Ask question

Ask question

All categories

4 years ago

14

The weight of an African elephant is 6 × 106 grams; the weight of a tiger is 3 × 105 grams. How many times heavier is the elepha

nt than the tiger?
A. 2 grams
B.20 grams
C.200 grams
D. 2,000 grams

1 answer:

Otrada [13]4 years ago

6 0

I got 20. 6 x 10^6/3 x 10^5 = 2 x 10^1 = 2 x 10 = 20

You might be interested in

Simplify the expression 5+(4+x)

photoshop1234 [79]

5 + 4x
I think that’s right

4 0

3 years ago

Read 2 more answers

A painting without its frame has a width 2.5 times its height, h, in inches. Its frame is 3 in. wide all along its perimeter. Wh

Lena [83]

Answer:

You will not be able to find the area of the framed painting without finding the length, which you did not provide. Assuming you meant “3 inches wide” to actually mean the length then the surface area of the painting is 9 inches squared. However, to find the height, you must divide 3 by 2.5 as the width was said to be 2.5 times the height of the painting. This will give you an answer of 1.2 inches in height.

7 0

3 years ago

If 124 is subtracted from the square of a number the result is 200 what is the number

BartSMP [9]

Answer:

x=18

Step-by-step explanation:

let the number =x

then

x²-124=200

x²=200+124

x²=324

x=18

7 0

3 years ago

Is it A B or C????????????????????????

Andre45 [30]

I am not sure but I think it's C

5 0

3 years ago

Read 2 more answers

(1 point) Find the length traced out along the parametric curve x=cos(cos(4t))x=cos⁡(cos⁡(4t)), y=sin(cos(4t))y=sin⁡(cos⁡(4t)) a

Mazyrski [523]

The length of a curve $C$ given parametrically by $(x(t),y(t))$ over some domain $t\in[a,b]$ is

$\displaystyle\int_C\mathrm ds=\int_a^b\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt$

In this case,

$x(t)=\cos(\cos4t)\implies\dfrac{\mathrm dx}{\mathrm dt}=-\sin(\cos4t)(-\sin4t)(4)=4\sin4t\sin(\cos4t)$

$y(t)=\sin(\cos4t)\implies\dfrac{\mathrm dy}{\mathrm dt}=\cos(\cos4t)(-\sin4t)(4)=-4\sin4t\cos(\cos4t)$

So we have

$\displaystyle\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2=16\sin^24t\sin^2(\cos4t)+16\sin^24t\cos^2(\cos4t)=16\sin^24t$

and the arc length is

$\displaystyle\int_0^1\sqrt{16\sin^24t}\,\mathrm dt=4\int_0^1|\sin4t|\,\mathrm dt$

We have

$\sin(4t)=0\implies4t=n\pi\implies t=\dfrac{n\pi}4$

where $n$ is any integer; this tells us $\sin(4t)\ge0$ on the interval $\left[0,\frac\pi4\right]$ and $\sin(4t)$ on $\left[\frac\pi4,1\right]$ . So the arc length is

$=\displaystyle4\left(\int_0^{\pi/4}\sin4t\,\mathrm dt-\int_{\pi/4}^1\sin4t\,\mathrm dt\right)$

$=-\cos(4t)\bigg_0^{\pi/4}-\left(-\cos(4t)\bigg_{\pi/4}^1\right)$

$=(\cos0-\cos\pi)+(\cos4-\cos\pi)=\boxed{3+\cos4}$

7 0

3 years ago