What is the scale factor from the original design to the enlarged design?

Mathematics

2 answers:

bazaltina [42]1 year ago

5 0

The answer is 2 due to the fact that 2 is even, and can be added more than once.

Ket [755]1 year ago

3 0

2. Each of the larger image is multiplied by two times the amount of its previous size.

You might be interested in

Write the equation of a horizontal line that passes through the point (-2,2)

Serjik [45]

Y=2
If u r finding the equation of horizontal lines it is y=
If u r finding the equation of horizontal lines it is x=
Then just see what point the horizontal or verticals line I passing through, in this case it is positive 2

5 0

2 years ago

The price of a small pizza increased from $6 to $9 what is the percent of increase in the cost of the pizza? Round to the neares

horrorfan [7]

Answer:

50%

Step-by-step explanation:

Since it was $6, half of $6 is $3. So, $6 to $9 is a $3 difference, which is 1/2. 1/2 is a 50% increase.

7 0

2 years ago

Determine whether each integral is convergent or divergent. If it is convergent evaluate it. (a) integral from 1^(infinity) e^(-

AVprozaik [17]

Answer:

a) So, this integral is convergent.

b) So, this integral is divergent.

c) So, this integral is divergent.

Step-by-step explanation:

We calculate the next integrals:

$\int_1^{\infty} e^{-2x} dx=\left[-\frac{e^{-2x}}{2}\right]_1^{\infty}\\\\\int_1^{\infty} e^{-2x} dx=-\frac{e^{-\infty}}{2}+\frac{e^{-2}}{2}\\\\\int_1^{\infty} e^{-2x} dx=\frac{e^{-2}}{2}\\$

So, this integral is convergent.

$\int_1^{2}\frac{dz}{(z-1)^2}=\left[-\frac{1}{z-1}\right]_1^2\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\frac{1}{1-1}+\frac{1}{2-1}\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\infty\\$

So, this integral is divergent.

$\int_1^{\infty} \frac{dx}{\sqrt{x}}=\left[2\sqrt{x}\right]_1^{\infty}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=2\sqrt{\infty}-2\sqrt{1}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=\infty\\$