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jarptica [38.1K]

3 years ago

14

Figure 1 is dilated to get Figure 2. What is the scale factor? can you simplify it to please

2 answers:

JulijaS [17]3 years ago

6 0

It would be times 1/3

12345 [234]3 years ago

6 0

It’s would be 1/3 explanation

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Tonya and Leo each bought a cell phone at the same time. The trade-in values, in dollars, of the cell phones are modeled by the

Paha777 [63]

Answer:

The answer is Tonya's phone had the greater initial trade-in value.

Leo's phone decreases at an average rate slower than the trade in value of Tonya's phone.

Step-by-step explanation:

Given

Tonya

$f(x) = 490 * 0.88^x$

Leo

$x \to g(x)$

$0 \to 480$

$2 \to 360$

$4 \to 470$

Solving (a): The phone with greater initial value

The initial value is when x = 0. So, we have:

$f(x) = 490 * 0.88^x$

$f(0) = 490 * 0.88^0$

$f(0) = 490 * 1$

$f(0) = 490$

From Leo's table

$g(0) = 480$

By comparison;

$f(0) > g(0)$

i.e.

$490 > 480$

<em>So: Tonya's had the greater initial trade-in value</em>

Solving (b): The phone with lesser rate

An exponential function is:

$y = ab^x$

Where:

$b \to rate$

For Tonya

$b = 0.88$

For Leo, we have:

$(x_1,y_1) = (0,480)$

$(x_2,y_2) = (2,360)$

So, the equation becomes:

$y = ab^x$

$480 = ab^0$ and $360 = ab^2$

Solving $480 = ab^0$ , we have:

$480 = a * 1$

$480 = a$

$a= 480$

$360 = ab^2$ becomes

$360 = 480 * b^2$

Divide both sides by 480

$0.75 = b^2$

Take square roots

$0.87 = b$

$b=0.87$ -- Leo's rate

By comparison; Leo's rate is slower i.e. 0.87 < 0.88

6 0

3 years ago

Find dy/dx if y =x^3+5x+2/x²-1

stiks02 [169]

<u>Differentiate using the Quotient Rule</u> –

$\qquad$ $\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\$

According to the given question, we have –

f(x) = x^3+5x+2
g(x) = x^2-1

Let's solve it!

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2) \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5) - ( x^3+5x+2) 2x}{(x^2-1)^2 }\\$

$\qquad$ $\pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\$

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\$

$\qquad$ $\pink{\therefore \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}} = \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\$

7 0

2 years ago