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3 years ago

What is the missing value of n that will make the equation true? 3^8 . 3^n= 3^14

Mathematics

1 answer:

marissa [1.9K]3 years ago

7 0

Answer:

n = 6

Step-by-step explanation:

${3}^{8} . {3}^{n} = {3}^{14} \\ {3}^{8 + n} = {3}^{14} \\ 8 + n = 14 \\ n = 14 - 8 \\ n = 6$

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How do you solve -k/5 times 7/5

Firdavs [7]

Answer:

$\frac{-7}{25}k$

Step-by-step explanation:

$\frac{-k}{5} *\frac{7}{5}$

just multiply across

-k*7=-7k

5*5=25

$\frac{-7}{25}k$

hope this helps

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3 years ago

Find the slope of the line passing through the pair of given points. (5, 7) and (2,7)

gulaghasi [49]

Answer:

m = 0

Step-by-step explanation:

$m = \frac{7 - 7}{5 -2} \\m = \frac{0}{3}\\m = 0$

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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Deana bought 12 packages of batteries. Some of the packages contained 4 batteries and some contained 6 batteries. She bought a t

NARA [144]

Answer:

She bought 7 battery packs containing 6 batteries

Step-by-step explanation:

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7 0

3 years ago

Find the equation of the circle with a diameter whose end points are (-1,-2) and (-3,2)

Sergeeva-Olga [200]

Answer:

The equation of the circle with a diameter whose end points are (-1,-2) and (-3,2) is

$x^{2} +y^{2}+4x-1=0$

Step-by-step explanation:

Explanation:-

Step 1:-

The equation of the circle having center and radius is

$(x-h)^2+(y-k)^2=r^2$

here center is (h,k) and radius is r

Given diameter whose end points are (-1,-2) and (-3,2)

The diameter of the circle is passing through the center of the circle

so center of the circle = midpoint of two end points

$(\frac{-1 +(-3) }{2} ,\frac{-2+2 }{2} )$

$(-2,0)$

therefore center (h,k) = (-2,0)

Step 2:-

we have to find the radius of the circle

the radius of the circle = the distance from center to the one end point

i.e., C P = r

Given one end point is P(-3,2) and center C(-2,0)

The distance formula of two points are

$\sqrt{(x_{2}-x_{1} ) ^{2}+ (y_{2}-y_{1} ) ^{2}}$

$r=\sqrt{{(-3)-(-1) ) ^{2}+ (2-(-2)) ^{2}}$

$r=\sqrt{5}$

Step 3:-

center (h,k) = (-2,0) and

radius $r=\sqrt{5}$

The standard form of circle equation

$(x-h)^2+(y-k)^2=r^2$

$(x-(-2))^2+(y-0)^2=\sqrt{5} ^2$

on simplification is

$x^{2} +y^{2}+4 x-1=0$

5 0

3 years ago