All categories

3 years ago

Please help me with any of the questions

Mathematics

1 answer:

Svetlanka [38]3 years ago

7 0

1.) There are 6 numbers you can roll on a standard dice: 1, 2, 3, 4, 5, 6
There are 3 odd and 3 even. Keep in mind that an odd plus an odd will always be even and an even plus an even will always equal an even. Only if you add an odd and an even will you get an odd.

The chances of getting an even sum are much higher which is why you should choose even.

3.) The formula for finding the area of an equilateral triangle is (√3)/4(a^2) where a is just a side length.

Because the perimeter is 12 and each side of an equilateral triangle is equal, we can divide 12 by 3 sides and get each side is 4. Plug in 4 for a.

(√3)/4(4^2) = (√3)/4(16)

Because (√3)/4 = (1 x √3)/4,
(√3)/4(16) is equal to (16 x √3)/4

You have to divide (16 x √3) by 4. 16/4 = 4, and you only need to divide once to. We get 4√3 as the final answer.

You might be interested in

Which inequality statement describes the two numbers: -13 and a number 28 units to the right of -13?

kkurt [141]

So you have -13, then a number that's somewhere to the right of -13. imagine a number line: the negative values are on the left side of the zero, the positive values are on the right. if you're moving to the RIGHT of -13, that means that the value will be greater than -13, or in other words, -13 will be less than the new value because you moved right.

to find the number 28 units to the right of -13, you simply need to add these two numbers: -13 + 28. you add them because you're moving 28 units in the POSITIVE direction, aka you're going UP, so you want to add. -13 + 28 = 15.

now read the statements your question gave you. C and D are just straight-up false--a positive number is never less than a negative number. those are out immediately. now, the numbers you're working with are -13 and 15, so you can immediately ignore B as an answer choice, but still: A is correct because it shows the correct inequality. -13 is less than the number 28 units to the right of it, that number being positive 15.

7 0

3 years ago

Read 2 more answers

describe a system of equations in which subtraction would be the most efficient way of solving the system.

Sati [7]

Read https://www.wyzant.com/resources/answers/2997/how_do_you_figure_out_how_to_determine_the_best_method... and I think that will answer your question (:

7 0

3 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

Brianna visits 20 houses every day in her

Nimfa-mama [501]

Answer:

6 days

Step-by-step explanation:

5 0

3 years ago

(1) a) 5270 cl= ? dal

Ivanshal [37]

Answer:

Check below

Step-by-step explanation:

1) These metric volume units can be easily converted by dividing or multiplying by 10 and its multiple. Like this: each step up on the ladder multiply by 10. Each step down divide by 10

$a)5270 \:cl=527\:dl\\b)kl=5.3 \times 10=530 hl\\c)58000ml=5.8 dal =(\frac{58000}{10000})\\d)0.39 dal=3.9 l\\e)7.8 dal=780 dl\\f)57.2 ml=5.72 cl$ .

2) When it comes to area, the "ladder scheme" remains valid but now we'll multiply or divide by

$10^3$

Bear in mind these useful relations:

$1litre =\frac{1}{10^3}m^{3}$

$1 cm^{3} =1 ml$

$a) 3.7 l=3700 cm^{3}\\b)29.6 dm^{3}\\29.6 dm^{3} \rightarrow 29.6\times10^{3} cm^3=29.6\times10^{3} ml = 29.6\times10^{2} cl \:or\:2960 cl\\c)8.1l=8100 cm^{3}\\d)5l=\frac{5}{10^{-3}} m^{3}\\e)5.2 cl=52ml=52 cm^{3}=52\times10^{3}=52000 \:mm^{3}\\f)375 ml=375 cm^{3}\\g) 1300 cm^{3}=1300 ml=13 dl\\h) 8.6 ml= 8.6 cm^{3}\times\: 10^{3}=8600 mm^{3}\\i) 51.8\times 10^{11} l \rightarrow 51.8\times 10^{-11}km^{3}$

6 0

3 years ago