If a student ate 3/4 of their meals away from home, what percent of the total day is spent other than at home?

Robbie went parasailing. his sail reached 141 feet. it descended 22 feet before staying in the same place for 10 minutes. what w

mash [69]

141 minus 22 equals 119

5 0

2 years ago

In ΔGHI, the measure of ∠I=90°, IG = 6.8 feet, and HI = 2.6 feet. Find the measure of ∠G to the nearest degree.

Luda [366]

Answer:

$\angle G=20.8\approx 21^{\circ}$

Step-by-step explanation:

Given: In ΔGHI, $\angle I$ =90°, IG = 6.8 feet, and HI = 2.6 feet

To find: $\angle G$

Solution:

Trigonometry defines relationship between the sides and angles of the triangle.

For any angle $\theta$ ,

$tan\theta$ = side opposite to $\theta$ /side adjacent to $\theta$

In ΔGHI,

$tan G=\frac{HI}{GI}$

Put $HI=2.6 \,\,feet\,,\,GI=6.8\,\,feet$

So,

$tan G=\frac{2.6}{6.8}=0.38$

Therefore, $\angle G=20.8\approx 21^{\circ}$

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3 0

2 years ago

James purchased five bonds of face value of $1,000 that paid 5% annual interest rate. the total annual interest income of james

Trava [24]

Given:
5 bonds of face value of 1,000 that paid 5% annual interest rate.

5 bonds x 1,000 = 5,000
5,000 x 5% x 1 year = 250

The total annual interest income of James is 250. Each bond earns 50 per annum.

8 0

2 years ago

Read 2 more answers

In one year, the McAlister family drove their car 15,680 miles. To the nearest thousand, how many miles did they drive their car

Archy [21]

The answer is 16,000

8 0

2 years ago

Read 2 more answers

One method of slowing the growth of an insect population without using pesticides is to introduce into the population a number o

Zielflug [23.3K]

To be clear, the given relation between time and female population is an integral:
 $t = \int { \frac{P+S}{P[(r - 1)P - S]} } \,
dP$

The problem says that r = 1.2 and S = 400, therefore substituting:
 $t = \int { \frac{P+400}{P[(1.2 - 1)P - 400]}
} \, dP$ 

= $
\int { \frac{P+400}{P(0.2P - 400)} } \, dP$

In order to evaluate this integral, we need to write this rational function in a simpler way:
 $\frac{P+400}{P(0.2P - 400)} = \frac{A}{P} +
\frac{B}{(0.2P - 400)}$ 

where we need to evaluate A and B. In order to do so, let's calculate the LCD:
 $\frac{P+400}{P(0.2P - 400)} = \frac{A(0.2P -
400)}{P(0.2P - 400)} + \frac{BP}{P(0.2P - 400)}$ 

the denominators cancel out and we get:
P + 400 = 0.2AP - 400A + BP
 = P(0.2A + B) - 400A

The two sides must be equal to each other, bringing the system:
 $\left \{ {{0.2A + B = 1} \atop {-400A =
400}} \right.$ 

Which can be easily solved:
 $\left \{ {{B=1.2} \atop {A=-1}} \right.$ 

Therefore, our integral can be written as:
 $t = \int { \frac{P+400}{P(0.2P - 400)} } \,
dP = \int {( \frac{-1}{P} + \frac{1.2}{0.2P-400} )} \, dP$ 
= $- \int { \frac{1}{P} \, dP +
1.2\int { \frac{1}{0.2P-400} } \, dP$ 
= $- \int { \frac{1}{P} \, dP +
6\int { \frac{0.2}{0.2P-400} } \, dP$ 
= - ln |P| + 6 ln |0.2P - 400| + C

Now, let’s evaluate C by considering that at t = 0 P = 10000:
0 = - ln |10000| + 6 ln |0.2(10000) - 400| + C
C = ln |10000| - 6 ln |1600|
C = ln (10⁴) - 6 ln (2⁶·5²)
C = 4 ln (10) - 36 ln (2) - 12 ln (5) 
 Therefore, the equation relating female population with time requested is:
t = - ln |P| + 6 ln |0.2P - 400| + 4 ln (10) - 36 ln (2) - 12 ln (5)

8 0

2 years ago