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gogolik [260]
3 years ago
7

4/35 x 10 in simplest form

Mathematics
2 answers:
zavuch27 [327]3 years ago
8 0

Answer:

1 1/7 or 1.14285714286

Step-by-step explanation:

I hope this helped!

Anastaziya [24]3 years ago
7 0

Answer:

8/7

Step-by-step explanation:

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Are the triangles congruent?<br><br>a-yes<br>b-no<br>c-not enough information<br>​
ioda

the right answer is yes

5 0
3 years ago
A football is thrown from the top of the stands, 50 feet above the ground at an initial velocity of 62 ft/sec and at an angle of
Anvisha [2.4K]

a. i. The parametric equation for the horizontal movement is x = 43.84t

ii. The parametric equation for the vertical movement is y = 50 + 43.84t

b. the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

<h2>a. Parametric equations</h2>

A parametric equation is an equation that defines a set of quantities a functions of one or more independent variables called parameters.

<h3>i. Parametric equation for the horizontal movement</h3>

The parametric equation for the horizontal movement is x = 43.84t

Since

  • the angle of elevation is Ф = 45° and
  • the initial velocity, v = 62 ft/s,

the horizontal component of the velocity is v' = vcosФ.

So, the horizontal distance the football moves in time, t is x = vcosФt

= vtcosФ

= 62tcos45°

= 62t × 0.7071

= 43.84t

So, the parametric equation for the horizontal movement is x = 43.84t

<h3>ii Parametric equation for the vertical movement</h3>

The parametric equation for the vertical movement is y = 50 + 43.84t

Also, since

  • the angle of elevation is Ф = 45° and
  • the initial velocity, v = 62 ft/s,

the vertical component of the velocity is v" = vsinФ.

Since the football is initially at a height of h = 50 feet, the vertical distance the football moves in time, t relative to the ground is y = 50 + vsinФt

= 50 + vtcosФ

= 50 + 62tsin45°

= 50 + 62t × 0.7071

= 50 + 43.84t

<h3>b. Location of football at maximum height relative to starting point</h3>

The location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Since the football reaches maximum height at t = 1.37 s

The x coordinate of its location at maximum height is gotten by substituting t = 1.37 into x = 48.84t

So, x = 43.84t

x = 43.84 × 1.37

x = 60.0608

x ≅ 60.1 ft

The y coordinate of the football's location at maximum height relative to the ground is y = 50 + 48.84t

The y coordinate of the football's location at maximum height relative to the starting point is y - 50 = 48.84t

So,  y - 50 = 48.84t

y - 50 = 43.84 × 1.37

y - 50 = 60.0608

y - 50 ≅ 60.1 ft

So, the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Learn ore about parametric equations here:

brainly.com/question/8674159

5 0
2 years ago
Wat is 1+1 ummmmmmmmmmmmmmmm i dont know
Lisa [10]

Answer:

3

Step-by-step explanation:

7 0
3 years ago
A teacher is four time as old as a student .in 20 years, the student 's age will be half of the teacher 's age. How old are they
notka56 [123]
Let the present age of student be x
Present of teacher will be 4x

After 20 years,
Age of student = x + 20
Age of teacher = 4x + 20

According to the given condition after 20 years,
x + 20 = (4x + 20)/2
x + 20 = 2x + 10
2x - x = 20 - 10
x = 10

So student's present age is 10 years while teacher's is 4 x 10 i.e 40 years.

Hope This Helps You!
5 0
3 years ago
the volume of a cube is increasing at a constant rate of 824 cubic centimeters per minute. At the instant when the volume of the
Maurinko [17]

Answer:

DA/dt  = 75.27 cm²

Step-by-step explanation:

Cube Volume    =   V(c)   =  683  cm³

DV(c) /dt    =  824 cm³

V(c,x) = x³    

Then

DV(c,x)/ dt   =  3x² Dx/dt

( DV(c,x)/ dt )/ 3x²  =  Dx/dt       (1)

Now  as  V(c,x) = x³    when    V(c,x)  =  683 cm³     x = ∛683

x  =  8.806     ( from excel)

And by subtitution of this value in equation (1)

Dx/dt   =  ( DV(c,x)/ dt )/ 3x²   ⇒  Dx/dt  =  824 / 3*x²

  Dx/dt   = 824 /3*77.55

Dx/dt   =  824/232,64

Dx/dt   = 3,542

Then we got  Dx/dt  where x is cube edge. The area of the face is x² then

the rate of change of the suface area is  

DA/dt  = ( Dx/dt )²*6         ( 6 faces of a cube)

DA/dt  = (3.542)²*6

DA/dt  = 75.27 cm²

6 0
2 years ago
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