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Rama09 [41]
2 years ago
14

Answer me pleaseeee. What is 3*3*3*3*3*3*3*3*8*8*0*8*8*9*8*6? Gk test​

Mathematics
2 answers:
olchik [2.2K]2 years ago
6 0

Answer:

0

Step-by-step explanation:

Since we're multiplying by 0, the entire product becomes 0.

givi [52]2 years ago
4 0

Answer:

it's 0.

Step-by-step explanation:

3*3*3*3*3*3*3*3*8*8*0*8*8*9*8*6

= 0

there is a zero hidden . so it's zero.

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The graph below shows the solution to a system of inequalities: Solid line joining ordered pairs 0, 3.75 and 15, 0. Shade the po
algol [13]
(0,3.75)(15,0)
slope(m) = (0 - 3.75) / (15 - 0) = -3.75/15 = - 0.25 or -1/4

y = mx + b
slope(m) = -1/4
(15,0)...x = 15 and y = 0
now we sub
0 = -1/4(15) + b
0 = -15/4 + b
15/4 = b

y = -1/4x + 15/4
1/4x + y = 15/4....multiply by 4
x + 4y = 15.....and since it is a solid line, it contains an equal sign...and since it is shaded above the line, it is greater.
so ur inequality is : x + 4y > = 15 (thats greater then or equal)
4 0
3 years ago
Read 2 more answers
A particular isotope has a​ half-life of 74 days. If you start with 1 kilogram of this​ isotope, how much will remain after 150
shtirl [24]

\bf \stackrel{150~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &150\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{150}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{75}{37}}\implies \boxed{A\approx 0.24536}


\bf \stackrel{300~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &300\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{300}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{150}{37}}\implies \boxed{A\approx 0.060202}

6 0
3 years ago
a geometric series where the first term is -12, the last term is -972, and each term after the first is triple the previous term
Luba_88 [7]

Answer:

the geometric series is a(n) = -12(3)^(n-1)

Step-by-step explanation:

"Triple" denotes multiplication by 3.  Thus, the common factor here is 3.

The general formula for a geometric series is a(n) = a(1)(r)^(n-1), where a(1) is the first term, r is the common ratio.

Here, we have a(n)= (-12)(3)^(n-1) = -972.

We need to solve this for n, which represents the last term.

The first step towards solving for n is to divide both sides by -12:

3^(n-1) = 81

To solve for n-1, rewrite 81 as 3^4.  Then we have:

3^(n-1) = 3^4, implying that (n-1) = 4 and that n = 5.

Then we know that it is the 5th term that equals -972.

In summary, the geometric series is a(n) = -12(3)^(n-1).

8 0
3 years ago
The slope-intercept form of the equation of a line that passes through the point (4, -3) is y=-3x+5. What is the point-slope for
rewona [7]

Answer:

y+ 3 = -1/3 (x-4)

Step-by-step explanation:

Simply plug in the values using (4, -3)

4 = x₁

-3 = y₁

Now one rule to keep in mind when plugging it in, since a negative and a negative make a positive, you have to add three instead of subtracting it.

6 0
4 years ago
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What is the answer for h/8=3/12 in algebra
enot [183]
When you simplify 3/12, it's 1/4.
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Therefore h is 2

5 0
3 years ago
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