All categories

2 years ago

√(25 - 10 √(3) + 3) (not as a decimal)

Mathematics

1 answer:

nikitadnepr [17]2 years ago

6 0

Answer:

$\huge\boxed{\sqrt{25-10\sqrt3+3}=5-\sqrt3}$

Step-by-step explanation:

$\sqrt{25-10\sqrt3+3}=(*)\\\\25=5^2\\3=(\sqrt3)^2\\10\sqrt3=(2)(5)(\sqrt3)\\\\(*)=\sqrt{5^2-(2)(5)(\sqrt3)+(\sqrt3)^2}\\\\\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\=\sqrt{(5-\sqrt3)^2}\\\\\text{use}\ \sqrt{a^2}=|a|\\\\=|5-\sqrt3|=5-\sqrt3\\\\^{\text{because}\ 5>\sqrt3\to5-\sqrt3>0}$

You might be interested in

***PLEASE HELP WILL GIVE BRAINLIEST!!!***

Ede4ka [16]

in the number line, the end points are DG, and the point in between is O

DG = 88

DO = 5x + 12

OG = 2x

Set the equation. The two parts (DO & OG) are equal to the whole (DG)

2x + 5x + 12 = 88

Simplify. Combine like terms

(2x + 5x) + 12 = 88

7x + 12 = 88

Isolate the x. Remember to do the opposite of PEMDAS. Subtract 12 from both sides

7x + 12 (-12) = 88 (-12)

7x = 76

Isolate the x. Divide 7 from both sides:

7x/7 = 76/7

x = 76/7

----------------------------------------------------------------------------------------------------------------------------

Find DO. Plug in "76/7" for x:

DO = 5x + 12

DO = 5(76/7) + 12

Simplify. Remember to follow PEMDAS. Multiply 76 with 5

DO = 380/7 + 12

Next, divide 380 with 7

DO = 54.29 (rounded)

Finally, add

DO = 54.29 + 12

DO = 66.29

66.29 is your answer

hope this helps

4 0

3 years ago

A rescue shelter has a total of 30 occupants. The number of legs at the shelter totals 96 legs. There are spiders and frogs at t

Katarina [22]

Answer:

Hint- Spiders have 8 legs

hint- frogs have 2 legs

Step-by-step explanation:

7 0

3 years ago

Rewrite the following integral in spherical coordinates.

lora16 [44]

In cylindrical coordinates, we have $r^2=x^2+y^2$ , so that

$z = \pm \sqrt{2-r^2} = \pm \sqrt{2-x^2-y^2}$

correspond to the upper and lower halves of a sphere with radius $\sqrt2$ . In spherical coordinates, this sphere is $\rho=\sqrt2$ .

$1 \le r \le \sqrt2$ means our region is between two cylinders with radius 1 and $\sqrt2$ . In spherical coordinates, the inner cylinder has equation

$x^2+y^2 = 1 \implies \rho^2\cos^2(\theta) \sin^2(\phi) + \rho^2\sin^2(\theta) \sin^2(\phi) = \rho^2 \sin^2(\phi) = 1 \\\\ \implies \rho^2 = \csc^2(\phi) \\\\ \implies \rho = \csc(\phi)$

This cylinder meets the sphere when

$x^2 + y^2 + z^2 = 1 + z^2 = 2 \implies z^2 = 1 \\\\ \implies \rho^2 \cos^2(\phi) = 1 \\\\ \implies \rho^2 = \sec^2(\phi) \\\\ \implies \rho = \sec(\phi)$

which occurs at

$\csc(\phi) = \sec(\phi) \implies \tan(\phi) = 1 \implies \phi = \dfrac\pi4+n\pi$

where $n\in\Bbb Z$ . Then $\frac\pi4\le\phi\le\frac{3\pi}4$ .

The volume element transforms to

$dx\,dy\,dz = r\,dr\,d\theta\,dz = \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi$

Putting everything together, we have

$\displaystyle \int_0^{2\pi} \int_1^{\sqrt2} \int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}} r \, dz \, dr \, d\theta = \boxed{\int_0^{2\pi} \int_{\pi/4}^{3\pi/4} \int_{\csc(\phi)}^{\sqrt2} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta} = \frac{4\pi}3$

4 0

2 years ago

What is the solution to the following inequality?

tamaranim1 [39]

$\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the Inequalities.

so we get as,

(w+6)/2 < 5

so we get as,

w+6 < 10

=> w < 10-6

=> w < 4

so the answer is,

w < 4

4 0

2 years ago

Solve the compound inequality and graph<br> Please help! :)

frozen [14]

Answer:

Step-by-step explanation:

x+4≥10 and x-6>-15

x≥10-4 and x>-15+6

x≥ 6 and x>-9

combining the two

x>-9

8 0

3 years ago