All categories

1 year ago

Use 3.1 < sqrt 10 <3.2 to find the possible values of each expression. 2+sqrt 10

1 answer:

4 0

The inequality which represents possible values of the expression 2+sqrt 10 by virtue of the given inequality; 3.1 < sqrt 10 < 3.2 as in the task content is; 5.1 < 2 + sqrt 10 < 5.2.

<h3>Which inequality correctly expresses the possible values of the expression; 2 + √10 as required in the task content?</h3>

It follows from the task content that the expression given is;

3.1 < sqrt 10 < 3.2

Since the given premises is an inequality, it follows that adding the same number to all parts of the inequality stills holds the inequality true.

Hence by adding 2 to all parts of the inequality, we have;

2 + 3.1 < 2 + sqrt 10 < 2 + 3.2

Therefore, we have;

5.1 < 2 + sqrt 10 < 5.2

Ultimately, 5.1 < 2 + sqrt 10 < 5.2 represent the possible values of the expression 2+sqrt 10 as given by the inequality 3.1 < sqrt 10 < 3.2.

You might be interested in

What is the solution to 6x + 1 = 4x -3

PSYCHO15rus [73]

6x +1 = 4x -3

6x = 4x -4

2x = -4

x =-2

your answer is -2

4 0

3 years ago

Read 2 more answers

An=−2+32(n−1)<br><br> What is the 21st term of the sequence?

OLEGan [10]

Answer:

a₂₁ = 638

Step-by-step explanation:

substitute n = 21 into the explicit formula

a₂₁ = - 2 + 32(21 - 1) = - 2 + 32(20) = - 2 + 640 = 638

5 0

2 years ago

QUESTION 69

bazaltina [42]

The line is parallel to x-axis

the correct answer -

y = –4

8 0

3 years ago

If parallelogram RSTV has diagonals then RSTV must also be a _______. square and a rectangle rhombus and a rectangle rhombus rec

andrew11 [14]

<span>a rectangle rhombus(parallelogram) since even a parallelogram has diagonals</span>

5 0

3 years ago

The base of a solid in the region bounded by the two parabolas y2 = 8x and x2 = 8y. Cross sections of the solid perpendicular to

Usimov [2.4K]

The two curves intersect at two points, (0, 0) and (8, 8):

$x^2=8y\implies y=\dfrac{x^2}8$

$y^2=\dfrac{x^4}{64}=8x\implies\dfrac{x^4}{64}-8x=0\implies\dfrac{x(x-8)(x^2+8x+64)}{64}=0$

$\implies x=0,x=8\implies y=0,y=8$

The area of a semicircle with diameter $d$ is $\dfrac{\pi d^2}8$ . The diameter of each cross-section is determined by the vertical distance between the two curves for any given value of $x$ between 0 and 8. Over this interval, $y^2=8x\implies y=\sqrt{8x}$ and $\sqrt{8x}>\dfrac{x^2}8$ , so the volume of this solid is given by the integral

$\displaystyle\frac\pi8\int_0^8\left(\sqrt{8x}-\dfrac{x^2}8\right)^2\,\mathrm dx=\frac{288\pi}{35}$

6 0

3 years ago