All categories

3 years ago

If the length and width decrease by a factor of 3, what would be the effect on the area of the square?

1 answer:

6 0

So, if the scale factor is three, then the dimensions of the new shape will be three times larger than ... What impact did this have on the perimeter? ... The area of a rectangle is length times width, so it's 5 x 3, which is 15. ... When you're working with scale factors, square the scale factor to determine the area of the new figure.

You might be interested in

Find the measure of the missing angle.

melamori03 [73]

Answer:

It's 155

Step-by-step explanation:

Since the whole angle is 180

And u got 25

So 180-25=155

4 0

2 years ago

Read 2 more answers

Mario invested $5100 at 13%

cestrela7 [59]

Answer:

Amount = $6614 and 19 cent

Step-by-step explanation:

Formula for compound interest is

A= p(1+r/n)^(nt)

Compounded daily

So n= 365*2= 730

T= 2

r= 0.13

P= 5100

A= p(1+r/n)^(nt)

A= 5100(1+0.13/730)^(730*2)

A= 5100(1+1.78082*10^-4)^(1460)

A= 5100(1.000178082)^1460

A= 5100(1.2969)

A= 6614.19

Amount = $6614 and 19 cent

6 0

3 years ago

F(x) = -3(x - 11)(x + 1)

il63 [147K]

Answer:

Step-by-step explanation:

8 0

3 years ago

Fritz is recording the decay of radioactive material. The table displays the

damaskus [11]

Answer:

Every week, the radioactivity decreases by a factor of 5

After 1 week, it goes from 5,000 to 1,000

2 weeks 1,000 / 5 =200

3 weeks 200 / 5 = 40

4 weeks 40 / 5 = 8

5 weeks 8 / 5 = 1.6

6 weeks 1.6 / 5 = 0.064

7 weeks 0.064 / 5 = 0.0128

8 weeks .0128 / 5 = 0.00256

9 weeks 0.00256 / 5 = 0.000512

10 weeks 0.000512 / 5 = 0.0001024

Step-by-step explanation:

3 0

2 years ago

Use the four-step definition of the derivative to find f'(x) if f(x) = −4x^3 −1.

guapka [62]

$\stackrel{de finition \textit{ of a derivative as a limit}}{\lim\limits_{h\to 0}~\cfrac{f(x+h)-f(x)}{h}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{[-4(x+h)^3-1]~~ - ~~[-4x^3-1]}{h} \\\\\\ \cfrac{[-4(x^3+3x^2h+3xh^2+h^3)-1]~~ ~~+4x^3+1}{h} \\\\\\ \cfrac{[-4x^3-12x^2h-12xh^2-4h^3-1]~~ ~~+4x^3+1}{h} \\\\\\ \cfrac{-4x^3-12x^2h-12xh^2-4h^3-1+4x^3+1}{h}\implies \cfrac{-12x^2h-12xh^2-4h^3}{h}$

$\cfrac{h(-12x^2-12xh-4h^2)}{h}\implies -12x^2-12xh-4h^2 \\\\\\ \lim\limits_{h\to 0}~-12x^2-12xh-4h^2\implies \lim\limits_{h\to 0}~-12x^2-12x(0)-4(0)^2 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \lim\limits_{h\to 0}~-12x^2~\hfill$

7 0

2 years ago