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3 years ago

6

If the scale factor of the model train to a real train is 1:43, what is the length of the model if the actual train is 645 inche

s? The length of the model would be

1 answer:

OleMash [197]3 years ago

6 0

The answer is 15 inches.
To get the answer divide the real length by 43.
645÷43=15

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We are standing on the top of a 1680 ft tall building and throw a small object upwards. At every second, we measure the distance

MAVERICK [17]

Answer:

a) The height of the small object 3 seconds after being launched is 2304 feet.

b) The small object ascends 128 feet between 5 seconds and 7 seconds.

c) The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) The object will take 21 seconds to hit the ground.

Step-by-step explanation:

The correct formula for the height of the small object is:

$h(t) = -16\cdot t^{2}+256\cdot t+1680$ (1)

Where:

$h$ - Height above the ground, measured in feet.

$t$ - Time, measured in seconds.

a) The height of the small object at given time is found by evaluating the function:

$h(3\,s)= -16\cdot (3\,s)^{2}+256\cdot (3\,s)+1680$

$h(3\,s) = 2304\,ft$

The height of the small object 3 seconds after being launched is 2304 feet.

b) First, we evaluate the function at $t = 5\,s$ and $t = 7\,s$ :

$h(5\,s)= -16\cdot (5\,s)^{2}+256\cdot (5\,s)+1680$

$h(5\,s) = 2560\,s$

$h(7\,s)= -16\cdot (7\,s)^{2}+256\cdot (7\,s)+1680$

$h(7\,s) = 2688\,s$

We notice that the small object ascends in the given interval.

$\Delta h = h(7\,s)-h(5\,s)$

$\Delta h = 128\,ft$

The small object ascends 128 feet between 5 seconds and 7 seconds.

c) If we know that $h = 2640\,ft$ , then (1) is reduced into this second-order polynomial:

$-16\cdot t^{2}+256\cdot t-960=0$ (2)

All roots of the resulting equation come from the Quadratic Formula:

$t_{1} = 10\,s$ and $t_{2}= 6\,s$

The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) If we know that $h = 0\,ft$ , then (1) is reduced into this second-order polynomial:

$-16\cdot t^{2}+256\cdot t +1680 = 0$ (3)

All roots of the resulting equation come from the Quadratic Formula:

$t_{1} = 21\,s$ and $t_{2} = -5\,s$

Just the first root offers a solution that is physically reasonable.

The object will take 21 seconds to hit the ground.

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3 years ago

HELP ASAP PLEASE****<br><br> For f(x)=4x+1 and g(x)=x^2-5, find (f+g)(x)

lara [203]

Use Photomath it will probably help

8 0

3 years ago

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Find the surface area of the following figure.

marin [14]

Answer:

144

Step-by-step explanation:

just split the shapes and find the area of each shape then add the answers all together

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3 years ago

RST has vertices R(0,0), S(10,5), and T(5,-5). R’S’T’ is the image of RST after a dilation with center (0,0) and scale factor 1/

AysviL [449]

The coordinates of the point S' is (2,1)

Explanation:

Given that the RST has vertices R(0,0), S(10,5) and T(5,-5)

The image R'S'T' is the image of RST after a dilation with center (0,0) and scale factor $\frac{1}{5}$

We need to determine the coordinates of the point S'

The coordinates of S' can be determined by multiplying both the x and y - coordinates of the point S with the scale factor $\frac{1}{5}$

Let us consider the x - coordinate of the point S and multiply it with the scale factor $\frac{1}{5}$

Thus, we have,

x - coordinate of the point S' is given by

$10 \times \frac{1}{5}=2$

Thus, the x - coordinate of the point S' is 2

Similarly, the y - coordinate of the point S' can be determined by multiplying the point S with the scale factor $\frac{1}{5}$

Thus, we have,

The y - coordinate of the point S' is given by

$5\times\frac{1}{5}=1$

Thus, the y - coordinate of the point S' is 1

Hence, the coordinates of the point S' is (2,1)

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3 years ago

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A value two standards deviation from the mean is more likely to occur than a value three standard deviations from the mean.

AnnZ [28]

Answer:

This is true. The closer to the mean that a value is, the more likely it is to occur, if that makes sense

Step-by-step explanation:

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