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2 years ago

8. Which statement can be represented by the equation shown? 9x-2=8x

Mathematics

1 answer:

Thepotemich [5.8K]2 years ago

3 0

Answer:

A or the 1st one

Step-by-step explanation:

Angelo is making 9 dollars per hour but, loses 2 dollars due to the bus fee. So in math form that would be:

9x - 2

And the second part is Tom makes 8 dollars per hour but doesn't have to pay for bus fee. In math form that would be;

So put that together that would be

9x - 2 = 8x

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A line segment has more then one midpoint true or false

Anika [276]

Answer:

false

Step-by-step explanation:

any line segment or distance has only one midpoint.

there is only one middle to everything.

how would that look like, if something had 2 middles ?

then they would not be a middle, and the middle between these 2 middles would then be the real middle ...

6 0

1 year ago

Read 2 more answers

Simplify each only using positive exponents:<br> 2x^-3 • 4x^2<br> 2x^4 • 4x^-3<br> 2x^3y^-3 • 2x

erica [24]

<h2>Answer:</h2>

$\frac{2}{x}$

$\frac{x}{2}$

$\frac{4x^4}{y^3}$

<h2>Step-by-step explanation:</h2>

a. 2x^-3 • 4x^2

To solve this using only positive exponents, follow these steps:

i. Rewrite the expression in a clearer form

2x⁻³ . 4x²

ii. The position of the term with negative exponent is changed from denominator to numerator or numerator to denominator depending on its initial position. If it is at the numerator, it is moved to the denominator. If otherwise it is at the denominator, it is moved to the numerator. When this is done, the negative exponent is changed to positive.

In our case, the first term has a negative exponent and it is at the numerator. We therefore move it to the denominator and change the negative exponent to positive as follows;

$\frac{1}{2x^3} . 4x^2$

iii. We then solve the result as follows;

$\frac{1}{2x^3} . 4x^2$ = $\frac{2}{x}$

Therefore, 2x⁻³ . 4x² = $\frac{2}{x}$

b. 2x^4 • 4x^-3

i. Rewrite as follows;

2x⁴ . 4x⁻³

ii. The second term has a negative exponent, therefore swap its position and change the negative exponent to a positive one.

$2x^4 . \frac{1}{4x^3}$

iii. Now solve by cancelling out common terms in the numerator and denominator. So we have;

$\frac{x}{2}$

Therefore, 2x⁴ . 4x⁻³ = $\frac{x}{2}$

c. 2x^3y^-3 • 2x

i. Rewrite as follows;

2x³y⁻³ . 2x

ii. Change position of terms with negative exponents;

$2x^3.\frac{1}{y^3} .2x$

iii. Now solve;

$\frac{4x^4}{y^3}$

Therefore, 2x³y⁻³ . 2x = $\frac{4x^4}{y^3}$

8 0

2 years ago

Which phrase is represented by the expression 5 % (36 + 9)?

AnnZ [28]

Answer:

36+9=45+5%=47.25

Step-by-step explanation:

3 0

3 years ago

The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nu

gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt} =-\lambda N$

$\Rightarrow \frac{dN}{N} =-\lambda dt$

Integrating both sides

$\int \frac{dN}{N} =\int\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$\Rightarrow ln|N_0|=-\lambda .0+c$

$\Rightarrow c= ln|N_0|$

$\therefore ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$ .......(1)

$\frac{N}{N_0}=e^{-\lambda t}$ .........(2)

Logarithm:

$ln|\frac mn|= ln|m|-ln|n|$
$ln|ab|=ln|a|+ln|b|$

$ln|e^a|=a$

$ln|a|=b \Rightarrow a=e^b$
$ln|1|=0$

130 days is the half-life of the given radioactive element.

For half life,

$N=\frac12 N_0$ , $t=t_\frac12=130$ days.

we plug all values in equation (1)

$ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130$

$\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130$

$\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130$

$\rightarrow -ln|2|=-\lambda \times 130$

$\rightarrow \lambda= \frac{-ln|2|}{-130}$

$\rightarrow \lambda= \frac{ln|2|}{130}$

We need to find the time when the sample remains 80% of its original.

$N=\frac{80}{100}N_0$

$\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t$

$\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}$

$\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}$

$\Rightarrow t\approx 42$

We can use the sample about 42 days.

7 0

3 years ago

Mx^{2} + n x^{2} when m = -5 and n = 3 what is the value of the expression

Vinil7 [7]

$|(-5)^2+3^2|=|25+9|=|34|=34$

8 0

1 year ago