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Ainat [17]
3 years ago
7

In a rhombus whose side measures 20 and the smaller angle is 38°. Find the length of the larger diagonal, to the nearest tenth.

Mathematics
1 answer:
Lemur [1.5K]3 years ago
4 0

A rhombus diagonals form right angles. The diagonals also cut in half the angles

use cosine: cos38 = \frac{a}{20\\\\}

a = 20cos38

a=15.8

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Given: MNOK is a trapezoid, MN=OK, m∠M=60°, NK ⊥ MN , MK=16cm Find: The midsegment of MNOK
Sergeu [11.5K]

Answer:

12 cm

Step-by-step explanation:

1. Consider right triangle MNK. In this triangle angle N is right and m∠M=60°, then m∠K=30°. Thus, this triangle is special 30°-60°-90° right triangle with legs MN and NK and hypotenuse MK=16 cm. The leg MN is opposite to the angle with measure of 30°, then this leg is half of the hypotenuse, MN=8 cm.

2. Consider right triangle MNH, where NH is the height of trapezoid drawn from the point N. In this triangle m∠M=60°, angle H is right, then m∠N=30°. Similarly, the leg MH is half of the hypotenuse MN, MH=4 cm.

3. Trapezoid MNOK is isosceles, because MN=OK=8 cm. This means that NO=MK-2MH=16-8=8 cm.

4. The midsegment of the trapezoid is

\dfrac{MK+NO}{2}=\dfrac{16+8}{2}=12\ cm.

4 0
3 years ago
Which of the following is the slope of the line passing through the
julia-pushkina [17]

Answer:

I think the slope is UNDEFINED

y²-y¹/x²-x¹

(-5)-(16)/(-12)-(-12)

-21/0

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6 0
3 years ago
Gravel is being dumped from a conveyor belt at a rate of 20 ft3/min, and its coarseness is such that it forms a pile in the shap
11Alexandr11 [23.1K]

Answer:

0.25 feet per minute

Step-by-step explanation:

Gravel is being dumped from a conveyor belt at a rate of 20 ft3/min. Since we are told that the shape formed is a cone, the rate of change of the volume of the cone.

\dfrac{dV}{dt}=20$ ft^3/min

\text{Volume of a cone}=\dfrac{1}{3}\pi r^2 h

Since base diameter = Height of the Cone

Radius of the Cone = h/2

Therefore,

\text{Volume of the cone}=\dfrac{\pi h}{3} (\dfrac{h}{2}) ^2 \\V=\dfrac{\pi h^3}{12}

\text{Rate of Change of the Volume}, \dfrac{dV}{dt}=\dfrac{3\pi h^2}{12}\dfrac{dh}{dt}

Therefore: \dfrac{3\pi h^2}{12}\dfrac{dh}{dt}=20

We want to determine how fast is the height of the pile is increasing when the pile is 10 feet high.

h=10$ feet$\\\\\dfrac{3\pi *10^2}{12}\dfrac{dh}{dt}=20\\25\pi \dfrac{dh}{dt}=20\\ \dfrac{dh}{dt}= \dfrac{20}{25\pi}\\ \dfrac{dh}{dt}=0.25$ feet per minute (to two decimal places.)

When the pile is 10 feet high, the height of the pile is increasing at a rate of 0.25 feet per minute

5 0
3 years ago
Evaluate W - X if w = 18 and x = 6.<br> Giving BRIANLIEST
xenn [34]

Answer:

12

Step-by-step explanation:

18 -6

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Hope that this helps you and have a great day :)

5 0
3 years ago
Help this person with this problem because we are so done right now pic attached
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-3 is in the opposite direction of V
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