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3 years ago

3^2x+1 +9 = 3^x+3 +3^3

Mathematics

1 answer:

alisha [4.7K]3 years ago

3 0

Answer:

-4

Step-by-step explanation:

Since they all have the same base;

2x+1+9=x+3+3
2x-x=3+3-1-9
x=6-1-9
x=6-10
x=-4

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Solve for x. Enter your answer as a whole number or reduced fraction. x = ___ a0

miv72 [106K]

Answer:

$x = \frac{1}{2}$

Step-by-step explanation:

$x = log_{e}\sqrt{e} \\ x = log_{e}e^{ \frac{1}{2} } \\ x = \frac{1}{2} log_{e}e \\ x = \frac{1}{2} \times 1 \\ x = \frac{1}{2}$

5 0

3 years ago

If the arc on a particular circle has a length of 3 units, and the degree measure of the arc is 30o, which is the circumference

lord [1]

The answer is not 12.

The ratio of the arc to the angle equals the ratio of the circumference to 360°.

Then 3/30° = circumferece / 360°

3/30 = x / 360 => x = 360 * 3/30 = 36 units.

Answer: 36 units

3 0

3 years ago

Please help with this question!

IgorC [24]

Answer:

QR = 5

Step-by-step explanation:

Since the parallelograms are similar, then the ratios of corresponding sides are equal, that is

$\frac{AB}{PQ}$ = $\frac{BC}{QR}$ , then

$\frac{9}{3}$ = $\frac{15}{QR}$ ( cross- multiply )

9QR = 45 ( divide both sides by 9 )

QR = 5

8 0

3 years ago

Jason is pulling a box across the room. He is pulling with a force of 19 newtons and his arm is making a 30 angle with the horiz

Cerrena [4.2K]

We are to solve for the vertical and horizontal component of the force he is pulling with.

1. The vertical component of Jason's force is 9.5Newtons.

2.The horizontal component of Jason's force is 2.67Newtons.

3. Therefore, the net force in the vertical direction is -12.5Newtons.

4.Therefore, Resultant, R is 72.11

Question 1:

If Jason is pulling with a force of 19 Newtons and his arm is at an angle of 30° with the horizontal.

By resolving Jason's force of pull in the vertical direction, Fy = 19 × Sin 30° = 9.5 Newtons.

The vertical component of Jason's force is 9.5 Newtons.

Question 2:

Also,If Jason is pulling with a force of 14 Newtons and his arm is at an angle of 79° with the horizontal.

By resolving Jason's force of pull in the horizontal direction, Fx = 14 × Cos79° = 2.67 Newtons.

The horizontal component of Jason's force is 2.67Newtons.

Question 3:

If Jason is pulling with a force of 23Newtons and his arm is at an angle of 30° with the horizontal.

By resolving Jason's force of pull in the vertical direction, Fy = 23 × Sin 30° = 11.5Newtons.

The vertical component of Jason's force of pull is, 11.5Newtons.

If the box weighs 24Newtons.

By treating up as the +ve vertical direction and down as the -ve vertical direction,

Therefore, the weight of the box acts in the -ve vertical direction, while Jason's vertical force component acts in the +ve vertical direction.

Therefore, the net force in the vertical direction is = 11.5Newtons + (-24Newtons)

Therefore, the net force in the vertical direction is -12.5Newtons.

Question 4:

If there are two forces at right angles to eachother, one of magnitude, 68 and the other of magnitude, 24.

The resultant force on the object can be obtained by Pythagoras theorem (Triangle law of forces).

Resultant, R = √(68²+24²) = √5200.

Therefore, Resultant, R = 72.11

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4 0

2 years ago

Read 2 more answers

I need help with this problem

Arada [10]

Answer:

65.56°

Step-by-step explanation:

We know that if we take dot product of two vectors then it is equal to the product of magnitudes of the vectors and cosine of the angle between them

That is let p and q be any two vectors and A be the angle between them

So, p·q=|p|*|q|*cosA

⇒ $cosA=\frac{u.v}{|u||v|}$

Given u=-8i-3j and v=-8i+8j

$|u|=\sqrt{(-8)^{2}+ (-3)^{2}} =8.544$

$|v|=\sqrt{(-8)^{2}+ (8)^{2}} =11.3137$

let A be angle before u and v

therefore, $cosA=\frac{u.v}{|u||v|}=\frac{(-8)*(-8)+(-3)*(8)}{8.544*11.3137} =\frac{40}{96.664}$

⇒ $A=arccos(\frac{40}{96.664} )=arccos(0.4138 )=65.56$

Therefore angle between u and v is 65.56°

5 0

3 years ago

3^2x+1 +9 = 3^x+3 +3^3​

3^2x+1 +9 = 3^x+3 +3^3