Water is flowing at the rate of 15 km/hr through a pipe of diameter 14 cm into a cuboidal pond

Please help! Its for my big test tomorrow!

Travka [436]

Answer:

# The solution x = -5

# The solution is x = 1

# The solution is x = 6.4

# The solution is x = 4

# The solution is 1.7427

# The solution is 0.190757

Step-by-step explanation:

* Lets revise some rules of the exponents and the logarithmic equation

# Exponent rules:

1- b^m × b^n = b^(m + n) ⇒ in multiplication if they have same base

we add the power

2- b^m ÷ b^n = b^(m – n) ⇒ in division if they have same base we

subtract the power

3- (b^m)^n = b^(mn) ⇒ if we have power over power we multiply

them

4- a^m × b^m = (ab)^m ⇒ if we multiply different bases with same

power then we multiply them ad put over the answer the power

5- b^(-m) = 1/(b^m) (for all nonzero real numbers b) ⇒ If we have

negative power we reciprocal the base to get positive power

6- If a^m = a^n , then m = n ⇒ equal bases get equal powers

7- If a^m = b^m , then a = b or m = 0

# Logarithmic rules:

1- $log_{a}b=n-----a^{n}=b$

2- $loga_{1}=0---log_{a}a=1---ln(e)=1$

3- $log_{a}q+log_{a}p=log_{a}qp$

4- $log_{a}q-log_{a}p=log_{a}\frac{q}{p}$

5- $log_{a}q^{n}=nlog_{a}q$

* Now lets solve the problems

# $3^{x+1}=9^{x+3}$

- Change the base 9 to 3²

∴ $9^{x+3}=3^{2(x+3)}=3^{2x+6}$

∴ $3^{x+1}=3^{2x+6}$

- Same bases have equal powers

∴ x + 1 = 2x + 6 ⇒ subtract x and 6 from both sides

∴ 1 - 6 = 2x - x

∴ -5 = x

* The solution x = -5

# ㏒(9x - 2) = ㏒(4x + 3)

- If ㏒(a) = ㏒(b), then a = b

∴ 9x - 2 = 4x + 3 ⇒ subtract 4x from both sides and add 2 to both sides

∴ 5x = 5 ⇒ divide both sides by 5

∴ x = 1

* The solution is x = 1

# $log_{6}(5x+4)=2$

- Use the 1st rule in the logarithmic equation

∴ 6² = 5x + 4

∴ 36 = 5x + 4 ⇒ subtract 4 from both sides

∴ 32 = 5x ⇒ divide both sides by 5

∴ 6.4 = x

* The solution is x = 6.4

# $log_{2}x+log_{2}(x-3)=2$

- Use the rule 3 in the logarithmic equation

∴ $log_{2}x(x-3)=2$

- Use the 1st rule in the logarithmic equation

∴ 2² = x(x - 3) ⇒ simplify

∴ 4 = x² - 3x ⇒ subtract 4 from both sides

∴ x² - 3x - 4 = 0 ⇒ factorize it into two brackets

∴ (x - 4)(x + 1) = 0 ⇒ equate each bract by 0

∴ x - 4 = 0 ⇒ add 4 to both sides

∴ x = 4

OR

∵ x + 1 = 0 ⇒ subtract 1 from both sides

∴ x = -1

- We will reject this answer because when we substitute the value

of x in the given equation we will find $log_{2}(-1)$ and this

value is undefined, there is no logarithm for negative number

* The solution is x = 4

# $log_{4}11.2=x$

- You can use the calculator directly to find x

∴ x = 1.7427

* The solution is 1.7427

# $2e^{8x}=9.2$ ⇒ divide the both sides by 2

∴ $e^{8x}=4.6$

- Insert ln for both sides

∴ $lne^{8x}=ln(4.6)$

- Use the rule $ln(e^{n})=nln(e)$ ⇒ ln(e) = 1

∴ 8x = ln(4.6) ⇒ divide both sides by 8

∴ x = ln(4.6)/8 = 0.190757

* The solution is 0.190757

7 0

3 years ago

Read 2 more answers

Help plz..And No links!! I repeat No links!!

patriot [66]

This is Right answer....

I hope you understand...

Mark me as brainliest....

Thanks

6 0

2 years ago

Read 2 more answers

Independent random samples of vehicles traveling past a given point on an interstate highway have been observed on monday versus

quester [9]

Hi!

To compare this two sets of data, you need to use a t-student test:

You have the following data:

-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph

-Wednesday n2=20; </span>x̄2=56,3 mph; s2=4,4 mph

You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.

$t= \frac{X1-X2}{ \sqrt{ \frac{(n1-1)* s1^{2}+(n2-1)* s2^{2} }{n1+n2-2}} * \sqrt{ \frac{1}{n1}+ \frac{1}{n2}} } =2,2510$

To calculate the degrees of freedom you need to use the following equation:

$df= \frac{ (\frac{ s1^{2}}{n1} + \frac{ s2^{2}}{n2})^{2}}{ \frac{(s1^{2}/n1)^{2}}{n1-1}+ \frac{(s2^{2}/n2)^{2}}{n2-1}}=33,89$ ≈34

The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10

So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.

5 0

3 years ago

What is the poinet-slope form of the equation for the line in the graph?

Ne4ueva [31]

Answer:

$y -5 = \frac{9}{13}(x - 7)$

Step-by-step explanation:

Given :

Two points are given in graph (-6, -4) and {7, 5).

The point-slope form of the equation of a straight line is:

$y -y_{1} = m(x - x_{1})$ ------------(1)

Let $(x_{1}, y_{1})=(7,5)$ and $(x_{2}, y_{2})=(-6,-4)$

The slope of the line $m=\frac{y_{2}- y_{1}}{x_{2}- x_{1}}$

Put all known value in above equation.

$m=\frac{-4- 5}{-6- 7}$

$m=\frac{-9}{-13}$

$m=\frac{9}{13}$

The slope of the line $m=\frac{9}{13}$

We know m, and also know that $(x_{1}, y_{1})=(7,5)$ , so we put these value in equation 1.

$y -5 = \frac{9}{13}(x - 7)$

Therefore, the equation of the line is $y -5 = \frac{9}{13}(x - 7)$ .

6 0

3 years ago

I am confused about how to do this problem and I don't think it should be hard, but I just don't know how to approach it, so if

solniwko [45]

Answer:

a) (0.555, 0) and (6, 0)

b) r = -3 and r = 1.8

c) (0.875, 0.676)

d) (0, 1.235)

Step-by-step explanation:

Set each term in the numerator and denominator equal to 0 and find r.

In the numerator:

r = 7/8, 5/9, or 6

In the denominator:

r = 9/5, 7/8, or -3

Zeros in the numerator that aren't in the denominator are r-intercepts.

Zeros in the denominator that aren't in the numerator are vertical asymptotes.

Zeros in both the numerator and the denominator are holes.

a) (0.555, 0) and (6, 0)

b) r = -3 and r = 1.8

c) Evaluate m(r) at r = 7/8. To do that, first divide out the common term (-8r + 7) from the numerator and denominator.

m(r) = (-9r+5)² (r−6)² / ( (-5r+9)² (r+3)² )

m(⅞) = (-9×⅞+5)² (⅞−6)² / ( (-5×⅞+9)² (⅞+3)² )

m(⅞) = (-23/8)² (-41/8)² / ( (37/8)² (31/8)² )

m(⅞) = (-23)² (-41)² / ( (37)² (31)² )

m(⅞) = 0.676

The hole is at (0.875, 0.676).

d) Evaluate m(r) at r = 0.

m(0) = (-9×0+5)² (0−6)² / ( (-5×0+9)² (0+3)² )

m(0) = (5)² (-6)² / ( (9)² (3)² )

m(0) = 1.235

The m(r)-intercept is (0, 1.235).

5 0

3 years ago