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3 years ago

Solve for x: 10^x = 9A. x = log⁡ 9B. x = log^9 ⁡10C. x = log ⁡10

Mathematics

2 answers:

romanna [79]3 years ago

4 0

Answer:

A. x = log 9

Step-by-step explanation:

10^x = 9

<u>Log both sides:</u>

log 10^x = log 9
x log 10 = log 9
x*1 = log 9
x = log 9

Correct choice is A

Tpy6a [65]3 years ago

4 0

Answer:

A x=9

it's right answer ok

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Can you please help me with 22.? (BE SERIOUS) (ITS DUE TOMORROW)

attashe74 [19]

pool has 45000 gallons. It loses 780 gallons per hour, so the amount at any given time:

45000 - 780h, where h is the number of hours.

The chemical could be added once the amount of gallons in the pool hits 15000. So the equation to solve:

45000 - 780h = 15000

solve:

30000 = 780h

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3 years ago

9. Plot the points A (-2,3), B(2,-3), C(2,3) ,D(-2,3) join the points and find the distance between each point.

pav-90 [236]

Answer:

AB=7.21 unit

BC=6 unit

CD=7.21 unit

AD= 6 unit

AC=4 unit

BD=4 unit

Step-by-step explanation:

Coordinates of A =(-2,3)

Coordinates of B = (2,-3)

Coordinates of C = (2,3)

Coordinates of D =(-2,-3)

Distance formula : $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$A=(x_1,y_1)=(-2,3)\\B=(x_2,y_2)=(2,-3)\\AB = \sqrt{(2+2)^2+(-3-3)^2}$

AB=7.21 unit

$B=(x_1,y_1)=(2,-3)\\C=(x_2,y_2)=(2,3)\\BC=\sqrt{(2-2)^2+(3+3)^2}$

BC=6

$C=(x_1,y_1)=(2,3)\\D=(x_2,y_2)=(-2,-3)\\CD=\sqrt{(-2-2)^2+(-3-3)^2}$

CD=7.21

$A=(x_1,y_1)=(-2,3)\\D=(x_2,y_2)=(-2,-3)\\AD=\sqrt{(-2+2)^2+(-3-3)^2}$

AD=6

$A=(x_1,y_1)=(-2,3)\\C=(x_2,y_2)=(2,3)$

$AC=\sqrt{(2+2)^2+(3-3)^2}$

AC=4

$B=(x_1,y_1)=(2,-3)\\D=(x_2,y_2)=(-2,-3)$

BD= $\sqrt{(-2-2)^2+(-3+3)^2}$

BD=4

4 0

3 years ago

Not? Is t(n)=2⋅3n

Vera_Pavlovna [14]

Answer:

Step-by-step explanation: I am used to describing arithmetic sequences like this:

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3 years ago

Bacteria usually reproduce by a process called binary fission. In this type of reproduction, one bacterium divides to

Fed [463]

Answer:

$\large \boxed{\text{0.0462 min}^{-1}}$

Step-by-step explanation:

One formula for the reproduction of bacteria is

$\dfrac{N}{N_{0}} = e^{kt}$

Data:

N = 2

N₀ = 1

t = 15 min

Calculation:

$\begin{array}{rcl}\dfrac{N}{N_{0}}& =& e^{kt}\\\\\dfrac{2}{1}& = & e^{k\times15 \text{ min}}\\\\2 & = & e^{15k \text{ min}}\\\ln 2 & = & 15k \text{ min}\\k & = & \dfrac{\ln2 }{\text{15 min}}\\\\& = & \textbf{0.0462 min}^{\mathbf{-1}}\\\\\end{array}\\\text{The value of the constant k is $\large \boxed{\textbf{0.0462 min }^{\mathbf{-1}}}$}$