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3 years ago

9

Thank you for the help!

1 answer:

horrorfan [7]3 years ago

7 0

Answer:

see explanation

Step-by-step explanation:

cosM = $\frac{adjacent}{hypotenuse}$ = $\frac{MN}{LM}$ = $\frac{8}{17}$

sinM = $\frac{opposite}{hypotenuse}$ = $\frac{LN}{LM}$ = $\frac{15}{17}$

tanM = $\frac{opposite}{adjacent}$ = $\frac{LN}{MN}$ = $\frac{15}{8}$

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Which fraction would you find marked on a ruler?

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3 years ago

Drag the tiles to the correct boxes to complete the pairs.

Elza [17]

Answer:

Part 1) -7 → $\dfrac{y^2-y+6}{-2(y+7)}$

Part 2) 3/2 → $\dfrac{y^2-2y+1}{2y-3}$

Part 3) 1 → $\dfrac{5y^2-6y+1}{-5(y-1)}$

Part 4) -1/4 → $\dfrac{y(y+5)}{4y+1}$

Step-by-step explanation:

<u><em>The complete question in the attached figure</em></u>

we know that

To find out the non permissible replacements for y, equate the denominator of each expression equal to 0.

step 1

we have

$\dfrac{y^2-2y+1}{2y-3}$

Equate (2y-3) equal to 0.

$2y-3=0$

solve for y

Adds 3 both sides.

$2y=3$

Divide both sides by 2.

$y=\dfrac{3}{2}$

therefore

3/2 is the non permissible replacement for y.

step 2

we have

$\dfrac{y(y+5)}{4y+1}$

Equate (4y+1) equal to 0.

$4y+1=0$

Subtract 1 both sides

$4y=-1$

Divide by 4 both sides

$y=-\dfrac{1}{4}$

therefore

-1/4 is the non permissible replacement for y.

step 3

we have

$\dfrac{5y^2-6y+1}{-5(y-1)}$

Equate -5(y-1) equal to 0.

$-5(y-1)=0$

Divide by -5 both sides

$y-1=0$

Adds 1 both sides

$y=1$

therefore

1 is the non permissible replacement for y.

step 4

we have

$\dfrac{y^2-y+6}{-2(y+7)}$

Equate -2(y+7) equal to 0.

$-2(y+7)=0$

Divide by -2 both sides

$y+7=0$

Subtract 7 both sides

$y=-7$

therefore

-7 is the non permissible replacement for y

7 0

3 years ago

If F(x) = f(x)g(x), where f and g have derivatives of all orders

user100 [1]

a. By the product rule,

$F=fg\implies F'=f'g+fg'$

$\implies F''=(f''g+f'g')+(f'g'+fg'')=f''g+2f'g'+fg''$

b. By the same rule,

$F'''=(f'''g+f''g')+2(f''g'+f'g'')+(f'g''+fg''')$

$F'''=f'''g+3f''g'+3f'g''+fg'''$

and

$F^{(4)}=(f^{(4)}g+f'''g')+3(f'''g'+f''g'')+3(f''g''+f'g''')+(f'g'''+fg^{(4)})$

$F^{(4)}=f^{(4)}g+4f'''g'+6f''g''+4f'g'''+fg^{(4)}$

c. You might recognize the coefficients as those that appear in the expansion of $(a+b)^n$ :

1, 1

1, 2, 1

1, 3, 3, 1

1, 4, 6, 4, 1

and so on; the general pattern (known as the general Leibniz rule) is

$F^{(n)}=\displaystyle\sum_{k=0}^n\binom nkf^{(n-k)}g^{(k)}$

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3 years ago