection point be Q.

⇒ NQ = 2R

Consider triangle NQM,

tan(∠NMQ) = \frac{NQ}{QM}

⇒ tan65° = \frac{NQ}{QM}

⇒ QM = \frac{2R}{2.1445}

QM = 0.9326 R .

⇒ MT = MQ + QT

          = 0.9326 R + 0.637 R  (as QT = SN)

⇒ MT = 1.5696 R

⇒ MK = 2×MT = 2×1.5696 R = 3.1392 R

Now, area of trapezium is (sum of parallel sides/ 2)×(distance between them).

⇒ A = (\frac{NP + MK}{2}) × (ST)

       = (\frac{1.274 R + 3.1392 R}{2}) × 2 R

       = 4.4132 R²

⇒ Area of trapezium = 4.4132 R²

5 0
3 years ago
For f(x) = -4x -9 evaluate for f(6.5)​
kolezko [41]

Answer:

f=−35

Step-by-step explanation:

<em>1.Simplify  4×6.5 to 26</em>

f=-26-9

<em>2. Simplify  -26-9 to  -35</em>

f=-35

7 0
3 years ago
Jerome photographed an aquarium that measured 12 inches by 18 inches by 24 inches. Lim photographed an aquarium that was 12 inch
8090 [49]

Answer:

the answer is 864 in3

Step-by-step explanation:

Jerome photographed an aquarium that measured 12 inches by 18 inches by 24 inches.

So V=LxWxH....

12 x 18 x 24= 5,184

Then...

Lim photographed an aquarium that was 12 inches by 30 inches by 12 inches.

So= V=LxWxH

12 x 30 x 12=4,320

And they need to see who has more so 5,184-4,320=864in3

I hope this helps also Into Math said it was correct :)

6 0
3 years ago
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