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Olenka [21]
3 years ago
8

Find the length of the altitude drawn to the hypotenuse.

Mathematics
1 answer:
loris [4]3 years ago
3 0

Answer:

A. 4 \sqrt{5}

Step-by-step explanation:

By geometric mean theorem:

Length of altitude

=  \sqrt{5 \times 16}  \\  =  \sqrt{5 \times  {4}^{2} }  \\  = 4 \sqrt{5}

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Jessica is eating a bag of sour skittles . Her friend ask her to share, and Jessica throws over an extra bag. Her friend does no
liraira [26]

Answer:

A parabola opens downward with vertex at (1,20)

Step-by-step explanation:

The function that models the height of bag of candies is

h(t) =  - 16 {t}^{2}  + 32t + 4

The graph of this function is a parabola that opens downward.

To find the maximum height of the candy bag, we complete the squares to obtain the vertex form of the function.

Factor -16 to get;

h(t) =  - 16( {t}^{2}  - 2t) + 4

Add and subtract the square of half the coefficient of t.

h(t) =  - 16( {t}^{2}  - 2t + 1)  + 16+ 4

Factor the perfect square trinomial:

h(t) =  - 16( {t - 1)}^{2}  + 20

The vertex is (1,20)

This means the maximum height after 1 second is 20 feet.

7 0
3 years ago
just help prepare one half of the bicycles and a bike shop last week if Jeff work on 5 bicycles how many bicycles did the shop r
Bond [772]
The shop would have repaired 10 because Jeff only repaired half so you would do 5 x 2 = 10 to double what Jeff did to show how many bikes the shop fixed.
6 0
3 years ago
Combine like terms to create an equivalent expression.​
Andrews [41]

Answer:

3w - 9/2 is the equivalent expression

5 0
3 years ago
Read 2 more answers
Use this information to complete activities 1.1
Llana [10]
14*4= 56 
so 56 would be the answer 
5 0
3 years ago
A force of 8 N makes an angle of π/4 radian with the y-axis, pointing to the right. The force acts against the movement of an ob
Grace [21]

Answer:

a) F= 4\sqrt{2} i + 4\sqrt{2} j

b) Θ=11.3

c) The work done is 20\sqrt{2}

Step-by-step explanation:

a) ||F||=8, α=π/4

Fx=||F||·sin(π/4)=8·\frac{\sqrt{2}}{2}

Fy=||F||·cos(π/4)=8·\frac{\sqrt{2}}{2}

F=Fx i + Fy j = 4\sqrt{2} i + 4\sqrt{2} j

b) We can find the value of Θ using the equation:

cos(Θ)=\frac{F.D}{||F||||D||}

where:

D= 3 i + 2 j

F=4\sqrt{2} i + 4\sqrt{2} j

The dot product is defined as the sum of the products of the components of each vector as:

F · D= (4\sqrt{2})*3+(4\sqrt{2})*2=20\sqrt{2}

||F||= 8

||D||= \sqrt{3^2+2^2} =\sqrt{13}

Hence:

Θ=arccos(\frac{20\sqrt{2} }{8\sqrt{13} })

Θ=arccos(0.981)

Θ= 11.3°

c) Work is equal to:

F · D= (4\sqrt{2})*3+(4\sqrt{2})*2=20\sqrt{2}=28.3

Other way of obtainig the work is:

||F||||D||cos(Θ)

where:

||F||= 8

||D||= \sqrt{3^2+2^2} =\sqrt{13}

Θ=11.3°

So,  ||F||||D||cos(Θ)=8×\sqrt{13}×cos(11.3°)=28.3

3 0
3 years ago
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