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mel-nik [20]
3 years ago
7

Trying to figure out daughters math homework

Mathematics
1 answer:
Nina [5.8K]3 years ago
8 0
(10² ÷ 5) × (3 + 7) = 200
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A tunnel is built in form of a parabola. The width at the base of tunnel is 7 m. On
uysha [10]

Given:

The width at the base of parabolic tunnel is 7 m.

The ceiling 3 m from each end of the base there are light fixtures.

The height to light fixtures is 4 m.

To find:

Whether it is possible a trailer truck carrying cars is 4 m wide and 2.8 m high is going to drive through the tunnel.

Solution:

The width at the base of tunnel is 7 m.

Let the graph of the parabola intersect the x-axis at x=0 and x=7. It means x and (x-7) are the factors of the height function.

The function of height is:

h(x)=ax(x-7)             ...(i)

Where, a is a constant.

The ceiling 3 m from each end of the base there are light fixtures and the height to light fixtures is 4 m. It means the graph of height function passes through the point (3,4).

Putting x=3 and h(x)=4 in (i), we get

4=a(3)((3)-7)

4=a(3)(-4)

\dfrac{4}{(3)(-4)}=a

-\dfrac{1}{3}=a

Putting a=-\dfrac{1}{3}, we get

h(x)=-\dfrac{1}{3}x(x-7)              ...(ii)

The center of the parabola is the midpoint of 0 and 7, i.e., 3.

The width of the truck is 4 m. If is passes through the center then the truck must m 2 m on the left side of the center and 2 m on the right side of the center.

2 m on the left side of the center is x=1.5.

A trailer truck carrying cars is 4 m wide and 2.8 m high is going to drive through the tunnel is possible if h(1.5) is greater than 2.8.

Putting x=1.5 in (ii), we get

h(1.5)=-\dfrac{1}{3}(1.5)(1.5-7)

h(1.5)=-(0.5)(-5.5)

h(1.5)=2.75

Since h(1.5)<2.8, therefore the trailer truck carrying cars is 4 m wide and 2.8 m high is going to drive through the tunnel is not possible.

6 0
3 years ago
This is my last question
mariarad [96]

Answer: B

Step-by-step explanation:because my teacher said she having trouble

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Csqrt%7B0.81%7D" id="TexFormula1" title="\sqrt{0.81}" alt="\sqrt{0.81}" align="absmiddle" cl
Mandarinka [93]

Answer: .9

Step-by-step explanation:

3 0
3 years ago
Picture shown please help !!!
natulia [17]
Let's factor the top and the bottom.
Top factors 4x + 20 = 4(x + 5)
Bottom x^2 + 2x - 15 = (x + 5)(x - 3)

The gcf is the largest value that will go into both the top and the bottom.  In our problem the ONLY thing that will go into top and bottom is x + 5 so that is your GCF. (c)
4 0
3 years ago
Does anybody know how to do time with exponential decay
My name is Ann [436]
<span>From the message you sent me:

when you breathe normally, about 12 % of the air of your lungs is replaced with each breath. how much of the original 500 ml remains after 50 breaths

If you think of number of breaths that you take as a time measurement, you can model the amount of air from the first breath you take left in your lungs with the recursive function

b_n=0.12\times b_{n-1}

Why does this work? Initially, you start with 500 mL of air that you breathe in, so b_1=500\text{ mL}. After the second breath, you have 12% of the original air left in your lungs, or b_2=0.12\timesb_1=0.12\times500=60\text{ mL}. After the third breath, you have b_3=0.12\timesb_2=0.12\times60=7.2\text{ mL}, and so on.

You can find the amount of original air left in your lungs after n breaths by solving for b_n explicitly. This isn't too hard:

b_n=0.12b_{n-1}=0.12(0.12b_{n-2})=0.12^2b_{n-2}=0.12(0.12b_{n-3})=0.12^3b_{n-3}=\cdots

and so on. The pattern is such that you arrive at

b_n=0.12^{n-1}b_1

and so the amount of air remaining after 50 breaths is

b_{50}=0.12^{50-1}b_1=0.12^{49}\times500\approx3.7918\times10^{-43}

which is a very small number close to zero.</span>
5 0
3 years ago
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