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3 years ago

Solve the inequality 2x + 5 2 27

Mathematics

2 answers:

Akimi4 [234]3 years ago

8 0

Step-by-step explanation:

Hope this help you

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devlian [24]3 years ago

5 0

Answer:

$2x + 5 \geqslant 27 \\ = 2x \geqslant 27 - 5 \\ = 2x \geqslant 22 \\ = x \geqslant 22 \div 2 \\ x \geqslant 11 \\ x[11 \: \infty ) \\ thank \: you$

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Write the expression for the situation.

alexdok [17]

Answer:

2 + 0.50m

Step-by-step explanation:

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3 years ago

Write 2^8 * 8^2 * 4^-4 in the form 2^n

Eva8 [605]

The given expression 2^8 * 8^2 * 4^-4 can be written in the exponential form 2^n as 2^6.

<h3>What are exponential forms?</h3>

The exponential form is a more convenient way to write repetitive multiplication of the same integer by using the base and its exponents.

<u>For example:</u>

If we have a*a*a*a, it can be written in exponential form as:

=a^4

where

a is the base, and
4 is the power.

The power in this format reflects the number of times we multiply the base by itself. The exponent is also known as the index or power.

From the information given:

We can write 2^8 * 8^2 * 4^-4 in form of 2^n as follows:

$\mathbf{= 2^8\times (2^3)^2 \times (2^2)^{-4} }$

$\mathbf{= 2^8\times (2^6) \times (2^{-8}) }$

$\mathbf{= 2^{8+6+(-8)}}$

$\mathbf{= 2^{6}}$

Therefore, we can conclude that by using the exponential form, the given expression 2^8 * 8^2 * 4^-4 in the form 2^n is 2^6.

Learn more about exponential forms here:

brainly.com/question/8844911

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4 0

2 years ago

Saturn's rings extend to about 120,000 kilometers above Saturn's equator. Saturn's diameter at its equator is also about 120,000

Allisa [31]

<u>Answer:</u>

1130400 km

<u>Step-by-step explanation:</u>

We know the formula for circumference of a circle:

<em>Circumference = $2\pi r$ </em>

So we need the radius of Saturn ring to calculate its circumference.

We are given Saturn's diamter which is 120,000km; and know that Saturn's ring extends for about 120,000 km from the circumference of the Saturn.

Therefore, the radius of Saturn's ring will be 120000 km plus half of the diameter of Saturn:

Radius of Saturn's ring = 120000 + (120000/2) = 180000 km

Circumference of Saturn's rings = 2 x 3.14 x 180000 = 1130400 km

3 0

4 years ago

Please help. Explain using complete sentences

solniwko [45]

It would be an over-estimate if u rounded ur numbers up and an under-estimate if u rounded ur numbers down

7 0

3 years ago

Read 2 more answers

Find the volume v of the described solid s. the base of s is an elliptical region with boundary curve 4x2 + 9y2 = 36. cross-sect

Tasya [4]

$4x^2+9y^2=36\iff\dfrac{x^2}9+\dfrac{y^2}4=1$

defines an ellipse centered at $(0,0)$ with semi-major axis length 3 and semi-minor axis length 2. The semi-major axis lies on the $x$ -axis. So if cross sections are taken perpendicular to the $x$ -axis, any such triangular section will have a base that is determined by the vertical distance between the lower and upper halves of the ellipse. That is, any cross section taken at $x=x_0$ will have a base of length

$\dfrac{x^2}9+\dfrac{y^2}4=1\implies y=\pm\dfrac23\sqrt{9-x^2}$
$\implies \text{base}=\dfrac23\sqrt{9-{x_0}^2}-\left(-\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac43\sqrt{9-{x_0}^2}$

I've attached a graphic of what a sample section would look like.

Any such isosceles triangle will have a hypotenuse that occurs in a $\sqrt2:1$ ratio with either of the remaining legs. So if the hypotenuse is $\dfrac43\sqrt{9-{x_0}^2}$ , then either leg will have length $\dfrac4{3\sqrt2}\sqrt{9-{x_0}^2}$ .

Now the legs form a similar triangle with the height of the triangle, where the legs of the larger triangle section are the hypotenuses and the height is one of the legs. This means the height of the triangular section is $\dfrac4{3(\sqrt2)^2}\sqrt{9-{x_0}^2}=\dfrac23\sqrt{9-{x_0}^2}$ .

Finally, $x_0$ can be chosen from any value in $-3\le x_0\le3$ . We're now ready to set up the integral to find the volume of the solid. The volume is the sum of the infinitely many triangular sections' areas, which are

$\dfrac12\left(\dfrac43\sqrt{9-{x_0}^2}\right)\left(\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac49(9-{x_0}^2)$

and so the volume would be

$\displaystyle\int_{x=-3}^{x=3}\frac49(9-x^2)\,\mathrm dx$
$=\left(4x-\dfrac4{27}x^3\right)\bigg|_{x=-3}^{x=3}$
$=16$

6 0

3 years ago