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2 years ago

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Drag each tile to the correct box. Not all tiles will be used. Drag the explanations used to solve the given equation in the cor

rect sequence. 5(6x-9)=125
Subtract 9 from each side
Divide each side by 5
Divide each side by 6
Equate the exponents of the equation
Add 9 to each side
Rewrite 125 as a power of 5

2 answers:

Westkost [7]2 years ago

8 0

Answer:

Rewrite 125 as a power of 5
Equate the exponents of the equation
Add 9 to each side
Divide each side by 6

sergejj [24]2 years ago

3 0

Answer:

1. Rewrite 125 as a power of 5.

2. Equate the exponents of the equation.

3. Add 9 to each side.

4. Divide each side by 6.

Step-by-step explanation:

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Perform the indicated operations. give answers in simplified form.<br><br> √9/25

tino4ka555 [31]

$\sqrt{\frac{9}{25}} = \frac{\sqrt{9}}{\sqrt{25}} = \frac{3}{5}$

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2 years ago

Can someone help me with all this? Lol idk any of this and E-learning sucks I need to show my work as well

Margarita [4]

Answer:

1a) 230 feet

1b) 3.75s

1c) t=7.54s

2) t=2s

3) t=2.32s

Step-by-step explanation:

The equation that models Brett's last home run is $h(t)=-16t^2+120t+5$

We need to complete the square to obtain the function in the vertex form:

$h(t)=-16(t^2-7.5t)+5$

$h(t)=-16(t^2-7.5t+3.75^2)+5+-16(-3.75)^2$

$h(t)=-16(t-3.75)^2+230$

1a) The vertex is (3.75,230).The maximum value is the y-coordinate of the vertex, which is 230 feet.

1b) The time it takes to reach the maximum value is the x-coordinate of the vertex. It reach the maximum heigth after t=3.75 seconds

1c) To find the time the ball hit the ground, we equate the h(t)=0 and solve for t.

$-16(t-3.75)^2+230=0$

$-16(t-3.75)^2=-230$

$(t-3.75)^2=14.375$

$t-3.75=\pm \sqrt{14.375}$

$t=3.75\pm \sqrt{14.375}$

$t=3.75\pm 3.79$

$t=3.75-3.79\:\:or\:t=3.75+3.79$

$t=3.75-3.79\:\:or\:t=3.75+3.79\\t=-0.04\:\:or\:t=7.54$

The time is positive, so the ball hit the ground after 7.54 seconds.

Question 2)

The function that models the amusement ride is $h(t)=-16t^2+64t+60$

We want to find the time it takes for the riders to reach the maximum height.

This time is given by: $t=-\frac{b}{2a}$

Comparing $h(t)=-16t^2+64t+60$ to $h(t)=at^2+bt+c$ we have a=-16, b=64, c=60.

We substitute to obtain:

$t=-\frac{64}{2*-16} \\t=-\frac{64}{-32} \\t=2$

Hence it took 2 seconds to rech the maximum height.

Question 3)

The equation that models the height f the catridge is $h(t)=-16t^2+35t+5$

To find the time that the catridge will land, we equate the function to zero and solve for t.

$-16t^2+35t+5=0$

This is a quadratic equation with =-16, b=35, an c=5

The solution is given by:

$t=\frac{-b\pm \sqrt{b^2-4ac} }{2a}$

We substitute the values to get:

$t=\frac{-35\pm \sqrt{35^2-4*-16*5} }{2*-16}$

This gives:

$t=\frac{35-\sqrt{1545} }{32} \:or\: t=\frac{35+\sqrt{1545} }{32}$

This simplifies to:

$t=-0.13\:or\: t=2.32$

The time it takes to land must be positive thefore t=2.32 seconds

7 0

3 years ago