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Alenkasestr [34]

3 years ago

7

A full grain silo empties at a constant rate. Write an expression to determine the amount after 5 seconds and make sure to be pr

ecise! Someone please help me! I need an answer fast! ;-;

1 answer:

Artist 52 [7]3 years ago

7 0

What is the constant rate

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The sum of two numbers is 24 and their difference is . what are the numbers

Darina [25.2K]

X + y = 24
x - y = 2
-------------add
2x = 26
x = 26/2
x = 13

x - y = 2
13 - y = 2
-y = 2 - 13
-y = - 11
y = 11

ur numbers are 11 and 13

7 0

4 years ago

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Which of the following are remote interior angles of one check all that apply?

RUDIKE [14]

Answer:

<4 and <3

Step-by-step explanation:

The remote interior angles are the two angles that are inside the triangle and away from the angle adjacent to the exterior angle.

5 is adjacent to 1

so 3 and 4 are the remote interior angles

6 0

3 years ago

G if v lies in the first quadrant and makes an angle π/3 with the positive x-axis and |v| = 8, find v in component form.

trasher [3.6K]

First, illustrate the problem into a diagram like the one shown in the attached picture. The component vectors are shown in red (y-component) and blue (x-component). Since this is a right triangle, let's apply the trigonometric functions of sine and cos.

y-component, vy:
sin (π/3) = vy/8
vy = 6.928

x-component, vx:
cos (π/3) = vx/8
vx = 4

6 0

4 years ago

MAKE IT 100% CORRECT

ahrayia [7]

I think <1 & <2 are vertical
<3 & <4 are adjacent

7 0

3 years ago

At the beginning of an experiment, a scientist has 300 grams of radioactive goo. After 150 minutes, her sample has decayed to 37

monitta

Answer:

Half-life of the goo is 49.5 minutes

$G(t)= 300e^{-0.014t}$

191.7 grams of goo will remain after 32 minutes

Step-by-step explanation:

Let $M_0\,,\,M_f$ denotes initial and final mass.

$M_0=300\,\,grams\,,\,M_f=37.5\,\,grams$

According to exponential decay,

$\ln \left ( \frac{M_f}{M_0} \right )=-kt$

Here, t denotes time and k denotes decay constant.

$\ln \left ( \frac{M_f}{M_0} \right )=-kt\\\ln \left ( \frac{37.5}{300} \right )=-k(150)\\-2.079=-k(150)\\k=\frac{2.079}{150}=0.014$

So, half-life of the goo in minutes is calculated as follows:

$\ln \left ( \frac{50}{100} \right )=-kt\\\ln \left ( \frac{50}{100} \right )=-(0.014)t\\t=\frac{-0.693}{-0.014}=49.5\,\,minutes$

Half-life of the goo is 49.5 minutes

$\ln \left ( \frac{M_f}{M_0} \right )=-kt\Rightarrow M_f=M_0e^{-kt}$

So,

$G(t)= M_f=M_0e^{-kt}$

Put $M_0=300\,\,grams\,,\,k=0.014$

$G(t)= 300e^{-0.014t}$

Put t = 32 minutes

$G(32)= 300e^{-0.014(32)}=300e^{-0.448}=191.7\,\,grams$

7 0

3 years ago