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Semenov [28]
3 years ago
7

Geometry please help

Mathematics
1 answer:
Afina-wow [57]3 years ago
5 0

Answer: 31

Step-by-step explanation:

Assuming this is a rhombus, then the diagonals are perpendicular, meaning angle MQN measures 90 degrees.

Now, if we consider triangle MNQ, we know that angles in a triangle add to 180 degrees, meaning NMQ measures 31 degrees.

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# 3 please I don't understand
dmitriy555 [2]
3/4 = /340 find how 4 equals 340 and multiply 3 by that
340/4=85 so 3X85 will be the answer which is 255 so 3/4 = 255/340
255/340 is your answer
6 0
4 years ago
Mrs. Jacobs buys 20 kilograms if rice at $0.84 per kilogram. she buys 700 grams of shrimp at $1.02 per 100 grams.
rusak2 [61]
0.84(20)=$16.80
700/100=7
7(1.02)=$7.14
$7.14+16.80=$23.94 in total
8 0
3 years ago
Tim had $150 to spend on 3 pairs of pants. After buying
Aneli [31]

Answer:D

Step-by-step explanation: 3 represents the pants, x is the price of the pants, and the 30 is the remaining money. All this combined =150.

5 0
3 years ago
On February 10, 1990, high tide in Boston was at midnight. The water level at high tide was 9.9 feet; later, at low tide, it was
Verizon [17]

Answer:

a) Model

H(t)=4.9cos(\frac{\pi}{6}t)+5

b) The tide will be at 6.5 ft high at 2:24 am (going down).

As the tide will rise again, the tide will be again at 6.5 ft high at 9:36 am (going up).

c) H(10 am) = 7.45

H(4 pm) = 2.55

Step-by-step explanation:

We can model this as a sine or cosine wave, for which we will calculate its parameters by the data given.

The model we will use is:

H(t)=Acos(\omega t+\phi)+B

t will be in hours, starting from midnight (t=0)

1) We know that at midnight happens the high tide. That means that

H(0)=Acos(\omega* 0+\phi)+B=A+B=9.9\\\\ \phi=0\\\\A+B=9.9

2) Six hours later (t=6) happens the low tide (0.1 ft)

H(0)=Acos(\omega*6)+B=-A+B=0.1\\\\\\ cos(\omega*6)=-1,\, \omega=\pi/6\\\\\\B=0.1+A=9.9-A\\\\2A=9.9-0.1\\\\A=9.8/2=4.9\\\\\\B=0.1+A=0.1+4.9=5

Now that we have calculated all the parameters, the model for the height of the tide is:

H(t)=4.9cos(\frac{\pi}{6}t)+5

On Feb 10, when is the water level 6.5 feet high?

H(t)=4.9cos(\frac{\pi}{6}t)+5=6.5\\\\cos(\frac{\pi}{6}t)=(6.5-5)/4.9=0.306\\\\t=\frac{6}{\pi}arcos(0.306)=2.4

The tide will be at 6.5 ft high at 2:24 am (going down).

As the tide will rise again, the tide will be again at 6.5 ft high at 9:36 am (going up).

On Feb 10, what is the water level at 10 am? 4pm?

At 10 am is t=10:

H(10)=4.9cos(\frac{\pi}{6}10)+5=4.9*0.5+5=2.45+5=7.45

At 4 pm, is t=12+4=16

H(16)=4.9cos(\frac{\pi}{6}16)+5=4.9*(-0.5)+5=-2.45+5=2.55

3 0
3 years ago
If 1=2 what conclusions can u draw about m 3 and m 4
BlackZzzverrR [31]

Answer: second one

Step-by-step explanation:

5 0
3 years ago
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