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Alja [10]
3 years ago
7

Solve for x: 30 > 6x

Mathematics
2 answers:
Eddi Din [679]3 years ago
7 0

Answer:

If it has the little line under the > sign then x = 5

Step-by-step explanation:

Pani-rosa [81]3 years ago
4 0

here is the answer to solve for x x<5

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Identify the range of values for x.
Harlamova29_29 [7]

Answer:   Overall, the steps for algebraically finding the range of a function are:

Write down y=f(x) and then solve the equation for x, giving something of the form x=g(y).

Find the domain of g(y), and this will be the range of f(x). ...

If you can't seem to solve for x, then try graphing the function to find the range.

Step-by-step explanation:

5 0
2 years ago
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A science experiment calls for millilitres of a 25% sulfuric acid solution what percentage of the solution is not sulfuric acid
iVinArrow [24]

Answer: Im first

Step-by-step explanation:

5 0
3 years ago
��� How does the value of sine and cosine of complementary angles relate
vlabodo [156]
Cosine is co added onto sine. Basically, cosine is the sine function moved over 90degrees or pi/2 (pi/2 on a unit circle is 90 degrees)

Sin(x)=cos(x+90) <--degrees
Sin(x)=cos(x+pi/2) <--radians

The above two equations for converting them is called a cofunction identity. There's many more identities to convert sines, cosines, tangents, cosecantes, secantes, and cotangents between each other. This is taught to you in PreCalculus.
7 0
3 years ago
A. Write the equation of the line that represents the linear approximation to the following function at the given point a.
anastassius [24]

Answer:

a) f(x) = 4 - x²

The linear approximation of the function at a=1 is

L(x) = 5 - 2x at a = 1

b) The graph of the function and the linear approximation at that point is attached to this solution.

The curve represent the real function,

f(x) = 4 - x²

The straight line represents the linear approximation of the function at a=1.

L(x) = 5 - 2x

The curve and the function evidently cross paths at x=1 and understandably so.

c) Using the linear approximation obtained at a = 1.

f(1.1) = 2.8

Using the actual function, the actual value of f(1.1) = 2.79

d) Percent error = 0.358%

Step-by-step explanation:

f(x) = 4 - x²

a) The linear approximation of the function at the given point is given as

L(x) = f(a) + f'(a) [x - a]

f(x) = 4 - x²

a = 1

f(a) = 4 - 1² = 3

f'(x) = -2x

f'(a) = -2(1) = -2

L(x) = f(a) + f'(a) [x - a]

L(x) = 3 + (-2)(x - 1)

L(x) = 3 -2x + 2

L(x) = 5 - 2x

L(x) = -2x + 5

f(x) = 4 - x²

L(x) = 5 - 2x at a = 1

b) The graph of the function and the linear approximation at that point is attached to this solution.

The curve represent the real function,

f(x) = 4 - x²

The straight line represents the linear approximation of the function at a=1.

L(x) = 5 - 2x

The curve and the function evidently cross paths at x=1 and understandably so.

c) Use the linear approx. to estimate the given fxn value.

f(1.1)

L(x) = 5 - 2x

L(1.1) = 5 - 2(1.1) = 2.8

Using the function, the actual value of f(1.1) = 4 - 1.1² = 2.79

d) Compute the percent error in your approximation, 100*Iapprox-exactI/IexactI, where the exact value is given by a calculator

Percent error

= 100% × (|approx - exact|)/exact

Approximated value = 2.8

Exact value = 2.79

Percent error = 100% × (2.8-2.79)/2.79

Percent error = 0.358%

Hope this Helps!!!

6 0
4 years ago
A biologist recorded a count of 337 bacteria present in a culture after 5 minutes and 699 bacteria present after 15 minutes.
hoa [83]

The initial population was 234.

<em><u>Explanation</u></em>

<u>Formula for the exponential growth</u> is:   A= P*e^r^t , where P is the initial amount, A is the final amount, r is the rate of growth and  t  is the time duration.

There was 337 bacteria after 5 minutes and 699 bacteria after 15 minutes. So, the equations will be......

337=P*e^5^r .................................. (1)\\ \\ 699=P*e^1^5^r ................................. (2)

Now dividing equation (2) by equation (1) , we will get .......

\frac{699}{337}=\frac{e^1^5^r}{e^5^r} \\ \\ e^1^0^r = \frac{699}{337}

<u>Taking 'natural log'</u> on both sides.........

ln (e^1^0^r) = ln (\frac{699}{337})\\ \\ 10r= 0.7295....\\ \\ r= 0.07295.... \approx 0.073

Now, plugging this r=0.073 into equation (1), we will get......

337= P*e^5^(^0^.^0^7^3^)\\ \\ 337= P*1.4405 \\ \\ P= 233.946 \approx 234

So, the initial population was 234.

7 0
3 years ago
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