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Rudik [331]
3 years ago
14

100 points!! Please solve question 50 in detail

Mathematics
2 answers:
djyliett [7]3 years ago
6 0

Answer:

2.62

Step-by-step explanation:

First, write the square root as exponent.

Move the denominator to the numerator and negate the exponent.

Use log product property.

Use log exponent property.

Substitute values.

Step-by-step explanation:

SCORPION-xisa [38]3 years ago
4 0

Answer:

2.62

Step-by-step explanation:

log_{b} \frac{b^{2}x^{\frac{5}{2} }}{\sqrt{y}}

First, write the square root as exponent.

log_{b} \frac{b^{2}x^{\frac{5}{2} }}{y^{\frac{1}{2}}}

Move the denominator to the numerator and negate the exponent.

log_{b}(b^{2}x^{\frac{5}{2}}y^{-\frac{1}{2}})

Use log product property.

log_{b}(b^{2}) + log_{b}(x^{\frac{5}{2}}) + log_{b}(y^{-\frac{1}{2}})

Use log exponent property.

2 log_{b}(b) + {\frac{5}{2}}log_{b}(x) - {\frac{1}{2}}log_{b}(y)

Substitute values.

2(1) + \frac{5}{2}(0.36) - \frac{1}{2}(0.56) \\2.62

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Help me plz with my hw
inna [77]
Circumference is found with the formula

c = pi × d

d is diameter and we will use 3.14 for pi.

The diameter is the measure across the center of the circle. In the first problem, you are given the radius, so we have to multiply by 2 to get the diameter. Then we can use the formula.

12.4 in × 2 = 24.8 in (that's the diameter)

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circumference is 77.872 inches.

Try the other problems on your own. They are just like this one. Just make sure they are giving you the diameter and not the radius. Post if you have problems.
6 0
2 years ago
2. A solar lease customer built up an excess of 6,500 kilowatt hours (kwh) during the summer using his solar
tia_tia [17]

9514 1404 393

Answer:

  a)  E = 6500 -50d

  b)  5000 kWh

  c)  the excess will last only 130 days, not enough for 5 months

Step-by-step explanation:

<u>Given</u>:

  starting excess (E): 6500 kWh

  usage: 50 kWh/day (d)

<u>Find</u>:

  a) E(d)

  b) E(30)

  c) E(150)

<u>Solution</u>:

a) The exces is linearly decreasing with the number of days, so we have ...

  E(d) = 6500 -50d

__

b) After 30 days, the excess remaining is ...

  E(30) = 6500 -50(30) = 5000 . . . . kWh after 30 days

__

c) After 150 days, the excess remaining would be ...

  E(150) = 6500 -50(150) = 6500 -7500 = -1000 . . . . 150 days is beyond the capacity of the system

The supply is not enough to last for 5 months.

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2 years ago
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Answer:

shouldn't it be 72 if not try 58

Step-by-step explanation:

5 0
2 years ago
The point (8,-4) lies on a circle. What is the length of the radius of this circle if the center is located at (5,-7)
Kamila [148]
Notice the picture, the radius is just the distance from the
center to a point on the circle, thus

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\bf \begin{array}{llll}&#10;d = &\sqrt{({{ 8}}-{{ 5}})^2 + ({{ -4}}-{{(-7)}})^2}\\&#10;&\qquad \uparrow \\&#10;&radius&#10;\end{array}&#10;\\\\\\&#10;d=\sqrt{(3)^2+(-4+7)^2}\implies d=\sqrt{3^2+3^2}\implies d=\sqrt{18}&#10;\\\\\\&#10;d=\sqrt{9\cdot 2}\implies d=\sqrt{3^2\cdot 2}\implies d=3\sqrt{2}



4 0
3 years ago
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