I have one question, i'll give ya'll 50 points!

Solve the first equation in the system for y.

Alja [10]

Answer:

y=1-2x

x=2

(x,y)=(2,-3)

Step-by-step explanation:

5 0

3 years ago

The ball followed a path modelled by the equation h = −0.001! + 0.5 + 2.5 where h is the height of the ball in feet and is the h

Mama L [17]

The heights the balls hit a fence at 350 ft distance are 65 feet, 38 feet and 30 feet, respectively

<h3>Represent the distance-height relationship for each player’s ball as an equation, in a table and on a graph. </h3>

Juan's equation is given as:

h = -0.001d^2 + 0.5d + 2.5

h =

Set d to multiples of 50 from 0 to 400.

So, the table of values of Juan's function is:

d (ft) h(ft)

0 2.5

50 25

100 42.5

150 55

200 62.5

250 65

300 62.5

350 65

400 42.5

See attachment for the graph of Juan's function

A quadratic function is represented as:

h = ad^2 + bd + c

Using the values on the table of values, we have:

c = 3 -- the constant value

So, the equation becomes

h = ad^2 + bd + 3

Using the two other values on the table of values, we have:

23 = a(50)^2 + b(50) + 3

38 = a(100)^2 + b(100) + 3

Using a graphing tool, we have:

a = -0.001

b = 0.45

So, Mark's equation is h(d) = -0.001d^2 + 0.45d + 3

See attachment for Mark's graph.

<u>Barry</u>

From the graph, we have the table of values of Barry's function to be:

d (ft) h(ft)

0 2.5

50 21

100 35

150 44

200 48

250 46

300 41

350 30

400 14

450 0

Using a graphing tool, we have the quadratic function to be:

y = -0.001x^2 +0.4x +2.5

<h3><u>The shortest and the greatest distance before hitting the ground</u></h3>

From the graphs, equations and tables, the distance travelled by the balls are:

Juan = 505 feet

Mark = 457 feet

Barry = 450 feet

This means that Juan's ball would travel the greatest distance while Barry's ball would travel the shortest.

<h3>The height the balls hit a fence at 350 ft distance</h3>

To do this, we set d = 350

From the graphs, equations and tables, the height at 350 ft by the balls are:

Juan = 65 feet

Mark = 38 feet

Barry = 30 feet

The above represents the height the balls hit the fence

Read more about quadratic functions at:

brainly.com/question/12446886

#SPJ1

4 0

2 years ago

A farmer owns 30 acres of land on which he wishes to grow corn and barely. The cost per acre for seedcorn is $30, and the cost p

Sunny_sXe [5.5K]

ok hola bro graicas por los punto qui :

6 0

3 years ago

Find the area of the composite figure.

Roman55 [17]

Answer:

The Area of the composite figure would be 76.26 in^2

Step-by-step explanation:

<u>According to the Figure Given:</u>

Total Horizontal Distance = 14 in

Length = 6 in

<u>To Find :</u>

The Area of the composite figure

<u>Solution:</u>

Firstly we need to find the area of Rectangular part.

So We know that,

$\boxed{ \rm \: Area \: of \: Rectangle = Length×Breadth}$

Here, Length is 6 in but the breadth is unknown.

To Find out the breadth, we’ll use this formula:

$\boxed{\rm \: Breadth = total \: distance - Radius}$

According to the Figure, we can see one side of a rectangle and radius of the circle are common, hence,

$\longrightarrow\rm \: Length \: of \: the \: circle = Radius$

Since Length = 6 in ;

$\longrightarrow \rm \: 6 \: in = radius$

Hence Radius is 6 in.

So Substitute the value of Total distance and Radius:

Total Horizontal Distance= 14
Radius = 6

$\longrightarrow\rm \: Breadth = 14-6$

$\longrightarrow\rm \: Breadth = 8 \: in$

Hence, the Breadth is 8 in.

Then, Substitute the values of Length and Breadth in the formula of Rectangle :

Length = 6
Breadth = 8

$\longrightarrow\rm \: Area \: of \: Rectangle = 6 \times 8$

$\longrightarrow \rm \: Area \: of \: Rectangle = 48 \: in {}^{2}$

Then, We need to find the area of Quarter circle :

We know that,

$\boxed{\rm Area_{(Quarter \; Circle) } = \cfrac{\pi{r} {}^{2} }{4}}$

Now Substitute their values:

r = radius = 6
π = 3.14

$\longrightarrow\rm Area_{(Quarter \; Circle) } = \cfrac{3.14 \times 6 {}^{2} }{4}$

Solve it.

$\longrightarrow\rm Area_{(Quarter \; Circle) } = \cfrac{3.14 \times 36}{4}$

$\longrightarrow\rm Area_{(Quarter \; Circle) } = \cfrac{3.14 \times \cancel{{36} } \: ^{9} }{ \cancel4}$

$\longrightarrow\rm Area_{(Quarter \; Circle)} =3.14 \times 9$

$\longrightarrow\rm Area_{(Quarter \; Circle) } = 28.26 \: {in}^{2}$

Now we can Find out the total Area of composite figure:

We know that,

$\boxed{ \rm \: Area_{(Composite Figure)} =Area_{(rectangle)}+ Area_{ (Quarter Circle)}}$

So Substitute their values:

$\rm Area_{(rectangle)}$ = 48
$\rm Area_{(Quarter Circle)}$ = 28.26

$\longrightarrow \rm \: Area_{(Composite Figure)} =48 + 28 .26$

Solve it.

$\longrightarrow \rm \: Area_{(Composite Figure)} =\boxed{\tt 76.26 \:\rm in {}^{2}}$

Hence, the area of the composite figure would be 76.26 in² or 76.26 sq. in.

$\rule{225pt}{2pt}$

I hope this helps!

3 0

2 years ago

2(x−3)=(x−1)+7

Oksanka [162]

Answer:

x = 12

(getting to the 20 character limit, please ignore this)

6 0

3 years ago

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