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gogolik [260]
3 years ago
11

The function h(t) = -16t^2 + 16t represents the height (in feet) of a horse (t) seconds after it jumps during a steeplechase.

Mathematics
1 answer:
Vladimir79 [104]3 years ago
7 0
A.) To find the maximum height, we can take the derivative of h(t). This will give us the rate at which the horse jumps (velocity) at time t.

h'(t) = -32t + 16

When the horse reaches its maximum height, its position on h(t) will be at the top of the parabola. The slope at this point will be zero because the line tangent to the peak of a parabola is a horizontal line. By setting h'(t) equal to 0, we can find the critical numbers which will be the maximum and minimum t values.

-32t + 16 = 0

-32t = -16

t = 0.5 seconds

b.) To find out if the horse can clear a fence that is 3.5 feet tall, we can plug 0.5 in for t in h(t) and solve for the maximum height.

h(0.5) = -16(0.5)^2 + 16(-0.5) = 4 feet

If 4 is the maximum height the horse can jump, then yes, it can clear a 3.5 foot tall fence.

c.) We know that the horse is in the air whenever h(t) is greater than 0. 

-16t^2 + 16t = 0

-16t(t-1)=0

t = 0 and 1

So if the horse is on the ground at t = 0 and t = 1, then we know it was in the air for 1 second.
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Create a matrix for this system of linear equations, The determinant of the coefficient matrix is
marin [14]

Answer:  The determinant of the coefficient matrix is -15 and x = 3, y = 4, z = 1.

Step-by-step explanation:  The given system of linear equations is :

2x+y+3z=13~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\x+2y=11~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)\\\\3x+z=10~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)

We are given to find the determinant of the coefficient matrix and to find the values of x, y and z.

The determinant of the co-efficient matrix is given by

D=\begin{vmatrix}2 & 1 & 3\\ 1 & 2 & 0\\ 3 & 0 & 1\end{vmatrix}=2(2-0)+1(0-1)+3(0-6)=4-1-18=-15.

Now, from equations (ii) and (iii), we have

x+2y=11~~~~~\Rightarrow y=\dfrac{11-x}{2}~~~~~~~~~~~~~~~~~~~~~~~~~~(iv)\\\\\\3x+z=10~~~~~~\Rightarrow z=10-3x~~~~~~~~~~~~~~~~~~~~~~~~~(v)

Substituting the value of y and z from equations (iv) and (v) in equation (i), we get

2x+y+3z=13\\\\\Rightarrow 2x+\dfrac{11-x}{2}+3(10-3x)=13\\\\\Rightarrow 4x+11-x+60-18x=26\\\\\Rightarrow -15x+71=26\\\\\Rightarrow -15x=26-71\\\\\Rightarrow -15x=-45\\\\\Rightarrow x=3.

From equations (iv) and (v), we get

y=\dfrac{11-3}{2}=4,\\\\z=10-3\times3=1.

Thus, the determinant of the coefficient matrix is -15 and x = 3, y = 4, z = 1.

3 0
3 years ago
The 6th grade science teacher gives a test every 12 days and the math teacher gives a test every 9 days. Today, both the science
Y_Kistochka [10]

In 36 days will the science and math teachers both give tests on the same day again.

<u>Step-by-step explanation:</u>

Here we have , The 6th grade science teacher gives a test every 12 days and the math teacher gives a test every 9 days. Today, both the science and math are giving tests. We need to find In how many days will the science and math teachers both give tests on the same day again . Let's find out:

It's given that  science teacher gives a test every 12 days and the math teacher gives a test every 9 days . In order to find the number of days after which they'll give test at same day  , we will find LCM of 12 & 9 . i.e.

⇒ 12(1)=12\\12(2)=24\\12(3)=36       and  ,

⇒ 9(1)=9\\9(2)=18\\9(3)=27\\9(4)=36

Hence , LCM of 12 and 9 is 36 . Therefore , In 36 days will the science and math teachers both give tests on the same day again.

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4 years ago
I WILL GIVE BRAINLIEST IMMEDIATELY!! I REPEAT, IMMEDIATELY!!!!
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Answer:

B

Step-by-step explanation:

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3 years ago
Simplify 4^9/4^3<br> can someone help me with this?
zaharov [31]

Answer:

4096

Step-by-step explanation:

Need to know how you got it? Comment on this answer.

4 0
3 years ago
This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the ex
olga nikolaevna [1]

Answer:

maximum value , f max = 169 (taking the most out of the 4th power)

minimum value , f min = 169/3 (taking the least out of the 4th power)

Step-by-step explanation:

A) Since both function and restriction are symmetrical with respect to x,y and z, there is no reason for one to be more important than the others and therefore one solution would be x=y=z=λ and thus

x2 + y2 + z2 = 13  → 3λ² = 13 →λ² = 13/3

and f would be

f (x, y, z) = x4 + y4 + z4 = 3λ⁴ = 3*(13/3)²=13²/3=169/3

since x⁴ increases faster than 3*x² , f(x,y,z) would be a minimum

and the maximum value would be obtained taking the most out of x⁴, thus doing 2 coordinates =0 ( can be x=0 and y=0) and

z²= 13

f (x, y, z) = x4 + y4 + z4 = 13² = 169

B) strictly, using Lagrange multipliers

f (x, y, z) = x4 + y4 + z4

g (x, y, z) = x2 + y2 + z2 - 13

F(x,y,z) = f (x, y, z) -λ*g (x, y, z)

such that

Fx (x,y,z)=  fx(x, y, z) -λ*gx (x, y, z) = 0 → 4*x³ - λ*2*x = 0 → 2*x*(2*x² -λ) = 0

thus x=0 or x²= λ/2

Fy (x,y,z)=  fy(x, y, z) -λ*gy (x, y, z)= 0 → 4*y³ - λ*2*y = 0 → 2*y*(2*y² - λ) = 0

thus y=0 or y²= λ/2

Fz (x,y,z)=  fz(x, y, z) -λ*gz (x, y, z)= 0 → 4*z³ - λ*2*z = 0→ 2*z*(2*z² - λ) = 0

thus z=0 or z²= λ/2

g (x, y, z) = 0  → x2 + y2 + z2 = 13 → 3*(λ/2) = 13 → λ=13*2/3

thus  x²=y²=z²= λ/2 =13/3

f min = f (x, y, z) = x4 + y4 + z4 = 3*(13/3)²=169/3

for the x=0 , y=0 → z²= 13

f max = f (x, y, z) = x4 + y4 + z4 = 13² = 169

3 0
3 years ago
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