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3 years ago

What is equivalent to 3^-5

Mathematics

1 answer:

4vir4ik [10]3 years ago

4 0

Answer:

The answer is the choose: (D)

${3}^{ - 5} = \frac{1}{ {3}^{5} }$

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3 years ago

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Joe says that x

irga5000 [103]

The equations x+5x=16
and x+5x+25=16
No, the equations are not equal. In order for an equation to be equal the equation x+5x=16 contains 25 less than the second equation.

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3 years ago

A rocket is launched straight up from the ground with an initial velocity of 192 feet per second. The equation for the height of

Ivenika [448]

The equation for the height of the rocket at time t given

$h= -16t^2+192t$

We have to find the time t, when the rocket reaches 560 feet.

That means we have to find t when h = 560 ft. we will place 560 in the place of h to find t now.

$h= -16t^2+192t$

$560 = -16t^2+192t$

In the right side, we can check -16 is the common factor. So we will take out -16 from the rigbht side.

$560 = -16(t^2 - 12t)$

To get rid of -16 from the right side and move it to left side, we will divide both sides by -16.

$560/-16 = -16(t^2-12t)/-16$

$-35 = t^2 -12t$

Now we will move -35 to the righ side by adding 35 to both sides.

$-35+35 = t^2-12t+35$

$0 = t^2 -12t+35$

$t^2-12t+35 = 0$

We will factorize thee left side to find the values of t now. We need to find a pair of factors of 35 that by adding them we will get -12.

The pair of factors of 35 are -5 and -7 and by adding -5-7 we will get -12.

$t^2-12t+35 =0$

$(t-5)(t-7) =0$

So by using zero product property we will get

$t-5 =0$

$t-5+5 = 0+5$

$t=5$

Also $t-7 =0$

$t-7+7 = 0+7$

$t=7$

So we have got the rocket reaches at 560ft when t = 5 seconds and also when t = 7 seconds.

Now part b.

When the rocket completes its trajectory and hits the ground then the height or h = 0. So we will place h = 0 there in the equation.

$h= -16t^2+192t$

$0= -16t^2 + 192 t$

$0 = -16(t^2-12t)$

$-16(t^2-12t) = 0$

We will move -16 to the other side by dividing it to both sides.

$-16(t^2-12t)/-16 = 0/-16$

$t^2-12t = 0$

We will take out the common factor t from the left side. By taking out t we will get,

$t(t-12) = 0$

We will use zero product property now. By using that we will get,

$t = 0$

ans also $t-12 = 0$

$t-12+12 = 0+12$

$t = 12$

When the rocket completes its trajectory and hits the ground the time t can not be 0. When t =0, the rocket starts the trajectory.

So when the rocket completes its trajectory and hits the ground ,

then t = 12seconds.

So we have got the required answers.

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3 years ago

Find the side length of X

7nadin3 [17]

Answer:

7√3

Step-by-step explanation:

We can use pythagoreans theorem to solve this

Since, we know one side, and the hypotenuse, we can solve for the other side.

Pythagoreans theorem: a²+b²=c²

Where a and b are two sides, and c is the hypotenuse (the side opposite of the right angle)

In this triangle, 7 is the side, and 14 is the hypotenuse.

I will plug in the values into pythagoreans theorem, and then simplify:

$a^{2}+b^{2} =c^{2} \\\\7^{2} +x^{2} =14^{2} \\\\49+x^{2} =196\\\\x^{2} =147\\\\x=\sqrt{147} \\\\x=\sqrt{49*3 }\\\\ x=\sqrt{7^{2} *3 }\\\\x=7\sqrt{3}$

So x = 7√3

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3 years ago

Hal forgot the number of people at the basketball game.He does remember that the number had four digits and a 3 in the tens plac

Greeley [361]

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