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2 years ago

PLEASE HELP NOW!!!!!!!!!

1 answer:

4 0

Answer: In the unlikely event that the owner is really, truly, okay with this, you will still have the reputation as a thief, someone who is unable to do anything original and has to steal in order to pretend to create something.

That reputation will follow you. And in the equally unlikely event, you should get fame, that reputation will ruin everything you gained and no one will ever trust that anything you write is actually yours.

Step-by-step explanation: Hope it's Helpful

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Area of the following kite?

MakcuM [25]

Answer:

i think A=20

A=pq/2 =4·10/2 =20

Hope i helped

-lvr

6 0

3 years ago

Read 2 more answers

Sin tita= 0.6892.find the value Of tita correct to two decimal places

Anvisha [2.4K]

Answer:

$\theta \approx 6.28n + 2.38, \quad n \in \mathbb{Z}$

$\theta \approx 6.28n + 0.76, \quad n \in \mathbb{Z}$

Considering $\theta \in (0, 2\pi]$

$\theta \approx 2.38$

$\theta \approx 0.76$

Step-by-step explanation:

$\sin(\theta)=0.6892$

We have:

$\sin (x)=a \Longrightarrow x=\arcsin (a)+2\pi n \text{ or } x=\pi -\arcsin (a)+2\pi n \text{ as } n\in \mathbb{Z}$

Therefore,

$\theta= \arcsin (0.6892)+2\pi n, \quad n \in \mathbb{Z}$

$\theta = \pi -\arcsin (0.6892)+2\pi n, \quad n\in \mathbb{Z}$

---------------------------------

$\theta \approx 6.28n + 2.38, \quad n \in \mathbb{Z}$

$\theta \approx 6.28n + 0.76, \quad n \in \mathbb{Z}$

4 0

3 years ago

Given a rectangle with length of (2x+9)cm and width of (3x+1)cm.Two squares, each with sides x cm is removed from the rectangle.

il63 [147K]

Answer: The length is 13cm and the width is 7cm

Step-by-step explanation:

For a rectangle of length L and width W, the area is:

A = W*L

In this case we have:

L = (2*x + 9) cm

W=(3*x + 1) cm

Then the area of the rectangle is:

A = (2*x + 9)*(3*x + 1) cm^2

A = (6*x^2 + 2*x + 27*x + 9) cm^2

A = (6*x^2 + 29*x + 9) cm^2

now we remove two squares with sides of x cm

The area of each one of these squares is (x cm)*(x cm) = x^2 cm^2

Then the area of the figure will be:

area = (6*x^2 + 29*x + 9) cm^2 - (2*x^2 ) cm^2

area = (4*x^2 + 29*x + 9) cm^2

Now we know that the area of this shape is 83 cm^2, then we need to solve:

83 cm^2 = (4*x^2 + 29*x + 9) cm^2

0 = (4*x^2 + 29*x + 9) cm^2 - 83 cm^2

0 = (4*x^2 + 29*x - 74) cm^2

Then we need to solve:

0 = 4*x^2 + 29*x - 74

Here we can use Bhaskara's equation, the solutions of this equation are given by:

$x = \frac{-29 \pm \sqrt{29^2 - 4*4*(-74)} }{2*4} = \frac{-29 \pm 45}{8}$

Then the two solutions are:

x = (-29 - 45)/8 = -9.25 (for how the length and width are defined, we can not have x as a negative number, then this solution can be discarded).

The other solution is:

x = (-29 + 45)/8 = 2

x = 2

Then the length and width of the rectangle are:

Length = (2*2 + 9)cm = 13 cm

Width = (3*2 + 1)cm = 7cm

4 0

3 years ago

What is 2x+y=3 and x+y=3 1.{-1.5,0} 2.{0,-1.5] 3.{0,3} 4.{3,0}

Yakvenalex [24]

The answer would be 3. {0,3}.

Remember: {x,y}

2(0)+3=3
0+3=3

0+3=3

4 0

3 years ago

Is true that the equation of a line in slope-intercept form is y = mx + b where m is the y-intercept?

Elan Coil [88]

No, b is the y-intercept, m is the slope

3 0

3 years ago