Does anybody know how to do this ?

Help me please will give brainliest

kolezko [41]

C. 6

would be the answer

8 0

2 years ago

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What is the value of 3 to the power 2 over 3 to the power 4? A. 1 over 81 B. 1 over 27 C. 1 over 9 D. 1 over 3

rodikova [14]

One way you could solve this is to just multiply the top and bottom out so that you get 9/81, reducing it by 9/9 to get 1/9 or option C.
Another way would be to do $3^{(2-4)}$ since dividing numbers with exponents would be subtracting the bottom exponent from the top exponent, provided that the base number (in this case 3) is the same for both. For this method, you would get $3^{-2}$ , which is equal to 1/9 or .1 repeating, the same answer that you'd get with the first method.

6 0

3 years ago

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When Judy worked 4 hours and Ben worked 1 hour, their combined pay was $75. When Judy worked 2 hours and Ben worked 6 hours, the

arsen [322]

Answer:

$15

Step-by-step explanation:

let the hourly rate of pay for Ben = B

let the hourly rate of pay for Judy = J

Judy worked 4 hours and Ben worked 1 hour, their combined pay was $75;

4J + B = 75 ----- i

Judy worked 2 hours and Ben worked 6 hours, their combined pay was $120

2J + 6B = 120 --- ii

Now, we should solve the expression;

Multiply equation ii by 2 and equation i by 1;

4J + B = 75 ----- i x 1 ; 4J + B = 75 -- iii

2J + 6B = 120 --- ii x 2 ; 4J + 12B = 240 ---iv

Now,

equation iv - iii;

11B = 165

B = $15

So, solving for J;

4J + B = 75

4J + 15 = 75

4J = 75 - 15 = 60

J = $15

3 0

3 years ago

An engine cylinder 13.7 cm deep is being bored such that the radius increases by 0.13 mm divided by min. How fast is the volume

zzz [600]

Answer:

Step-by-step explanation:

Given that an engine cylinder 13.7 cm deep is being bored such that the radius increases by 0.13 mm divided by min.

We convert everything in mm for uniformity

We have formula for volume as

$V=\pi r^2 h$

Given that $\frac{dr}{dt} =0.13$

h is constant

$V= 3.14 (r^2) (13.7)\\\frac{dV}{dt} =6.28 (137) r dr/dt\\= 6.28(137)\frac{ (97)}{2} 0.13$

= 6954.67 mm/sec^3

6 0

3 years ago

If the initial velocity stays the same, but the height the projectile is launched from changes , how would the amount of time ch

pickupchik [31]

Answer:

Part A

The height to which the projectile reaches is given by the following formula;

$\Delta y = y-y_0 = v_{0y} \times t-\dfrac{1}{2} \times g \times t^2$

Where;

$v_{0y}$ = The initial velocity of the projectile = Stays the same

y = The height the projectile reaches

y₀ = The height from which the projectile is launched

g = The acceleration due to gravity

t = The time taken

Given that the projectile has the same initial velocity, the variation of Δy with time, 't', will be the same when the project is launched from a different height, and the time change that it takes for the projectile to reach the maximum height will be the same for the previous launch height

Part B

From the change in eight of the projectile, Δy = y - y₀, the maximum height reached, 'y', is therefore given as follows;

y = Δy + y₀

The maximum height will increase by 'y₀', where 'y₀' is the difference between new height and the previous height, $y_i$

That is, if y₀ > $y_i$ , the maximum height reached increases and if y₀ < $y_i$ , y₀ will be added as y = Δy - y₀ and the maximum height reached decreases

Step-by-step explanation:

6 0

2 years ago