Please help me Find x, y, and z

Evaulate the function at the indicated values h(t)=t+(3/t)i find h(x-1)

alex41 [277]

h(t)=t34−3t14

Note that the domain of h is [0,∞].

By differentiating,

h'(t)=34t−14−34t−34

by factoring out 34,

=34(1t14−1t34)

by finding the common denominator,

=34t12−1t34=0

⇒t12=1⇒t=1

Since h'(0) is undefined, t=0 is also a critical number.

Hence, the critical numbers are t=0,1.

I hope that this was helpful.

6 0

2 years ago

1+secA/sec A = sin^2 A / 1-cos A

Fofino [41]

Answer: see proof below

Step-by-step explanation:

$\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos A}$

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

Proof LHS → RHS

$\text{LHS:}\qquad \qquad \dfrac{1+\sec A}{\sec A}$

$\text{Identity:}\qquad \qquad \dfrac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}$

$\text{Simplify:}\qquad \qquad \dfrac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}\\\\\\.\qquad \qquad \qquad =\dfrac{1+\cos A}{1}$

$\text{Multiply:}\qquad \qquad \dfrac{1+\cos A}{1}\cdot \bigg(\dfrac{1-\cos A}{1-\cos A}\bigg)\\\\\\.\qquad \qquad \qquad =\dfrac{1-\cos^2 A}{1-\cos A}$

$\text{Identity:}\qquad \qquad \dfrac{\sin^2 A}{1-\cos A}$

$\text{LHS = RHS:}\quad \dfrac{\sin^2 A}{1-\cos A}=\dfrac{\sin^2 A}{1-\cos A}\quad \checkmark$

3 0

3 years ago

Provides data on sleep deprivation rates of Californians and Oregonians. The proportion of California residents who reported ins

Bond [772]

Answer:

Step-by-step explanation:

Download docx

8 0

2 years ago

Convert to the product of multiplication a^2-b^2-4b-4

kirill [66]

$a^2-b^2-4b-4$ in form of multiplication is $(a-b)(a+b)-4(b+1)$ .

Step-by-step explanation:

We know that factors are the numbers that are being multiplied together . As 6532*7 = 45724 .Here 6532 & 7 are the factors and 45724 is product of the factors .

Here , we need to convert to the product of multiplication

a^2-b^2-4b-4 or , $a^2-b^2-4b-4$ :

⇒ $a^2-b^2-4b-4$

We know by identity that , $p^2-q^2 = (p-q)(p+q)$

⇒ $(a-b)(a+b)-4b-4$

Taking common term out as -4 :

⇒ $(a-b)(a+b)-4(b+1)$

Therefore , $a^2-b^2-4b-4$ in form of multiplication is $(a-b)(a+b)-4(b+1)$ .

7 0

2 years ago

Suppose the diameter of a circle is \color{green}{6}6start color green, 6, end color green units. What is its circumference?

inessss [21]

Answer:

C = 6pi units

or

C is approximately 18.84 units

Step-by-step explanation:

The circumference of a circle is given by

C = pi *d

The diameter is 6 units

C = 6*pi units

If we approximate pi by 3.14

C = 6*3.14

C = 18.84 units

3 0

3 years ago