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Rasek [7]
3 years ago
5

Meyer used 6 loads of gravel to cover 2/5 of his driveway. how many loads of gravel will he need to cover the entire driveway?

Mathematics
2 answers:
valentinak56 [21]3 years ago
7 0
1<span>5</span> loads of gravel to cover the entire driveway

2/5 = 6/15


yulyashka [42]3 years ago
3 0
6loads cover 2/5 so you have to take 6÷2/5 and you get 15loads of gravel
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Select the graph for the solution of the open sentence. Click until the correct graph appears.
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5 0
3 years ago
Evaluate the function below for x = 4.<br> F(x) = x3 + 2x2 + 1
pav-90 [236]

Answer:

77

Step-by-step explanation:

Plug in:

F(4) = (4)3 + 2(4)2 + 1

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F(4) = 12 + 64 + 1

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4 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 <u>923.558 m²</u>

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} ≈ <u>923.558 m²</u>

Learn more about figures circumscribed by a circle here:

brainly.com/question/16478185

6 0
3 years ago
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Answer:

Step-by-step explanation:

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Compose the quadratic condition in standard shape, ax2 + bx + c = 0. Recognize the values of a, b, c. Write the Quadratic Equation. At that point substitute within the values of a, b, c. Simplify. Check the arrangements.

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