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3 years ago

I NEED HELP ON MATH PLS

Mathematics

1 answer:

kicyunya [14]3 years ago

7 0

Answer:

5/2 or 2½ or 2.5

Step-by-step explanation:

20/8 = 2.5

10/4 = 2.5

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The question is on the picture

Kamila [148]

answer:

ABCD is not congruent to KLMN

ABCD cannot be mapped onto KLMN

3 0

2 years ago

Find the derivative.

krek1111 [17]

Answer:

$\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$

General Formulas and Concepts:

<u>Algebra I</u>

Terms/Coefficients

Expanding/Factoring

<u>Calculus</u>

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle f(x) = \frac{\sqrt{x}}{e^x}$

<u>Step 2: Differentiate</u>

Derivative Rule [Quotient Rule]: $\displaystyle f'(x) = \frac{(\sqrt{x})'e^x - \sqrt{x}(e^x)'}{(e^x)^2}$
Basic Power Rule: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}(e^x)'}{(e^x)^2}$
Exponential Differentiation: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{(e^x)^2}$
Simplify: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{e^{2x}}$
Rewrite: $\displaystyle f'(x) = \bigg( \frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x \bigg) e^{-2x}$
Factor: $\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

7 0

2 years ago

dsp73

<h2> The answer is B. 600 </h2>

4 0

2 years ago

Maths functions question

yarga [219]

Answer:

a) OA = 1 unit

b) OB = 3 units

c) AB = √10 units

Step-by-step explanation:

<u>Given function</u>:

$g(x)=2^x$

Point A is the y-intercept of the exponential curve (so when x = 0).

To find the y-value of Point A, substitute x = 0 into the function:

$\implies g(0)=2^0=1$

Therefore, A (0, 1) so OA = 1 unit.

If BC = 8 units then the y-value of Point C is 8.

The find the x-value of Point C, set the function to 8 and solve for x:

$\begin{aligned}f(x) & = 8 \\\implies 2^x & = 8\\2^x & = 2^3\\\implies x &= 3\end{aligned}$

Therefore, C (3, 8) so Point B is (3, 0). Therefore, OB = 3 units.

From parts (a) and (b):

A = (0, 1)
B = (3, 0)

To find the length of AB, use the distance between two points formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\textsf{where }(x_1,y_1) \textsf{ and }(x_2,y_2)\:\textsf{are the two points.}$

Therefore:

$\implies \sf AB=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}$

$\implies \sf AB=\sqrt{(3-0)^2+(0-1)^2}$

$\implies \sf AB=\sqrt{(3)^2+(-1)^2}$

$\implies \sf AB=\sqrt{9+1}$

$\implies \sf AB=\sqrt{10}\:\:units$

5 0

1 year ago

How do you find a if given c and 0 (angle) in a right triangle where a is the opposite, c is the hypotenuse, b is adjacent and 0

Yakvenalex [24]

C can be found using pythagoras theorem. c2=a2+b2. Now, b is not given, but we know that cos(theta)=b/c=>b=c*cos(theta). Substituting b in the above relation, c2=a2+c2(cos(theta))^2=>c2=a2/(1-cos((theta))^2). c is the squareroot of c2. Hence c=sqrt(2/(1-cos((theta))^2))

4 0

3 years ago