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Tema [17]
4 years ago
10

Round 75.03078 to the nearest ten-thousandths

Mathematics
1 answer:
weeeeeb [17]4 years ago
5 0
75.0309 is the answer 
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What is the solution to the linear equation?<br><br> 4b+6=2-b+4
Morgarella [4.7K]
To start we need to put all the b's on the same side.

4b + 6 = 2 - b + 4
To move the b on the right side to the left side, add b to both sides.
4b + 6 + b = 2 - b + 4 + b

Simplify. 4b + b = 5b and -b + b = 0 so...
The equation is now:  5b + 6 = 2 + 4

Now move 6 to the right side so we can isolate b.
5b + 6 - 6 = 2 + 4 - 6
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b = 0/5
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3 years ago
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Completing the square: x^2+2x-7=8x
tia_tia [17]

Answer: x = 7, x = -1

Step-by-step explanation: Move the term to the left side

x^2 + 2x - 7 = 8x

x^2 + 2x - 7 - 8x = 0

Combine like terms

x^2 + 2x - 7 - 8x = 0

x^2 - 6x - 7 = 0

x^2 - 6x - 7 = 0

a = 1

b = -6

c = -7

x = -(-6) + (-6)^2 - 4 x 1(-7)/ 2 x 1

Evaulate the Exponent

Multiply the numbers

Add the numbers

Evaluate the square root

Multiply the numbers x = 6 ± 8/2

To solve for the unknown variable, separate into two equations: one with a plus and the other with a minus.

x = 6 + 8/2

x = 6 - 8/2

Rearrange and isolate the variable to find each solution

x = 7

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4 0
3 years ago
The volume of a spherical balloon is increasing at the constant rate of 8 cubic feet per minute. How fast is the radius incresin
OverLord2011 [107]

Answer:

Rate of increase of radius = 0.0064 ft/sec

Rate of increase of surface area = 1.61 ft^2/sec

Step-by-step explanation:

Given that:

Rate of change of volume of a spherical balloon = 8 cubic feet per minute

\dfrac{dV}{dt} = 8 ft^3/min

Radius, r = 10 feet

To find:

The rate of change of radius at this moment and rate of change of surface area at this moment?

Solution:

First of all, let us have a look at the formula:

1.\ V = \dfrac{4}{3}\pi r^3\\2.\ A =4\pi r^2

Now, differentiating the volume and area, we get:

\dfrac{dV}{dt} = \dfrac{4}{3}\times 3 \pi r^2 \dfrac{dr}{dt}\\\Rightarrow \dfrac{dV}{dt} = 4 \pi r^2 \dfrac{dr}{dt}

\dfrac{dA}{dt} = 4\pi \times 2 r \dfrac{dr}{dt}\\\Rightarrow \dfrac{dA}{dt} = 8\pi r \dfrac{dr}{dt}

\dfrac{dV}{dt} = 8 = 4\pi r^2 \dfrac{dr}{dt}\\\Rightarrow 8 = 4\times 3.14 \times 10^2 \dfrac{dr}{dt}\\\Rightarrow 2 =  3.14 \times 10^2 \dfrac{dr}{dt}\\\Rightarrow \dfrac{dr}{dt} = 0.0064\ ft/sec

\dfrac{dA}{dt} = 8\times \pi \times r \dfrac{dr}{dt}\\\Rightarrow \dfrac{dA}{dt} = 8\times 3.14 \times 10 \times 0.0064\\\Rightarrow \dfrac{dA}{dt} = \bold{1.61 \ ft^2/sec}

7 0
2 years ago
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