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prisoha [69]
3 years ago
8

Write an irrational number that can be added to the irrational number the square root of 15 to get a rational sum of 2

Mathematics
1 answer:
aniked [119]3 years ago
7 0

Answer:

Step-by-step explanation:

if u is an irrational number added to a rational number (sqrt 15) to yield 2, then that would mean u would = root -1.78.....

which means you wouldn't have a real solution. if you want a complex answer, it would be sqrt(1.78)i

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-BARSIC- [3]

Answer:

the answer is 30

8 0
3 years ago
Read 2 more answers
What is the inequalities of 4(-4x-3)&gt;36
adelina 88 [10]

Answer:

x<-3

Step-by-step explanation:

-16x-12>36

-16x>48

x<-3

4 0
3 years ago
use the formulas for lowering powers to rewrite the expression in terms of the first power of cosine cos^4
Firdavs [7]

The expression cos⁴ θ in terms of the first power of cosine is <u>[ 3 + 2cos 2θ + cos 4θ]/8.</u>

The power-reducing formula, for cosine, is,

cos² θ = (1/2)[1 + cos 2θ].

In the question, we are asked to use the formulas for lowering powers to rewrite the expression in terms of the first power of cosine cos⁴ θ.

We can do it as follows:

cos⁴ θ

= (cos² θ)²

= {(1/2)[1 + cos 2θ]}²

= (1/4)[1 + cos 2θ]²

= (1/4)(1 + 2cos 2θ + cos² 2θ] {Using (a + b)² = a² + 2ab + b²}

= 1/4 + (1/2)cos 2θ + (1/4)(cos ² 2θ)

= 1/4 + (1/2)cos 2θ + (1/4)(1/2)[1 + cos 4θ]

= 1/4 + cos 2θ/4 + 1/8 + cos 4θ/8

= 3/8 + cos 2θ/4 + cos 4θ/8

= [ 3 + 2cos 2θ + cos 4θ]/8.

Thus, the expression cos⁴ θ in terms of the first power of cosine is <u>[ 3 + 2cos 2θ + cos 4θ]/8</u>.

Learn more about reducing trigonometric powers at

brainly.com/question/15202536

#SPJ4

5 0
2 years ago
A ball is projected in to the air. Its height at time t is given by the equation
Ksju [112]
The equation gives the height of the ball. That is, h is the height of the ball. t is the time. Since we are looking for the time at which the height is 8 (h=8), we need to set the equation equal to 8 and solve for t. We do this as follows:

h=-16 t^{2} +60t+1
8=-16 t^{2} +60t+1
16 t^{2} -60t-1+8=0
16 t^{2} -60t+7=0

This is a quadratic equation and as it is set equal to 0 we can solve it using the quadratic formula. That formula is:
t= \frac{-bplus minus \sqrt{ b^{2}-4ac } }{2a}
You might recall seeing this as "x=..." but since our equation is in terms of t we use "t-=..."

In order to use the formula we need to identify a, b and c.
a = the coefficient (number in front of) t^{2} = 16.
b = the coefficient of t = -60
c = the constant (the number that is by itself) = 7

Substituting these into the quadratic formula gives us:
t= \frac{-(-60)plus minus \sqrt{ (-60)^{2}-4(16)(7) } }{2(16)}
t= \frac{60plus minus \sqrt{3600-448 } }{32}
t= \frac{60plus minus \sqrt{3152 } }{32}

As we have "plus minus" (this is usually written in symbols with a plus sign over a minus sign) we split the equation in two and obtain:
t=  \frac{60+56.1426}{32} =3.63
and
t=  \frac{60-56.1426}{32} =.12

So the height is 8 feet at t = 3.63 and t=.12

It should make sense that there are two times. The ball goes up, reaches it's highest height and then comes back down. As such the height will be 8 at some point on the way up and also at some point on the way down.


3 0
3 years ago
A 150 no container can hold 23.6 g of aluminum what is the density of the aluminum?
Drupady [299]

Answer:

# The density of the aluminum is 59/375 ≅ 0.157 g/cm³

# The volume of the container is 4/17 ≅ 0.235 cm³

# The mass of iron is 1225 g

Step-by-step explanation:

* Lets explain how to solve the problem

- Density is a measurement that compares the amount of matter

 of an object to its volume [ Density = mass/volume]

- Density is found by dividing the mass of an object by its volume

- The unit of measuring density is gram per centimeter³ (milliliter)

* Lets solve the problems

# A container of volume 150 cm³ can hold 23.6 g of aluminum

∵ Density = mass/volume

∵ The mass = 23.6 g

∵ The volume = 150 cm³

∴ The density = 23.6/150 = 59/375 ≅ 0.157 g/cm³

* The density of the aluminum is 59/375 ≅ 0.157 g/cm³

# The density of mercury is 13.6 g/cm³ at 23 C°

- A container can hold 3.2 g of mercury at this temperature

∵ Density = mass/volume

∵ The mass = 3.2 g

∵ The density = 13.6 g/cm³

∴ 13.6 = 3.2/volume

- By using cross multiplication

∴ The volume = 3.2/13.6 = 4/17 cm³

* The volume of the container is 4/17 ≅ 0.235 cm³

# The density of the iron is 4.9 g/cm³

- The beaker has a volume 250 ml

∵ Density = mass/volume

∵ The ml = cm³

∴ The volume of the beaker is 250 cm³

∵ The density = 4.9 g/cm³

∴ 4.9 = mass/250

- Multiply both sides by 250

∴ mass = 1225 g

* The mass of iron is 1225 g

7 0
3 years ago
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