- 4+-3 - (-11) =<br> O<br> O 10<br> 0-18<br> 04

What is the area of the rhombus shown below? AC=17 BD=15 AB=11.2 Answers: A. 255 square units B. 190.4 square units C. 16 square

Lena [83]

Answer:

D. 127.5 square units

Step-by-step explanation:

AC=17

BD=15

AB=11.5

Diagonal (1&2) are given, base of the rhombus is also given.

Height is not given

Area of a rhombus given the diagonals

=1/2×d1×d2

Where,

d1=AC=17

d2=BD=15

Area of a rhombus=1/2×d1×d2

=1/2×17*15

=1/2×255

=127.5

D. 127.5 square units

4 0

3 years ago

Evaluate the integral of the quantity x divided by the quantity x to the fourth plus sixteen, dx . (2 points) one eighth times t

Anika [276]

Answer:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

Step-by-step explanation:

Given

$\int\limits {\frac{x}{x^4 + 16}} \, dx$

Required

Solve

Let

$u = \frac{x^2}{4}$

Differentiate

$du = 2 * \frac{x^{2-1}}{4}\ dx$

$du = 2 * \frac{x}{4}\ dx$

$du = \frac{x}{2}\ dx$

Make dx the subject

$dx = \frac{2}{x}\ du$

The given integral becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{x}{x^4 + 16}} \, * \frac{2}{x}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{1}{x^4 + 16}} \, * \frac{2}{1}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

Recall that: $u = \frac{x^2}{4}$

Make $x^2$ the subject

$x^2= 4u$

Square both sides

$x^4= (4u)^2$

$x^4= 16u^2$

Substitute $16u^2$ for $x^4$ in $\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16u^2 + 16}} \,\ du$

Simplify

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16}* \frac{1}{8u^2 + 8}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{2}{16}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

In standard integration

$\int\limits {\frac{1}{u^2 + 1}} \,\ du = arctan(u)$

So, the expression becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(u)$

Recall that: $u = \frac{x^2}{4}$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

4 0

2 years ago

How do you do this question?

DENIUS [597]

Answer:

z = -ln│5eᵗ + C│

Step-by-step explanation:

dz/dt + 5eᵗ⁺ᶻ = 0

dz/dt = -5eᵗ⁺ᶻ

dz/dt = -5eᵗeᶻ

-e⁻ᶻ dz = 5eᵗ dt

e⁻ᶻ = 5eᵗ + C

-z = ln│5eᵗ + C│

z = -ln│5eᵗ + C│

7 0

3 years ago

HELP, I'M GIVING A GOOD AMOUNT OF POINTS

jekas [21]

Answer:

I need to know what the question asks

Step-by-step explanation:

5 0

3 years ago

HELP ASAP!!! WILL GIVE BRAINLEIST!!!

netineya [11]

Answer:

The first question is (n*8) - 4

The second question is no.

Step-by-step explanation:

The first question, you add 8. The first is 1. 1 * 8 = 8 - 4 = 4.

Therefore, the equation is (n*8) - 4

The second question, you can plug in numbers. You can solve for a term. -100 + 17 = -83. -83 is not a multiple of -4. -100 is not a term.

8 0

3 years ago

- 4+-3 - (-11) = O O 10 0-18 04