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Lena [83]
3 years ago
6

Please help!!! 15 points problem below

Mathematics
1 answer:
MissTica3 years ago
5 0

We are given height function of the bridge

h(x) =-0.5(x-4)^2 +2.

Where height is the height of the bridge in feet and x is the distance between two bases.

We have height =0 on the base of the bridge. In order to find the both bases, we need to plug h(x) as 0 and solve for x.

-0.5(x-4)^2 +2 = 0

\mathrm{Subtract\:}2\mathrm{\:from\:both\:sides}

-0.5\left(x-4\right)^2+2-2=0-2

-0.5\left(x-4\right)^2=-2

\mathrm{Multiply\:both\:sides\:by\:}10

-0.5\left(x-4\right)^2\cdot \:10=-2\cdot \:10

-5\left(x-4\right)^2=-20

\mathrm{Divide\:both\:sides\:by\:}-5

\frac{-5\left(x-4\right)^2}{-5}=\frac{-20}{-5}

\left(x-4\right)^2=4

Taking square root on both sides, we get

x-4=\sqrt{4}

\:x-4=\sqrt{4} \ and \ x-4=-\sqrt{4}

x-4=2 \\x-4=-2

x=6,\:x=2

We got horizontal distances 2 feet and 6 feet.

Therefore, correct option is D.

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marissa [1.9K]
Answer is 2 miles
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8 0
2 years ago
Last time, one out of every 4 students in our class was absent. This time, after Jen came back, only one out of every 5 students
VMariaS [17]

Answer: There are 20 students and 4 students are not here today.

Step-by-step explanation:

If the number of students = 4

Let the number of students be 'x'.

Fraction of students absent = \dfrac{x}{4}

If the number of students = 5

Fraction of students absent = \dfrac{x}{5}

And Jen came back,

So, the fraction of student absent is also written as

\dfrac{x}{4}-1

According to question, it becomes,

\dfrac{x}{5}=\dfrac{x}{4}-1\\\\\dfrac{x}{5}=\dfrac{x-4}{4}\\\\4x=5(x-4)\\\\4x=5x-20\\\\4x-5x=-20\\\\-x=-20\\\\x=20

Hence, there are 20 students in our school.

And number of students are not here today is \dfrac{20}{5}=4

3 0
3 years ago
08.01) Which of the following statements shows a characteristic of a statistical question? (4 points) Select one: a. The questio
Alexeev081 [22]

Answer:

the distribution can be broken into categories

7 0
3 years ago
Read 2 more answers
Plz some one help me
sveta [45]
9 games

You can get this by multiplying each percentage by the total number of games to see what they have to expect.

.55*180 = 99 games (red team)

.60*180 = 108 games (blue team)

Now subtract the blue team expectation from the red team.

108 - 99 = 9 games. 
6 0
3 years ago
4.) What is the exact value of sinθ when θ lies in Quadrant II and cosθ=−513
dimaraw [331]

Answer:

Part 4) sin(\theta)=\frac{12}{13}

Part 10) The angle of elevation is 40.36\°

Part 11) The angle of depression is 78.61\°

Part 12) arcsin(0.5)=30\°  or arcsin(0.5)=150\°

Part 13) -45\°  or 225\°

Step-by-step explanation:

Part 4) we have that

cos(\theta)=-\frac{5}{13}

The angle theta lies in Quadrant II

so

The sine of angle theta is positive

Remember that

sin^{2}(\theta)+ cos^{2}(\theta)=1

substitute the given value

sin^{2}(\theta)+(-\frac{5}{13})^{2}=1

sin^{2}(\theta)+(\frac{25}{169})=1

sin^{2}(\theta)=1-(\frac{25}{169})  

sin^{2}(\theta)=(\frac{144}{169})

sin(\theta)=\frac{12}{13}

Part 10)

Let

\theta ----> angle of elevation

we know that

tan(\theta)=\frac{85}{100} ----> opposite side angle theta divided by adjacent side angle theta

\theta=arctan(\frac{85}{100})=40.36\°

Part 11)

Let

\theta ----> angle of depression

we know that

sin(\theta)=\frac{5,389-2,405}{3,044} ----> opposite side angle theta divided by hypotenuse

sin(\theta)=\frac{2,984}{3,044}

\theta=arcsin(\frac{2,984}{3,044})=78.61\°

Part 12) What is the exact value of arcsin(0.5)?

Remember that

sin(30\°)=0.5

therefore

arcsin(0.5) -----> has two solutions

arcsin(0.5)=30\° ----> I Quadrant

or

arcsin(0.5)=180\°-30\°=150\° ----> II Quadrant

Part 13) What is the exact value of arcsin(-\frac{\sqrt{2}}{2})

The sine is negative

so

The angle lies in Quadrant III or Quadrant IV

Remember that

sin(45\°)=\frac{\sqrt{2}}{2}

therefore

arcsin(-\frac{\sqrt{2}}{2}) ----> has two solutions

arcsin(-\frac{\sqrt{2}}{2})=-45\° ----> IV Quadrant

or

arcsin(-\frac{\sqrt{2}}{2})=180\°+45\°=225\° ----> III Quadrant

5 0
3 years ago
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