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Nookie1986 [14]
3 years ago
6

Which of the following pairs is not correctly written in scientific notation

Mathematics
1 answer:
kow [346]3 years ago
4 0
B in the answer <span>not correctly written in scientific notation</span>
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3^x+2=8^x-1 Solve for x.
zubka84 [21]

I assume the equation is supposed to be

3^{x+2}=8^{x-1}

Then we can write

9\cdot3^x=\dfrac18\cdot8^x\implies\left(\dfrac38\right)^x=\dfrac1{72}

Take the logarithm of base 3/8 on both sides:

\log_{3/8}\left(\dfrac38\right)^x=\log_{3/8}\dfrac1{72}

\implies x\log_{3/8}\dfrac38=-\log_{3/8}72

\implies x=-\log_{3/8}72

- - -

If the equation is actually 3^x+2=8^x-1, I'm afraid it cannot be solved exactly.

5 0
3 years ago
Options 1,2,3 or4?<br><br> Help!<br> Last one I swear!
lapo4ka [179]

Answer:

option 3 = 3.75

Step-by-step explanation:

x is 3.75

.

8 0
3 years ago
Read 2 more answers
Write the equation of a line that is perpendicular to the line y= - 1/3x +8 and passes through the point (0,-4)
pantera1 [17]
The answer would be y = 3x-4
8 0
3 years ago
Plz help me I will give you 40 points and Brainly
Amiraneli [1.4K]

Answer:

1) slope 3/2, y-intercept 7/2

2y-3x=7

2) slope 1/5, y-intercept 3

5y-x=15

3) slope 1/5, y-intercept 4/5

-x+5y=4

4) slope 5/2, y-intercept 7/2

2y=5x+7

Step by step explanation:

1) 2y-3x=7

2y=3x+7

y=3/2y+7/2

2)5y-x=15

5y=x+15

y=1/5x+3

3)-x+5y=4

5y=x+4

y=1/5x+4/5

4)2y=5x+7

y=5/2x+7/2

7 0
2 years ago
A line passes through the point (3,2) and (6,2) a write an equation for the line in point-slope form rewrite the equation in sta
NeTakaya
(3, –2) and (6, 2)First, we must establish slope...m=(y-y1)/(x-x1)m=(-2-2)/(3-6)m=-4/-3m=4/3Point slope formula is...y-y1=m(x-x1)Let's select either coordinate.  I randomly select the first (3, –2)...y--2=4/3(x-3)Subtracting a negative number is the same as adding a positive number...y+2=4/3(x-3)This corresponds to the first answer.Standard form...Ax+By=Cy+2=(4/3)x-4y=(4/3)x-6-(4/3)x+y=-6Multiply both sides by 3...-4x+3y=-18This also corresponds to the first answer. How about the second coordinate, (6, 2)...y-y1=m(x-x1)y-2=(4/3)(x-6)Let's convert it into standard form..y-2=(4/3)x-8y=(4/3)x-6-(4/3)x+y=-6Multiply both sides by 3...-4x+3y=-18
8 0
3 years ago
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