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3 years ago

Which whole number is closest to the value of square root of 54?

Mathematics

2 answers:

Digiron [165]3 years ago

5 0

The answer is 7 because 7 • 7 is 49 which is the to the value of the square root of 54.
Hope this helps!

Blababa [14]3 years ago

3 0

The answer for this problem is 7

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A garden is shaped in the form of a regular heptagon (seven-sided), MNSRQPO. A circle with center T and radius 25m circumscribes

Alenkinab [10]

The relationship between the sides MN, MS, and MQ in the given regular heptagon is $\dfrac{1}{MN} = \dfrac{1}{MS} + \dfrac{1}{MQ}$

The area to be planted with flowers is approximately 923.558 m²

The reason the above value is correct is as follows;

The known parameters of the garden are;

The radius of the circle that circumscribes the heptagon, r = 25 m

The area left for the children playground = ΔMSQ

Required;

The area of the garden planted with flowers

Solution:

The area of an heptagon, is;

$A = \dfrac{7}{4} \cdot a^2 \cdot cot \left (\dfrac{180 ^{\circ}}{7} \right )$

The interior angle of an heptagon = 128.571°

The length of a side, S, is given as follows;

$\dfrac{s}{sin(180 - 128.571)} = \dfrac{25}{sin \left(\dfrac{128.571}{2} \right)}$

$s = \dfrac{25}{sin \left(\dfrac{128.571}{2} \right)} \times sin(180 - 128.571) \approx 21.69$

$The \ apothem \ a = 25 \times sin \left ( \dfrac{128.571}{2} \right) \approx 22.52$

The area of the heptagon MNSRQPO is therefore;

$A = \dfrac{7}{4} \times 22.52^2 \times cot \left (\dfrac{180 ^{\circ}}{7} \right ) \approx 1,842.94$

$MS = \sqrt{(21.69^2 + 21.69^2 - 2 \times 21.69 \times21.69\times cos(128.571^{\circ})) \approx 43.08$

By sine rule, we have

$\dfrac{21.69}{sin(\angle NSM)} = \dfrac{43.08}{sin(128.571 ^{\circ})}$

$sin(\angle NSM) =\dfrac{21.69}{43.08} \times sin(128.571 ^{\circ})$

$\angle NSM = arcsin \left(\dfrac{21.69}{43.08} \times sin(128.571 ^{\circ}) \right) \approx 23.18^{\circ}$

∠MSQ = 128.571 - 2*23.18 = 82.211

The area of triangle, MSQ, is given as follows;

$Area \ of \Delta MSQ = \dfrac{1}{2} \times 43.08^2 \times sin(82.211^{\circ}) \approx 919.382^{\circ}$

The area of the of the garden plated with flowers, $A_{req}$ , is given as follows;

$A_{req}$ = Area of heptagon MNSRQPO - Area of triangle ΔMSQ

Therefore;

$A_{req}$ = 1,842.94 - 919.382 ≈ 923.558

The area of the of the garden plated with flowers, $A_{req}$ ≈ 923.558 m²

Learn more about figures circumscribed by a circle here:

brainly.com/question/16478185

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Alekssandra [29.7K]

It should be in the third quadrant

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Mars2501 [29]

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erica [24]

Answer:

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Step-by-step explanation:

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PLEASE HELP ASAP ILL GIVE BRAINLIEST!!!!! find measure of arc MK

Gwar [14]

Answer:

Arc length MK = 15.45 units (nearest hundredth)

Arc measure = 58.24°

Step-by-step explanation:

Calculate the measure of the angle KLN (as this equals m∠KLM which is the measure of arc MK)

ΔKNL is a right triangle, so we can use the cos trig ratio to find ∠KLM:

$\sf \cos(\theta)=\dfrac{A}{H}$

where:

$\theta$ is the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

Given:

$\theta$ = ∠KLM
A = LN = 8
H = KL = 15.2

$\implies \sf \cos(KLM)=\dfrac{8}{15.2}$

$\implies \sf \angle KLM=\cos^{-1}\left(\dfrac{8}{15.2}\right)$

$\implies \sf \angle KLM=58.24313614^{\circ}$

Therefore, the measure of arc MK = 58.24° (nearest hundredth)

$\textsf{Arc length}=2 \pi r\left(\dfrac{\theta}{360^{\circ}}\right) \quad \textsf{(where r is the radius and}\:\theta\:{\textsf{is the angle)}$

Given:

r = 15.2
∠KLM = 58.24313614°

$\implies \textsf{Arc length MK}=2 \pi (15.2)\left(\dfrac{\sf \angle KLM}{360^{\circ}}\right)$

$\implies \textsf{Arc length MK}=\sf 15.45132428\:units$

6 0

2 years ago