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2 years ago

Find the product of the sum of 57 and 16 and the difference of 61 and 31

Mathematics

1 answer:

Sergio039 [100]2 years ago

3 0

Step-by-step explanation:

???

what is the problem ? you don't know how to add, subtract or multiply yet ?

sum of 57 and 16 : 57 + 16 = 73

difference of 61 and 31 : 61 - 31 = 30

the product of the sum and the difference :

73 × 30 = 2190

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Answer:

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3 years ago

Please help me ASAP will mark best answer brainliest

IrinaK [193]

Answer:

4.) Silvia had the greater variation in her race results for the season because she has a higher standard deviation.

Step-by-step explanation:

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4 0

4 years ago

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

6 0

3 years ago

Help please picture attached

Naya [18.7K]

Answer:

-10

Step-by-step explanation:

3/5(b)=-6

Divide -6 by 3/5, you get -30/3.

4 0

3 years ago

HELP TUTOR… ss z z z z

kondor19780726 [428]

Answer:

A)σ = 15.37 miles per hour

B) σ = 55.10 miles per hour

C) 39.73 miles per hour

Step-by-step explanation:

Formula for standard deviation is;

σ = √(Σ(x - μ)²)/(n - 1)

A) For large wild cats;

μ = Σx/n = (70 + 40 + 30 + 40 + 35 + 35 + 35 + 35 + 10)/9 = 36.67

σ = √((70 - 36.67)² + (40 - 36.67)² + (30 - 36.67)² + (40 - 36.67)² + (35 - 36.67)² + (35 - 36.67)² + (35 - 36.67)² + (35 - 36.67)² + (10 - 36.67)²)/(9 - 1)

σ = √236.1139

σ = 15.37

B) For various birds;

μ = Σx/n = (218 + 106 + 97 + 56 + 66 + 39 + 55 + 32 + 54 + 20 + 25 + 25)/12

μ = 66.0833

σ = √((218 - 66.0833)² + (106 - 66.0833)² + (97 - 66.0833)² + (56 - 66.0833)² + (66 - 66.0833)² + (39 - 66.0833)² + (55 - 66.0833)² + (32 - 66.0833)² + (54 - 66.0833)² + (20 - 66.0833)² + (25 - 66.0833)² + (25 - 66.0833)²)/(12 - 1)

σ = √3035.72

σ = 55.10

C) difference in standard deviations = 55.10 - 15.37 = 39.73 miles per hour

8 0

3 years ago

Find the product of the sum of 57 and 16 and the difference of 61 and 31​

Find the product of the sum of 57 and 16 and the difference of 61 and 31