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olga2289 [7]
3 years ago
7

Please help answer this

Mathematics
2 answers:
zmey [24]3 years ago
8 0
5 because 4/2 is 2 and 10/2 is 5
zysi [14]3 years ago
6 0

Answer:

the missing length is 5

Step-by-step explanation:

Since we are given similar triangles, we can simply use proportions to solve the problem:

we say that 2 is to side of length 4, the same as m is to side lengths 10. And in math terms:

\frac{2}{4} =\frac{m}{10} \\m=\frac{2*10}{4} \\m=\frac{20}{4} = 5

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A zombie apocalypse follows a/an ___________ pattern.Immersive Reader
MariettaO [177]

Answer:

B. Quadratic​​​​​​​​​​

8 0
2 years ago
Help me pls this is urgent ;>
postnew [5]

Answer:

\frac{ |a + x| }{2}  -  \frac{ |a - x| }{2}  \\  \\  =  \frac{ | - 2 - 6| }{2}  -  \frac{ | - 2 + 6| }{2}  \\  \\  =  \frac{ | - 8| }{2}  -  \frac{ |4| }{2}  \\  \\  =  \frac{8}{2}  -  \frac{4}{2}  \\  \\  = 4 - 2 = 2

5 0
2 years ago
Find [5(cos 330 degrees + I sin 330 degrees)]^3
earnstyle [38]
Given a complex number in the form:
z= \rho [\cos \theta + i \sin \theta]
The nth-power of this number, z^n, can be calculated as follows:

- the modulus of z^n is equal to the nth-power of the modulus of z, while the angle of z^n is equal to n multiplied the angle of z, so:
z^n = \rho^n [\cos n\theta + i \sin n\theta ]
In our case, n=3, so z^3 is equal to
z^3 = \rho^3 [\cos 3 \theta + i \sin 3 \theta ] = (5^3) [\cos (3 \cdot 330^{\circ}) + i \sin (3 \cdot 330^{\circ}) ] (1)
And since 
3 \cdot 330^{\circ} = 990^{\circ} = 2\pi +270^{\circ}
and both sine and cosine are periodic in 2 \pi,  (1) becomes
z^3 = 125 [\cos 270^{\circ} + i \sin 270^{\circ} ]

6 0
3 years ago
Square root of 256 is
miskamm [114]

Answer:

16

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
If 9a + 9b + 9c = 59a+9b+9c=5 ,<br><br>what is 72a + 72b + 72c72a+72b+72c?
Gekata [30.6K]

Answer:

40

Step-by-step explanation:

9a + 9b + 9c = 5

Multiply the equation by 8

8*(9a + 9b + 9c )= 5*8

Distribute

72a +72b +72 c = 40

6 0
3 years ago
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