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mariarad [96]
3 years ago
5

Please help need it for school and need 20 characters which is like really but pls help solve

Mathematics
1 answer:
vampirchik [111]3 years ago
3 0
a^{log_a(b)}=b also
ln=log_e



e^{2x-3}=11
take the ln of both sides
2x-3=ln(11)
add 3 to both sides
2x=3+ln(11)
dividie both sides by 2 or times 1/2
x=1.5+0.5ln(11) or x= \frac{3}{2}+ln \sqrt{11}
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(a) The parametric equations x = f(t) and y = g(t) give the coordinates of a point (x, y) = (f(t), g(t)) for appropriate values
Gnesinka [82]

Answer:

Step-by-step explanation:

a). Given a parametric equation, we are describing a set of coordinates based on the value of t. The variable t is called the parameter.

b) we have the following equations. x=t y=t^2, so in order for us to know where the object is at t=t' we must replace t with the specific value t'. Hence, when t=0 the object is at (0,0^2) = (0,0) (the origin). When t=6, the object is at (6,6^2) = (6,36).

c). To eliminate the parameter, we replace the parameter in one equation by using the second equation. Recall that we have that x=t. Then, by replacing in the second equation, we have the following

y=t^2 = (x)^2 = x^2

where x\geq 0

5 0
3 years ago
How to solve this kind of question? <br><br><br> Show steps!
melomori [17]
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3 years ago
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KengaRu [80]

We are given coordinates of the triangle: A(2,2), B(7,1) and C(8,-4).

We need to rotated 90° counterclockwise about the origin.

In order to find the new coordinates of rotatation 90°counterclockwise about the origin, we can apply rule (h, k) ---> (-k,h).

Where (h,k) are the coordinates of original image on axes and (-k,h) are the coordinates of rotated image.

In resulting coordinates of the image first swap the x and y coordinates of the original image and then make the sign opposite of each x-coordinate.

On applying rule (h, k) ---> (-k,h), we get

A(2,2) --> A'(-2,2)

B(7,1) --> B'(-1,7)

C(8,-4) --> C'(4,8)

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3 years ago
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Nookie1986 [14]
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3 0
3 years ago
Estimate the following quantities without using a calculator. Then find a more precise result, using a calculator if necessary.
zubka84 [21]

Explanation:

a. It is so easy to double a number that no approximation is necessary.

  31,000 × 200 = 3.1×10⁴ × 2×10² = 6.2×10⁶ exactly

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b. 6.2×10³ × 5.2×10⁶ ≈ 6×5×10⁹ = 3×10¹⁰ approximately

The approximation can be refined a bit by taking the ".2" into account:

  6.2×10³ × 5.2×10⁶ ≈ (6×5 + .2×(6+5))×10⁹

  = 32.2×10⁹ = 3.22×10¹⁰ approximately

Actual product: 3.224×10¹⁰.

For most purposes, the approximation is an adequate approximation, as it it within 10% of the actual value.

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c. 9×10⁶ -2.3×10⁴ ≈ 9×10⁶ approximately

A better approximation is to actually subtract an approximation of the smaller number:

  9×10⁶ -2.3×10⁴ ≈ (9 - 0.02)×10⁶ ≈ 8.98×10⁶ approximately

The actual value uses all of the digits of the smaller number:

  9×10⁶ -2.3×10⁴ ≈ (9 - 0.023)×10⁶ = 8.977×10⁶ exactly

As in part B, either approximation is adequate for most purposes, as the difference from the actual value is less than .3%.

_____

The accuracy required of an approximation, hence the work you expend improving accuracy, should depend on the need in the final application of the number. Often, approximations are used for budget or resource planning purposes where some "slop" is allowed or even expected.

They can also be used in engineering applications, where the error needs to be on the side of more safety (rather than less).

7 0
3 years ago
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