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3 years ago

Does half of one pizza represent the same amount as half of another pizza? Justify your answer.

Mathematics

2 answers:

zhenek [66]3 years ago

8 0

Answer:

yes

Step-by-step explanation:

well there both half of a pizza

erik [133]3 years ago

5 0

No it does not represent the same amount since both of the pizzas may be of different sizes and size is not defined in the question. #answerwithquality #BAL

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X+2y=6 and x-y=3 solve by linear equations by substitution

nevsk [136]

X+2y=6
x-y=3x=6-2y
x=3+y

6-2y=3+y
3=3y
y=1

x=3+1
x=4
Answer: (4;1)

3 0

3 years ago

The perimeter of a rectangle is equal to 163 yard. The width is 8 more than twice the length. Find the length and width.

Sedaia [141]

Answer:

l = 29.4 yards

w = 66.8 yards

General Formulas and Concepts:

Order of Operations: BPEMDAS
Perimeter of a Rectangle: P = 2w + 2l

Step-by-step explanation:

Step 1: Define

P = 163 yards

w = 2l + 8

l = l

Step 2: Set up equation

P = 2w + 2l

163 = 2(2l + 8) + l

Step 3: Solve for l

Distribute 2: 163 = 4l + 16 + l
Combine like terms: 163 = 5l + 16
Subtract 16 on both sides: 147 = 5l
Divide both sides by 5: 147/5 = l
Rewrite: l = 147/5
Evaluate: l = 29.4 yards

Step 4: Find w

Define: w = 2l + 8
Substitute: w = 2(29.4) + 8
Multiply: w = 58.8 + 8
Add: w = 66.8 yards

4 0

3 years ago

Find the differential coefficient of <img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^

Gemiola [76]

Answer:

$\rm \displaystyle y' = 2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x}$

Step-by-step explanation:

we would like to figure out the differential coefficient of $e^{2x}(1+\ln(x))$

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

$\displaystyle y = {e}^{2x} \cdot (1 + \ln(x) )$

to do so distribute:

$\displaystyle y = {e}^{2x} + \ln(x) \cdot {e}^{2x}$

take derivative in both sides which yields:

$\displaystyle y' = \frac{d}{dx} ( {e}^{2x} + \ln(x) \cdot {e}^{2x} )$

by sum derivation rule we acquire:

$\rm \displaystyle y' = \frac{d}{dx} {e}^{2x} + \frac{d}{dx} \ln(x) \cdot {e}^{2x}$

Part-A: differentiating $e^{2x}$

$\displaystyle \frac{d}{dx} {e}^{2x}$

the rule of composite function derivation is given by:

$\rm\displaystyle \frac{d}{dx} f(g(x)) = \frac{d}{dg} f(g(x)) \times \frac{d}{dx} g(x)$

so let g(x) [2x] be u and transform it:

$\displaystyle \frac{d}{du} {e}^{u} \cdot \frac{d}{dx} 2x$

differentiate:

$\displaystyle {e}^{u} \cdot 2$

substitute back:

$\displaystyle \boxed{2{e}^{2x} }$

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

$\displaystyle \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)$

let

$f(x) \implies \ln(x)$
$g(x) \implies {e}^{2x}$

substitute

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \frac{d}{dx}( \ln(x) ) {e}^{2x} + \ln(x) \frac{d}{dx} {e}^{2x}$

differentiate:

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \boxed{\frac{1}{x} {e}^{2x} + 2\ln(x) {e}^{2x} }$

Final part:

substitute what we got:

$\rm \displaystyle y' = \boxed{2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} }$

and we're done!

6 0

3 years ago

Show that the equation x^3+6x-5=0 has a solution between x=0 and x=1

Mnenie [13.5K]

Answer:

Since the value of f(0) is negative and the value of f(1) is positive, then there is at least one value of x between 0 and 1 for which f(x) =0.

Step-by-step explanation:

The equation f(x) given is:

$f(x) = x^3+6x-5$

For x = 0. the value of the expression is:

$f(0) = 0^3+0-5\\f(0) = -5$

For x = 1, the value of the expression is:

$f(1) = 1^3+6-5\\f(1)=2$

Since the value of f(0) is negative and the value of f(1) is positive, then there is at least one value of x between 0 and 1 for which f(x) =0.

In other words, there is at least one solution for the equation between x=0 and x=1.

6 0

3 years ago

What is 1/6 and 1/8 close to 0, 1 , 1/2

Alex_Xolod [135]

It is closer to ) hope i helped

7 0

4 years ago

Read 2 more answers