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3 years ago

What are the composite number less then 22

Mathematics

1 answer:

agasfer [191]3 years ago

7 0

Answer: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21

Step-by-step explanation:

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Given the function f ( x ) = x^2 + 7 x + 10/ x^2 + 9 x + 20 Describe where the function has a hole and how you found your answer

ryzh [129]

${x}^{2} + 7x + \frac{10}{ {x}^{2} } + 9x + 20 \\ = {x}^{2} + 16x + \frac{10}{ {x}^{2} } + 20 \\ x = 0 \\ \frac{10}{ {x}^{2} } = \frac{10}{0}$

= undefined

It is undefined as you cannot divide by zero. Hence, there is a hole at x = 0.

8 0

3 years ago

2+2<br> Show your work and the first answer gets a brainliest

bazaltina [42]

Answer:

Four/4

Step-by-step explanation:

Add both numbers together.

2+2

8 0

3 years ago

Please help and explain how you did it

jenyasd209 [6]

To determine the lengths of the sides from shortest to longest, you need to calculate the corresponding angles. The higher angles will correspond to longer sides.

To find the angles, you have to solve for x. You’re already given that angle A is 76. To find the others, you know that angle C is 180-(16x+16) since it’s supplemental to the exterior angle. Then, you know the sum of the angles of the entire triangle is 180, so add up A, B, and C

A+B+C=180
76+6x+(180-16x-16)=180
240-10x=180
-10x=-60
x=6

So to find angle B, you use 6x or 6(6)=36.

To find angle C, you use 180-(16x-16) or 180-16(6)-16=68

So now match up the angles with their corresponding sides to find the length from shortest to longest.

Angle A (76) corresponds with BC
Angle B (36) corresponds with AC
Angle C (68) corresponds with AB

Again, the higher the degree, the longer the corresponding side, so AC is shortest, AB is next, and BC is the longest.

5 0

3 years ago

Find the solution of the initial value problem y Superscript prime prime Baseline plus 4 y Superscript prime Baseline plus 5 y e

sergiy2304 [10]

Answer:

$y(t) = - 5 {e}^{2\pi - 2t} \cos(t)$

Step-by-step explanation:

The given initial value problem is

$y''+4y'+5=0$

$y( \frac{ \pi}{2} ) = 0$

$y'( \frac{ \pi}{2} ) = 5$

The corresponding characteristic equation is

${m}^{2} + 4m + 5 = 0$

$m = - 2 \pm \: i$

The general solution becomes:

$y(t) = A {e}^{ - 2t} \cos(t) + B {e}^{ - 2t} \sin(t)$

We differentiate to get:

$y'(t) = - A {e}^{ - 2t} \sin(t) - 2 A {e}^{ - 2t} \cos(t) + B {e}^{ - 2t} \cos(t) - 2B {e}^{ - 2t} \sin(t)$

We apply the initial conditions to get;

$y( \frac{\pi}{2} ) = A {e}^{ - 2\pi} \cos( \frac{\pi}{2} ) + B {e}^{ - 2\pi} \sin( \frac{\pi}{2} )$

$A {e}^{ - 2\pi} (0 ) + B {e}^{ - 2\pi} ( 1) = 0$

$B = 0$

Also;

$y'( \frac{\pi}{2} ) = - A {e}^{ - 2\pi} \sin( \frac{\pi}{2} ) - 2 A {e}^{ - 2\pi} \cos( \frac{\pi}{2} ) + B {e}^{ - 2\pi} \cos( \frac{\pi}{2} ) - 2B {e}^{ - 2\pi} \sin( \frac{\pi}{2} )$