Evaluate p\4 when p=12

Which of the following happens when unbalanced forces act on an object?

tigry1 [53]

Answer:

C. The object moves in the direction of the stronger force.

Step-by-step explanation:

A.) The object remains stationary.

- Cannot be since an outside force is applied, the net force is uneven, making the object move

B.) The object moves with the same velocity.

The object at rest has the velocity of 0m/s and since an outside force is exerted the object moved.

C.) This statement is correct since the object moved towards the direction the force is applied.

D.) Another wrong statement; unbalanced force is applied, the object moved.

3 0

3 years ago

Complete the identity.<br> 1) sec^4 x + sec^2 x tan^2 x - 2 tan^4 x = ?

Alecsey [184]

Answer:

See Explanation

Step-by-step explanation:

<em>Question like this are better answered if there are list of options; However, I'll simplify as far as the expression can be simplified</em>

Given

$sec^4 x + sec^2 x tan^2 x - 2 tan^4 x$

Required

Simplify

$(sec^2 x)^2 + sec^2 x tan^2 x - 2 (tan^2 x)^2$

Represent $sec^2x$ with a

Represent $tan^2x$ with b

The expression becomes

$a^2 + ab- 2 b^2$

Factorize

$a^2 + 2ab -ab- 2 b^2$

$a(a + 2b) -b(a+ 2 b)$

$(a -b) (a+ 2 b)$

Recall that

$a = sec^2x$

$b = tan^2x$

The expression $(a -b) (a+ 2 b)$ becomes

$(sec^2x -tan^2x) (sec^2x+ 2 tan^2x)$

..............................................................................................................................

In trigonometry

$sec^2x =1 +tan^2x$

Subtract $tan^2x$ from both sides

$sec^2x - tan^2x =1 +tan^2x - tan^2x$

$sec^2x - tan^2x =1$

..............................................................................................................................

Substitute 1 for $sec^2x - tan^2x$ in $(sec^2x -tan^2x) (sec^2x+ 2 tan^2x)$

$(1) (sec^2x+ 2 tan^2x)$

Open Bracket

$sec^2x+ 2 tan^2x$ ------------------This is an equivalence

$(secx)^2+ 2 (tanx)^2$

Solving further;

................................................................................................................................

In trigonometry

$secx = \frac{1}{cosx}$

$tanx = \frac{sinx}{cosx}$

Substitute the expressions for secx and tanx

................................................................................................................................

$(secx)^2+ 2 (tanx)^2$ becomes

$(\frac{1}{cosx})^2+ 2 (\frac{sinx}{cosx})^2$

Open bracket

$\frac{1}{cos^2x}+ 2 (\frac{sin^2x}{cos^2x})$

$\frac{1}{cos^2x}+ \frac{2sin^2x}{cos^2x}$

Add Fraction

$\frac{1 + 2sin^2x}{cos^2x}$ ------------------------ This is another equivalence

................................................................................................................................

In trigonometry

$sin^2x + cos^2x= 1$

Make $sin^2x$ the subject of formula

$sin^2x= 1 - cos^2x$

................................................................................................................................

Substitute the expressions for $1 - cos^2x$ for $sin^2x$

$\frac{1 + 2(1 - cos^2x)}{cos^2x}$

Open bracket

$\frac{1 + 2 - 2cos^2x}{cos^2x}$

$\frac{3 - 2cos^2x}{cos^2x}$ ---------------------- This is another equivalence

8 0

3 years ago

Completa los datos que hacen folta en la tabla pitagoricacolorea de rojo los multiplos de 8 , azul los 5 ,verde los de 10 y de a

lakkis [162]

Answer:

i don't understand your language broo

6 0

3 years ago

What is the value of x

BaLLatris [955]

Answer:

25

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

I will give a brainlest, i promise, please help me with this.

valkas [14]

Answer:

1) b and m

2) m∠8=m∠6

3) 160°

4)x=60°

Step-by-step explanation:

1) all straight lines sum to 180°

subtract the angles given from 180°

the other angle for b is 25°, while the other angle for m is 155°

so we can see that the angles for both lines are the same, hence they are parallel.

2) ∠8 should be the same as ∠6, ∠10 should be the same as ∠3, ∠7 same as ∠5 and ∠9 same as ∠4

in the options we are only given '∠8 should be the same as ∠6' as the correct answer, so we take that.

3) from the image we can see that both horizontal lines are parallel to each other, so both angles on the lines should be same, so ∠CET would be (2x-16)°

(2x-16)°+(7x+20)°=180°

we get x=20(nearest whole number)

∠CED=7x+20=7(20)+20=160°

4) since we need to show that they are parallel,

(2x+30)°=(4x-90)°

2x-4x=-90-30

-2x=-120

x=60

we then plug the x value into the two equations, in which we get 150° for both the angles [2(60)+30=4(60)-90] ⇒ (150=150)

I hope u understand it the way I put it.

8 0

2 years ago