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2 years ago

8

What did everyone say after the bumpy airplane landing

1 answer:

Nikitich [7]2 years ago

8 0

<span>the thing that everyone say after the bumpy airplane landing was : it's not the airplane's fault and it's not the pilot's fault so it's probably the asphalt To be fair, there is no way for the passangers to know what may have caused the bump in landing. Everyone only created their own assumptions based om their personal perspective</span>

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???whats the answers

Aleks04 [339]

11, 3, -3, -12
Hope this helps :)

3 0

3 years ago

See attachment below

KatRina [158]

Answer: Third Option

$x=1.469743$

Step-by-step explanation:

We have the following exponential equation

$3^{x+1}=15$

We must solve the equation for the variable x

To clear the variable x apply the $log_3$ function on both sides of the equation

$log_3(3^{x+1})=log_3(15)$

Simplifying we get the following:

$x+1=log_3(15)$

To simplify the expression $log_3 (15)$ we apply the base change property

$log_b(y)=\frac{log(y)}{log(b)}$

This means that:

$log_3 (15)=\frac{log(15)}{log(3)}$

Then:

$x+1=\frac{log(15)}{log(3)}$

$x=\frac{log(15)}{log(3)}-1$

$x=1.469743$

6 0

2 years ago

-3x+2-5x-7=4x+2 what is the first step solve this<br> equation?

Triss [41]

Answer: x = -0.58

Step-by-step explanation:

-3x+2-5x-7=4x+2

First step, collect like terms

-3x -5x-4x = 2-2+7

-12x = 7

x = 7/-12

x = -0.58

5 0

2 years ago

Someone help me with this<br> Find the 10th term of the sequence -63, -93, -123

Katarina [22]

Answer:

-333

Step-by-step explanation:

It seems like it is going down by -30 each term.

6 0

3 years ago

Please I need help with this!!!!!!!

inysia [295]

Answer:

Step-by-step explanation:

1). Step 4:

$x=5^{\frac{4}{3}}=(5^4)^{\frac{1}{3}}$

$x=\sqrt[3]{5^4}$ [Since, $a^{\frac{1}{3}}=\sqrt[3]{a}$ ]

$x=\sqrt[3]{5\times 5\times 5\times 5}$

Step 5:

$x=\sqrt[3]{(5)^3\times 5}$

$x=\sqrt[3]{5^3}\times \sqrt[3]{5}$

2). He simplified the expression by removing exponents from the given expression.

3). Let the radical equation is,

$(3x-1)^{\frac{1}{5}}=2$

Step 1:

$(3x-1)^{\frac{1}{5}\times \frac{5}{1} }=2^{\frac{5}{1}}$

Step 2:

$(3x-1)=2^5$

Step 3:

$3x=32+1$

Step 4:

$x=11$

4). By substituting $x=11$ in the original equation.

$(3\times 11-1)^{\frac{1}{5}}=(32)^\frac{1}{5}$

$=(2^5)^\frac{1}{5}$

$=2$

There is no extraneous solution.

3 0

2 years ago