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3 years ago

15

Arithmetic sequence of -10,8,26

1 answer:

BlackZzzverrR [31]3 years ago

5 0

Answer:

the sequence is going up by 18

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The measure of one base angle of an isosceles triangle is 35 degrees. What is the measure of the vertex angle?

Elis [28]

Answer:

110 degrees

Step-by-step explanation:

The measures of the two base angles of an isosceles triangle are the same so 180 - 2(35) = 110 which is the vertex angle.

8 0

3 years ago

Can people help me please to find the volume about question number 10 please and tell me the answer step by step please I need h

Alecsey [184]

#10
Volume of a cylinder: $\pi$ r²h

Plug the numbers in.

$\pi$ 3²15 = <span>424
</span>
The volume of the pipe is 424 cm³.

7 0

3 years ago

What value should be added to both sides for completing the square in this equation?

Mila [183]

Hey there!

x^2 + 4x = 12

SUBTRACT 12 to BOTH SIDES

x^2 + 4x - 12 = 12 - 12

SIMPLIFY IT

x^2 + 4x - 12 = 0

FACTOR the LEFT SIDE of your EQUATION

(x - 2)(x + 6) = 0

• EQUATION #1: x - 2 = 0

OR

• EQUATION #2: x + 6 = 0

SIMPLIFY IT

• EQUATION #1 answer: x = 2

OR

• EQUATION #2 answer: x = -6

OVERALL ANSWER: x = 2 or x = -6

YOUR ANSWER: x = 2 (Option A.)

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

3 0

2 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

Find the measure of the angle indicated.

jarptica [38.1K]

Where is the pic ?i can’t see it

6 0

3 years ago