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Free_Kalibri [48]
3 years ago
9

Anyone know this ???

Mathematics
1 answer:
sineoko [7]3 years ago
8 0
Make into improper fraction

5 and 1/6=5+1/6=30/6+1/6=31/6

3 and 2/4=3+2/4=3+1/2=6/2+1/2=7/2

5 and 1/6 times 2 and 2/4=
31/6 times 7/2=
217/12=
216/12+1/12=
18+1/12=
18 and 1/12
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Solve for X y and z brainliest Round to nearest tenth
kykrilka [37]

Answer:

x=1

Z=35.4965

assuming you want y to be the height, y=5.916

Step-by-step explanation:

x/height=height/35

35x=y^2

using Pythagorean theorem

y^2+x^2=6^2

y^2+35^2=z^2

substituting for y^2

35x+x^2=6^2

x^2+35x-36=0

(x+36)(x-1)=0

x=1

y^2+x^2=6^2

substitute

y^2+1=36

y^2=35

y=5.916

y^2+35^2=z^2

substitute

35+1225=z^2

z^2=1260

z=35.4965

4 0
2 years ago
54. The triangles in the diagram below are congruent.
Reika [66]

Answer:

Angle N should be 64

Step-by-step explanation:

A triangle adds up to 180. So triangle JKL already has 60 towards 180. Triangle ONM already has 56 towards 180. So you have to find the other value needed in order to make 180. 56+60= 116 so we need to find the other angle to make 180. To do so just subtract 180-116 and that gives tou the answer 64 so that would be Angle N

7 0
3 years ago
Please help this is due today!
miskamm [114]

Answer:

C. )

Step-by-step explanation:

3 0
3 years ago
A tank contains 10 liters of pure water. Saline solution with a variable concentration 5 grams of salt per liter is pumped into
algol [13]

Answer:

dQ(t)/dt = 20 - 2Q(t)/5 , Q(0) = 0

Step-by-step explanation:

The mass flow rate dQ(t)/dt = mass flowing in - mass flowing out

Since 5 g/L of salt is pumped in at a rate of 4 L/min, the mass flow in is thus 5 g/L × 4 L/min = 20 g/min.

Let Q(t) be the mass present at any time, t. The concentration at any time ,t is thus Q(t)/volume = Q(t)/10. Since water drains at a rate of 4 L/min, the mass flow out is thus, Q(t)/10 g/L × 4 L/min = 2Q(t)/5 g/min.

So, dQ(t)/dt = mass flowing in - mass flowing out

dQ(t)/dt = 20 g/min - 2Q(t)/5 g/min

Since the salt just begins to be pumped in, the initial mass of salt in the tank is zero. So Q(0) = 0

So, the initial value problem is thus

dQ(t)/dt = 20 - 2Q(t)/5 , Q(0) = 0

3 0
2 years ago
No links or files please im doing two at a time to get it done quicker
andrew-mc [135]

Answer: 5.6 cm

Step by Step: I used the same formula as the question before and I got the answer for this question. When solving questions like this, you can work backwards from the problem and then solve

3 0
2 years ago
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