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3 years ago

13

Evaluate square root of 16 minus x end root when x = 8.

1 answer:

My name is Ann [436]3 years ago

7 0

$\sqrt{16-x}$
x=8
16-8=8
$\sqrt{8}$

Use a calculator to find the square root of 8.

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The answer is B) 2 and 3

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If in State A, you pay $6,000 per year in taxes, but state B you pay $1500 in taxes, why won't you leave State A?

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3 years ago

F(x)=2x^3+ax^2-37x-60<br><br> The binomial x-4 is a factor of f(x). Find the value of a.

jekas [21]

Given:

The function is

$f(x)=2x^3+ax^2-37x-60$

The binomial $x-4$ is a factor of f(x).

To find:

The value of $a$ .

Solution:

If $x-c$ is a factor of f(x), then $f(c)=0$ .

It is given that, $x-4$ is a factor of f(x), then $f(4)=0$ .

We have,

$f(x)=2x^3+ax^2-37x-60$

Substituting x=4, we get

$f(4)=2(4)^3+a(4)^2-37(4)-60$

$f(4)=2(64)+a(16)-148-60$

$f(4)=128+16a-208$

$f(4)=16a-80$

Now,

$f(4)=0$

$16a-80=0$

$16a=80$

$a=\dfrac{80}{16}$

$a=5$

Therefore, the value of a is 5.

6 0

2 years ago

2. When Joey dives off a diving board, the equation of his pathway can be modeled by h = -16- + 15 + 12.

Sergeu [11.5K]

Answer:

a) The maximum height is approx 15.5 unit.

b) The time it will take for Joey to reach the water is 1.45 hour.

Step-by-step explanation:

Given : When Joey dives off a diving board, the equation of his pathway can be modeled by $h(t)= -16t^2+15t + 12$

To find : a) Find Joey's maximum height.

b) Find the time it will take for Joey to reach the water.

Solution :

Modeled $h(t)= -16t^2+15t + 12$ ....(1)

a) To find maximum height

Derivate (1) w.r.t. t,

$h'= -32t+15$

For critical point put h'=0,

$-32t+15=0$

$t=\frac{15}{32}$

$t=0.46875$

Derivate again w.r.t. t,

$h''= -32$

It is maximum at t=0.46875.

Substitute t in equation (1),

$h(0.46875)= -16(0.46875)^2+15(0.46875)+12$

$h(0.46875)= -3.515625+7.03125+12$

$h(0.46875)= 15.515625$

The maximum height is approx 15.5 unit.

b) To find the time it will take for Joey to reach the water.

Put h=0 in equation (1),

$-16t^2+15t + 12=0$

Apply quadratic formula, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Here, a=-16 , b=15, c=12

$t=\frac{-15\pm\sqrt{(15)^2-4(-16)(12)}}{2(-16)}$

$t=\frac{-15\pm\sqrt{225+768}}{-32}$

$t=\frac{-15\pm\sqrt{993}}{-32}$

$t=\frac{-15+\sqrt{993}}{-32},\frac{-15-\sqrt{993}}{-32}$

$t=−0.515,1.453$

Reject negative value.

The time is t=1.45.

The time it will take for Joey to reach the water is 1.45 hour.

5 0

3 years ago