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3 years ago

V=1/3bh for B please help and explain!!!

1 answer:

4 0

To solve for V=1/3bh we need to do this:

v=bh/3
v * 3 = bh
3v = bh
b = 3v/h

Explanation: first reduce the fraction 1/3bh to bh/3, next multiply both sides by 3, then regroup the terms, and lastly divide both sides by h and u are done.

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What is the value of x4 + 5.2y, when x = 2 and y = 3? (10 point) Question 12 options: 1) 23.6 2) 24.6 3) 27.6 4) 39.6

Firdavs [7]

Answer:

23.6

Step-by-step explanation:

7 0

3 years ago

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What is the output value of (4x+2)(x-5)/2 using input value of x=2

Firdavs [7]

Answer:

-15

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Step-by-step explanation:

Step 1: Define

(4x + 2)(x - 5) / 2

x = 2

Step 2: Evaluate

Substitute: (4 · 2 + 2)(2 - 5) / 2
Parenthesis - Multiply: (8 + 2)(2 - 5) / 2
Parenthesis - Add: 10(2 - 5) / 2
Parenthesis - Subtract: 10(-3) / 2
Multiply: -30 / 2
Divide: -15

6 0

3 years ago

Read 2 more answers

How many solutions does the equation have?

rjkz [21]

Answer:

it only has 1 :).........

6 0

3 years ago

Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

5 0

3 years ago

Three shipping companies want to compare the mean numbers of deliveries their drivers complete in a day.

Marianna [84]

Answer:

#1 is 10

# 2 is

6 + 7 + 8 + 9 + 10 + 10 + 10 + 12 + 14 + 14 = 100/10. The (Mean) = 10

(6- 10) + (7- 10) + (8- 10) + (9- 10) + (10- 10) + (10- 10) + (10- 10) + (12- 10) + (14- 10)

4 + 3 + 2 + 1 + 0 + 0 + 0 + 2 + 4

= 16/10 = 1 6/10

The (MAD) = 1 6/10

#3 Groups A & B

#4 sorry I can’t figure this out but I did 1/2/3

3 0

3 years ago