Ask question

Ask question

All categories

3 years ago

5

Scarface is one of the best movies here's my drawing but it's unfinished

1 answer:

saul85 [17]3 years ago

3 0

Answer:

nice work hope to see it once its finished!

Step-by-step explanation:

You might be interested in

Find the dy/dx for the equation below using implicit differentiation. Tq

KengaRu [80]

The derivative of an exponential function is

$\dfrac{d}{dx} a^x = a^x\log(a)$

Moreover, in this case we must apply the chain rule because the exponent is -x:

$\dfrac{d}{dx} 3^{-x} = 3^{-x}\log(3)\cdot (-1)=-3^{-x}\log(3)$

7 0

3 years ago

Estimate the square root of 20 to the nearest integer and to the nearest tenth

KengaRu [80]

√20 = 4.47213

√20 = 4 (nearest integer)

√20 = 4.5 (nearest tenth)

3 0

4 years ago

Read 2 more answers

Working alone, Elisa can sweep a porch in 11 minutes. Pranav can sweep the same porch in 13 minutes. How long would it take if t

GaryK [48]

Elisa, e = 11mins

Prenav,p = 13mins

time to complete, t

1 = t( 1/11 + 1/13) = t (24/143)

t = 143/24 = 5.96min

Answer is A 5.96mins

5 0

3 years ago

to make 5 apple pies you need about 2 pounds if apples how many pounds of apples do you need to make 20 apple pies

scoundrel [369]

8 apples

20 apples/5 apple pies +4
4* 2 pounds = 8 pounds

6 0

3 years ago

Read 2 more answers

If α and β are the zeroes of the polynomial 6y 2 − 7y + 2, find a quadratic polynomial whose zeroes are 1 α and 1 β .

ollegr [7]

Answer:

$2y^2-7y+6=0$

Step-by-step explanation:

We are given that $\alpha$ and $\beta$ are the zeroes of the polynomial $6y^2-7y+2$

$y^2-\frac{7}{6}y+\frac{1}{3}$

We have to find a quadratic polynomial whose zeroes are $1/\alpha$ and $1/\beta$ .

General quadratic equation

$x^2-(sum\;of\;zeroes)x+ product\;of\;zeroes$

We get

$\alpha+\beta=\frac{7}{6}$

$\alpha \beta=\frac{1}{3}$

$\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}$

$\frac{1}{\alpha}+\frac{1}{\beta}=\frac{7/6}{1/3}$

$\frac{1}{\alpha}+\frac{1}{\beta}=\frac{7}{6}\times 3=7/2$

$\frac{1}{\alpha}\times \frac{1}{\beta}=\frac{1}{\alpha \beta}$

$\frac{1}{\alpha}\times \frac{1}{\beta}=\frac{1}{1/3}=3$

Substitute the values

$y^2-(7/2)y+3=0$

$2y^2-7y+6=0$

Hence, the quadratic polynomial whose zeroes are $1/\alpha$ and $1/\beta$ is given by

$2y^2-7y+6=0$

4 0

3 years ago