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3 years ago

What is the square root of 12 divided by 2 plus 16?

Mathematics

2 answers:

slavikrds [6]3 years ago

4 0

Answer:

17.73205081

Step-by-step explanation:

Didn't use a calculator :) Please mark me as brainliest. Thanks!

iren [92.7K]3 years ago

4 0

Answer:

$\frac{2\sqrt{3} + 32}{2} \\$

Step-by-step explanation:

$\frac{\sqrt{12} }{2} + 16$

= $\frac{\sqrt{12} }{2} + \frac{32}{2}$

= $\frac{\sqrt{12} + 32}{2}$

= $\frac{2\sqrt{3} + 32}{2} \\$

* i'm not sure if it can be further broken down *

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Given:

Fynjy0 [20]

Answer:

Option B is correct.

$\triangle RST \cong \triangle TUR$

Step-by-step explanation:

Given: RSTU is a parallelogram.

By definition of parallelogram, two pairs of opposite sides are congruent in length.

⇒ $TU \cong RS$ and $TS \cong RU$

In triangle RST and triangle TUR

$RS \cong TU$ [Side]

$TS \cong RU$ [Side]

$RT \cong RT$ [Common side]

SSS(Side-Side-Side) postulates states that if three sides of one triangle are congruent to three sides of other triangle, then the triangles are congruent.

then, by SSS postulates we have;

$\triangle RST \cong \triangle TUR$

5 0

3 years ago

PLEASE HELP I WILL MARK BRAINLIST

shtirl [24]

The minimum is 91. The first quartile is 112. The median is 173. The third quartile is 215. The maximum is 253.

6 0

3 years ago

Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16

4vir4ik [10]

Answer:

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

$9 -16 (x^2 + y^2) = 0 \\ \\ 16 (x^2 + y^2) = 9 \\ \\ x^2+y^2 = \dfrac{9}{16}$

$x^2+y^2 = (\dfrac{3}{4})^2$

where:

0 ≤ r ≤ $\dfrac{3}{4}$ and 0 ≤ θ ≤ 2π

∴

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0} \ \ \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} ( 9r^2-16r^4}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi$

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

4 0

3 years ago

Which of the following lines is parallel to the line y = 4x - 6?

Dovator [93]

Answer:

d y=4x + 5

Step-by-step explanation:

Is parallel beacause they have the same slope, remember

y=ax + c

where a is the slope of the line

8 0

3 years ago

A long distance runner starts at the beginning of a trail and runs at a rate of 4 miles per hour. One hour later, a cyclist star

Scrat [10]

T = 1/4 hour will be the answer

7 0

3 years ago