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2 years ago

During the last schools football game, the Home Team scored 13 more points than the Visiting

Mathematics

1 answer:

Leya [2.2K]2 years ago

7 0

Answer:

Home 30, visitor 17.

Step-by-step explanation:

Visiting team score is x. Home score is then x+13. Total of both is 47 so...

x+(x+13)=47

2x+13=47

2x-34

x=17

x+13=30

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Write the decimal 40.99 in word form

Svetach [21]

Forty and ninety nine hundreths

8 0

3 years ago

Read 2 more answers

At sea, the distance D to the horizon is directly proportional to the square root of the elevation E of the observer. If a perso

Alla [95]

The problem says that the distance D to the horizon is directly proportional to the square root of the elevation E of the observer, so:

The person who is 36 feet above the water ⇒ √36=6

A person 49 feet above the water ⇒ √49=7

So, by proportions, you have:

(7/6)x7.4= 8.63 miles

Therefore, the answer is: a person 49 feet above the water can see 8.63 miles

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3 years ago

4TH TIME ASKING THIS!!! Please help me! Someone pleaseeee. I need the correct answers. I don’t want to fail

Alexeev081 [22]

Answer:

The functions are inverses; f(g(x)) = x ⇒ answer D

$h^{-1}(x)=\sqrt{\frac{x+1}{3}}$ ⇒ answer D

Step-by-step explanation:

* Lets explain how to find the inverse of a function

- Let f(x) = y

- Exchange x and y

- Solve to find the new y

- The new y = $f^{-1}(x)$

* Lets use these steps to solve the problems

∵ $f(x)=\sqrt{x-3}$

∵ f(x) = y

∴ $y=\sqrt{x-3}$

- Exchange x and y

∴ $x=\sqrt{y-3}$

- Square the two sides

∴ x² = y - 3

- Add 3 to both sides

∴ x² + 3 = y

- Change y by $f^{-1}(x)$

∴ $f^{-1}(x)=x^{2}+3$

∵ g(x) = x² + 3

∴ $f^{-1}(x)=g(x)$

∴ The functions are inverses to each other

* Now lets find f(g(x))

- To find f(g(x)) substitute x in f(x) by g(x)

∵ $f(x)=\sqrt{x-3}$

∵ g(x) = x² + 3

∴ $f(g(x))=\sqrt{(x^{2}+3)-3}=\sqrt{x^{2}+3-3}=\sqrt{x^{2}}=x$

∴ f(g(x)) = x

∴ The functions are inverses; f(g(x)) = x

* Lets find the inverse of h(x)

∵ h(x) = 3x² - 1 where x ≥ 0

- Let h(x) = y

∴ y = 3x² - 1

- Exchange x and y

∴ x = 3y² - 1

- Add 1 to both sides

∴ x + 1 = 3y²

- Divide both sides by 3

∴ $\frac{x + 1}{3}=y^{2}$

- Take √ for both sides

∴ ± $\sqrt{\frac{x+1}{3}}=y$

∵ x ≥ 0

∴ We will chose the positive value of the square root

∴ $\sqrt{\frac{x+1}{3}}=y$

- replace y by $h^{-1}(x)$

∴ $h^{-1}(x)=\sqrt{\frac{x+1}{3}}$

4 0

3 years ago

How do I write an equation for each parabolas?

fredd [130]

Step-by-step explanation:

In the first parabola it opens on the left and the equation of parabola can be expressed as,

in vertical component (y)² = (-) a (x-h)² + k

cause the parabola is horizontal and it opens on the left.

2. In the second parabola the vertex opens on the right and hence the equation cane be given as,

in vertical component (y)² = a (x-h)² + k

cause the parabola is horizontal and opens on the right.

3. the third equation is given as,

in horizontal component (x²) = (-) a (x-h)² + k

since the parabola is vertical and opens down.

4. the fourth equation is given as,

in the horizontal component (x)² = a (x-h)² + k

since the parabola is vertical and opens up.

7 0

2 years ago

An object is launched straight into the air. The projectile motion of the object can be modeled using h(t) = 96t – 16t2, where t

ratelena [41]

Answer with Step-by-step explanation:

We are given that height of projectile after t seconds is given by

$h(t)=96t-16t^2$

a.h(t)=144 ft

$144=96t-16t^2$

$16t^2-96t+144=0$

$t^2-6t+9=0$

$t^2-3t-3t+9=0$

$t(t-3)-3(t-3)=0$

$(t-3)(t-3)=0$

t-3=0

t=3

After 3 s, the height of the project will be 144 feet in the air.

b.h(t)=0

$96t-16t^2=0$

$16t(6-t)=0$

$16t=0\implies t=0$

$6-t=0\implies t=6$

At t=0, the initial position of projectile

At t=6 s , the projectile will hit the ground.

6 0

3 years ago