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zaharov [31]
2 years ago
10

Could 10.5\text{ cm}, 8.0\text{ cm},10.5 cm,8.0 cm,10, point, 5, start text, space, c, m, end text, comma, 8, point, 0, start te

xt, space, c, m, end text, comma and 4.0\text{ cm}4.0 cm4, point, 0, start text, space, c, m, end text be the side lengths of a triangle?
Mathematics
1 answer:
Reptile [31]2 years ago
3 0

Answer:

Yes, 10.5 cm, 8.0 cm and 4.0 cm can be the side lengths of a triangle.

Step-by-step explanation:

We have been given three lengths as 10.5 cm, 8.0 cm and 4.0 cm. We are asked to determine whether these side can be the side lengths of a triangle.

We will use triangle inequality theorem to solve our given problem.

Triangle inequality theorem states that sum of two sides of a triangle must be greater than 3rd side. Using triangle inequality theorem we will get 3 inequalities as:

10.5+8.0>4.0

18.5>4.0       True

8.0+4.0>10.5

12.0>10.5     True

10.5+4.0>8.0

14.5>8.0  True

Since all the three side lengths satisfy triangle inequality theorem, therefore, 10.5 cm, 8.0 cm and 4.0 cm can be the side lengths of a triangle.

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There are approximately 7.5 gallons of water in 1 cubic foot. If the water in the tank at time 80 minutes was transferred to an
icang [17]
<span>The volume of the cylinder: V=r²π*H
V=7.5ft²
r=3ft
H=?
7.5=3²*π*H
7.5=28.274H⇒H=7.5/28.2740
H=0.265ft
</span><span>
The water level will be 0.265ft.</span>
8 0
2 years ago
A small ball is released from rest from a point that is 40m above horizontal ground. The ball bounces on the ground and rebounds
Marina CMI [18]

Answer:

(a) 5.714s

(b) 0.625m

Step-by-step explanation:

Acceleration due to gravity=a=±9.8m/s/s (+ when falling, - when going upwards)

(a)

STAGE A: between start and first bounce

Initial speed=u=0m/s, Distance=s=40m, and Final velocity=v

v²=u²+2as

v²=0²+2(10)(40)

v²=794

v=28m/s (As the ball hits the ground)

v=u+at

28=0+9.8t_{a}

t_{a}=28/9.8=2.85714285714s

STAGE B: from the first bounce until it starts to fall

u=half the speed it hit the ground with=(28/2)=14m/s, and v=0m/s(when it starts to fall)

v=u+at

0=14-9.8t_{b}

t_{b}=14/9.8=1.42857142857s

STAGE C: between when it starts to fall after bounce 1, and bounce 2

We can assume that the time it takes to go from the ground to max height after bounce 1 is equal to the time it takes to fall from that same height to the ground. Therefore, t _{c}=1.42857142857s

The time from when the ball was released to when it hit the ground for the second time = t

t = t_{a} + t_{b} + t_{c}

t=2.85714285714+1.42857142857+1.42857142857

t=5.71428571428s≈5.714s

<u />

(b)

Before bounce 1, u=28m/s (see stage a)

After bounce 1, u= 28/2= 14m/s

After bounce 2, u= 14/2 = 7m/s

After bounce 3, u= 7/2 =3.5m/s

u=3.5m/s and v=0m/s(when it starts to fall)

v²=u²+2as

0²=3.5²+2(-9.8)s

19.6s=12.25

s=<u>0.625m (max height after third bounce)</u>

8 0
1 year ago
If the profits in your consulting business increase by 6​% one year and decrease by 3​% the following​ year, your profits are up
Naddika [18.5K]

Answer:

its 1.388 p

Step-by-step explanation:

i hope this helps

3 0
2 years ago
Malik’s solution to the equation , when , is shown below.
Gala2k [10]

Answer:

Malik substituted 60 for y instead of x

Step-by-step explanation:

<u><em>The complete question is</em></u>

Malik’s solution to the equation ,2/5x-4y=10 when x=60, is shown below.what error did Malik make first when solving the equation ?

Options are

malik did not multiply correctly. malik added 240 to each side of the equation. malik did not multiply correctly. malik substituted 60 for y instead of x

<u>Malik’s solution is</u>

\frac{2}{5}x-4(60)=10\\\frac{2}{5}x=10+240\\\frac{2}{5}x=250\\x=625

we have

\frac{2}{5}x-4y=10

For x=60

substitute the value of x in the equation and solve for y

\frac{2}{5}(60)-4y=10

24-4y=10

4y=24-10

4y=14

y=\frac{14}{4}\\\\ y=\frac{7}{2}

therefore

Malik substituted 60 for y instead of x

4 0
2 years ago
Read 2 more answers
A recently negotiated union contract allows workers in a shipping department 24 minutes for rest, 10 minutes for personal time,
Alik [6]

Answer:7.125 min

Step-by-step explanation:

Given

Resting time =24 min

Personal time=10 min

Delay=14 min

Observed Time=6min/cycle

Performance Rating=95%

Allowance Time=Rest time+Personal time+delay

Allowance time=24+10+14=48 min

Allowance %=\frac{Allowance Time}{Total Time}=\frac{48}{4\times 60}=20 \%

Allowance Factor=\frac{1}{1-0.20}=\frac{5}{4}

Calculate the Normal Time

Normal time=Observed time\timespersonal Time=6\times 0.95=5.7 min

Standard time=Normal time \times Allowance Factor

Standard time =NT\times Allowance factor=5.7\times 1.25=7.125 min

3 0
2 years ago
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