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DochEvi [55]
4 years ago
10

Need help again! ASAP!!

Mathematics
2 answers:
pickupchik [31]4 years ago
5 0

Answer:

B

Step-by-step explanation:

The formula is 1-decay, not just decay

so it has to be B the formula is:

a(1-r)^t

a, the original starting price, r, the rate of change and t for time

i hope this helps u pls give a brainliest and a thx ;)

leave a comment if i am wrong

Ivenika [448]4 years ago
4 0

Answer:

It is B, I couldn't see it well because it was upside down. But I know it is B.

Step-by-step explanation:

Hope It helped!

Please give branliest.

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Find the number of solutions of the equation 6x^2+6x+3=0 by using the discriminant
Neporo4naja [7]

Answer:

No real roots

Step-by-step explanation:

Given

6x^2 + 6x + 3 = 0

Required

The number of solution

The discriminant (d) for ax^2 + bx + c = 0 is:

d = b^2 - 4ac

So, we have:

d = 6^2 - 4*6*3

d = 36 - 72

d = -36

<em>Since </em>d<em>, then the equation has no real roots</em>

4 0
3 years ago
Write an equation of a line with slope 1/2 and crossing point (2,-3)
Yuliya22 [10]

Answer:

y= 1/2x +4

Step-by-step explanation:

plot a point on a graph at (2, -3) then insert ur slope (rise over run) and in this case that’s 1 up 2 to the right

7 0
3 years ago
WILL CROWN BRAINLIEST, GIVE THANKS, AND RATING!
ahrayia [7]
The range is 432

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3 years ago
Read 2 more answers
Can someone help me with this truth table. I just don't understand how to do it.
RUDIKE [14]

An implication (p\implies q) is true if either the premise (p) is false, or both the premise and conclusion (p\text{ and }q) are both true, and false otherwise.

\begin{array}{c|c|c|c|c|c|c}p&q&p\implies q&\neg q&(p\implies q)\land\neg q&\neg p&(p\implies q)\land\neg q\implies\neg p\\T&T&T&&&&\\T&F&F&&&&\\F&T&T&&&&\\F&F&T&&&&\end{array}

Negation (\neg q) is straightforward; if a statement is true, then its negation is false, and vice versa.

\begin{array}{c|c|c|c|c|c|c}p&q&p\implies q&\neg q&(p\implies q)\land\neg q&\neg p&(p\implies q)\land\neg q\implies\neg p\\T&T&T&F&&F\\T&F&F&T&&F\\F&T&T&F&&T\\F&F&T&T&&T\end{array}

A conjunction (p\land q) is true if both premises are true, and false otherwise.

\begin{array}{c|c|c|c|c|c|c}p&q&p\implies q&\neg q&(p\implies q)\land\neg q&\neg p&(p\implies q)\land\neg q\implies\neg p\\T&T&T&F&F&F\\T&F&F&T&F&F\\F&T&T&F&F&T\\F&F&T&T&T&T\end{array}

Finally, by the rules of implication, we can fill the last column:

\begin{array}{c|c|c|c|c|c|c}p&q&p\implies q&\neg q&(p\implies q)\land\neg q&\neg p&(p\implies q)\land\neg q\implies\neg p\\T&T&T&F&F&F&T\\T&F&F&T&F&F&T\\F&T&T&F&F&T&T\\F&F&T&T&T&T&T\end{array}

5 0
3 years ago
A ______________ consists of an a sample space and a list of events with their probabilities.
motikmotik

Sample space َََ َ َ َ َ َ َ َ َ َ

5 0
3 years ago
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