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3 years ago

11

The ratio of pens to pencils that Katy has on his desk is 6:5. If Katy has 15 pencils on his desk then how many pens does he hav

e?

2 answers:

Marina CMI [18]3 years ago

8 0

Answer:

18 pens

Step-by-step explanation:

6:5 so

5x3=15

6x3=18

polet [3.4K]3 years ago

6 0

Answer:

18

Step-by-step explanation:

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Simplify the rational expression. state any excluded values. 4x - 4/ x - 1

grin007 [14]

We need to simplify this expression:

$\frac{4x-4}{x-1}$

So, we will call this expression as:
$f(x) = \frac{4x-4}{x-1}$

We can write this equation like this:

$f(x) = \frac{4(x-1)}{(x-1)}$

So, if we simplify it, this can be written like this:

f(x) = 4 but given that the denominator can't be zero, then:
$x-1 \neq 0$ ∴ $x \neq 1$

Therefore:
f(x) = 4 if and only if $x \neq 1$

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3 years ago

Anyone can help me? Ive not done this..

kow [346]

Answer:

∠CED = 48°

Step-by-step explanation:

By adding all angles together, they’ll eventually equal ∠AED. Hence, sum of all given angles = ∠AED

∠AEB + ∠BEC + ∠CED = ∠AED

29°+12°+x = 89°

Solve for x and we will get measure of ∠CED:

41°+x = 89°

x = 89°-41°

x = 48°

Therefore, the measure of ∠CED is 48°

5 0

2 years ago

Who can help me d e f thanks

12345 [234]

d)

$y = (2ax^2 + c)^2 (bx^2 - cx)^{-1}$

Product rule:

$y' = \bigg((2ax^2+c)^2\bigg)' (bx^2-cx)^{-1} + (2ax^2+c)^2 \bigg((bx^2-cx)^{-1}\bigg)'$

Chain and power rules:

$y' = 2(2ax^2+c)\bigg(2ax^2+c\bigg)' (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} \bigg(bx^2-cx\bigg)'$

Power rule:

$y' = 2(2ax^2+c)(4ax) (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} (2bx - c)$

Now simplify.

$y' = \dfrac{8ax (2ax^2+c)}{bx^2 - cx} - \dfrac{(2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}$

$y' = \dfrac{8ax (2ax^2+c) (bx^2 - cx) - (2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}$

e)

$y = \dfrac{3bx + ac}{\sqrt{ax}}$

Quotient rule:

$y' = \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{\left(\sqrt{ax}\right)^2}$

$y'= \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{ax}$

Power rule:

$y' = \dfrac{3b \sqrt{ax} - (3bx+ac) \left(-\frac12 \sqrt a \, x^{-1/2}\right)}{ax}$

Now simplify.

$y' = \dfrac{3b \sqrt a \, x^{1/2} + \frac{\sqrt a}2 (3bx+ac) x^{-1/2}}{ax}$

$y' = \dfrac{6bx + 3bx+ac}{2\sqrt a\, x^{3/2}}$

$y' = \dfrac{9bx+ac}{2\sqrt a\, x^{3/2}}$

f)

$y = \sin^2(ax+b)$

Chain rule:

$y' = 2 \sin(ax+b) \bigg(\sin(ax+b)\bigg)'$

$y' = 2 \sin(ax+b) \cos(ax+b) \bigg(ax+b\bigg)'$

$y' = 2a \sin(ax+b) \cos(ax+b)$

We can further simplify this to

$y' = a \sin(2(ax+b))$

using the double angle identity for sine.

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2 years ago

Anything for this one?

Sergeeva-Olga [200]

Answer:

A. -2

Step-by-step explanation:

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Answer:

(C is your answer) -Raymond :3

Step-by-step explanation:

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