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3 years ago

Z =-19+3.14i what are the real and imaginary parts of z

Mathematics

1 answer:

Rom4ik [11]3 years ago

7 0

Answer:

gyuuiwhyggwftywgyuwgyugwyuwsyw7y7y

vtsftgygtwfsygyqgy

Step-by-step explanation:

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(24,215-23,500)/[3678/sqrt(85)]

Fantom [35]

Answer:

1.792

Step-by-step explanation:

(24,215-23,500)/[3678/sqrt(85)]

715/[3678/sqrt(85)]

715/398.94

1.792

Please give me brainliest :)

4 0

3 years ago

Please help me!! I am struggling... I will not accept nonsense answers!

telo118 [61]

Answer:

y = 110°

Step-by-step explanation:

The inscribed angle CHF is half the measure of its intercepted arc CDF

The 3 arcs in the circle = 360°, thus

arc CDF = 360° - 160° - 60° = 140°, so

∠ CHF = 0.5 × 140° = 70°

∠ CHF and ∠ y are adjacent angles and supplementary, thus

y = 180° - 70° = 110°

6 0

3 years ago

B. Find the angle measures.<br> 7x+<br> 7<br> 9x + 12<br> 490

Lubov Fominskaja [6]

Is 7x+7 and then what. Iam confused

5 0

3 years ago

The half-life of cesium-137 is 30 years. Suppose we have a

VMariaS [17]

(a) If $y(t)$ is the mass (in mg) remaining after $t$ years, then $y(t) = y(0) e^{kt}=100e^{kt}.$

$y(30) = 100 e^{30k} = \frac{1}{2}(100) \implies e^{30k} = \frac{1}{2} \implies k = -(\ln 2) /30 \implies \\ \\
y(t) = 100e^{-(\ln 2)t/30} = 100 \cdot 2^{-t/30}$

(b) $y(100) = 100 \cdot 2^{-100/30} \approx \text{9.92 mg}$

(c)
$100 e^{- (\ln 2)t/30} = 1\ \implies\ -(\ln 2) t / 30 = \ln \frac{1}{100}\ \implies\ \\ \\
t = -30 \frac{\ln 0.01}{\ln 2} \approx \text{199.3 years}$

3 0

3 years ago

For f(x) = -4x + 2, find f(x) when x = -1

Damm [24]

b...............................................................................

5 0

3 years ago