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3 years ago

Complete 5^9/5^3=5^a

Mathematics

2 answers:

Goryan [66]3 years ago

7 0

Answer:

6=a

Step-by-step explanation:

5^9/5^3=5^a

We know that b^c = b^d = b^(c-d)

5^9/5^3=5^a

5^(9-3) = 5^a

5^6 = 5^a

The bases are the same so the exponents are the same

6=a

Gemiola [76]3 years ago

6 0

Answer:

a=6

Step-by-step explanation:

5^9/5^3=5^a

5^(9-3)=5^a

9-3=a

6=a

a=6

I hope this helps!

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Provide reasons for the proof. Given line h is parallel to line k. line j is perpendicular to line h. Prove line j is perpendicu

dusya [7]

Answer: Provided.

Step-by-step explanation: We are given two lines 'h' and 'k' which are parallel to each other. Also, there is another line 'j' that is perpendicular to line 'h'.

We are to prove that line 'j' is perpendicular to line 'k'.

Let, m, n and p be the slopes of lines 'h', 'k' and 'j' respectively.

Now, since line 'h' and 'k' are parallel, so their slopes will be equal. i.e., m = n.

Also, lines 'h' and 'j' are perpendicular, so the product of their slopes is -1. i.e.,

m×p = -1.

Hence, we can write from the above two relations

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4 years ago

A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 4

Vladimir79 [104]

Answer:

The time interval when $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

The distance is 106.109 m

Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

$V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)$

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

We are also to that :

$V_Q(t) \geq 60$ between $0 \leq t \leq 4$

The schematic free body graphical representation of the above illustration was attached in the file below and the point when $V_Q(t) \geq 60$ is at 4 is obtained in the parabolic curve.

So, $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

Taking the integral of the time interval in order to determine the distance; we have:

distance = $\int\limits^{3.519}_{1.866} {V_Q(t)} \, dt$

= $\int\limits^{3.519}_{1.866} {45\sqrt{t} cos(0.063 \ t^2)} \, dt$

= By using the Scientific calculator notation;

distance = 106.109 m

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3 years ago

(PLEASE HELP) To feed 7 horses for 4 days, Mark needs 420 kg of grass. How many kilograms of grass does Mark need to feed 9 hors

DerKrebs [107]

Answer:

810kg is the answer

Step-by-step explanation:

Hope this helps:)

7 0

2 years ago

Read 2 more answers

In a fifth grade class, 4/5 of the girls have brown hair, of the brown-haired girls, 3/4

AfilCa [17]

Answer:

$\frac{3}{5}$

Step-by-step explanation:

In the fifth grade class, $\frac{4}{5}$ of the girls have brown hair. Again among those of the brown-haired girls, $\frac{3}{4}$ of them have long hair.