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3 years ago

1. What is the value of x in the figure?

Mathematics

1 answer:

irinina [24]3 years ago

5 0

Answer:

$x=6\sqrt{5}\\ \\y=3\sqrt{5}\\ \\z=6$

Step-by-step explanation:

The height of the right triangle drawn to the hypotenuse is geometric mean of two segments of hypotenuse, so

$z^2=12\cdot 3\\ \\z^2=36\\ \\z=6$

By the Pythagorean theorem,

$x^2=12^2+z^2\\ \\x^2=144+6^2\\ \\x^2=144+36\\ \\x^2=180\\ \\x=6\sqrt{5}$

and

$y^2=z^2+3^2\\ \\y^2=6^2+9\\ \\y^2=36+9\\ \\y^2=45\\ \\y=3\sqrt{5}$

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What is the y intercept of the function f(x)=2•3^x

kykrilka [37]

Answer:

$\boxed{Option \ A}$

Step-by-step explanation:

$f(x) = 2*3^x$

y intercept is when x = 0

So, Putting x = 0 in the above function

=> f(0) = $2*3^0$

=> f(0) = 2*1

=> f(0) = 2

So, y-intercept is (0,2)

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3 years ago

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Looking for an algebraic equAtion answer.. 2x + y = 1 4x - 2y = 6 Can you help?

Sergio [31]

X=1and y=−1 Your welcome

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3 years ago

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Karolina [17]

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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

A cookie factory uses 1/3 of a bag of flour in each batch of cookies. The factory used 2/3 of a bag of flour yesterday. How many

Mumz [18]

Answer:

They made 2 batches

Step-by-step explanation:

If the factory used ⅔, and 1 batch is ⅓, the factory made 1 batch for each 3rd they used. ⅔, 2 batches

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3 years ago