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3 years ago

Question 1

Mathematics

1 answer:

mamaluj [8]3 years ago

4 0

Answer:

The maximum volume of space that 1,000 BTUs will cool is 400 cubic foot.

Step-by-step explanation:

Given : A total of 1,000 BTUs are needed to cool a space. If 2.5 BTUs are needed to cool 1 cubic foot of space.

To find : What is the maximum volume of space that 1,000 BTUs will cool?

Solution :

Applying unitary method,

2.5 BTU are needed to cool 1 cubic foot of space.

So, 1 BTU are needed to cool $\frac{1}{2.5}$ cubic foot of space.

Then,1000 BTU are needed to cool $\frac{1}{2.5}\times 1000$ cubic foot of space.

i.e. 1000 BTU are needed to cool 400 cubic foot of space.

Therefore, the maximum volume of space that 1,000 BTUs will cool is 400 cubic foot.

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Suppose that $3$ is a factor of $a$, $a$ is a divisor of $12$, and $a$ is positive. What is the number of possible values of $a$

antiseptic1488 [7]

If $a$ is a divisor of 12, then

$a \in \{-12, -6, -4, -3, -2, -1, 1, 2, 3, 4, 6, 12\}$

If 3 is a factor (i.e. divisor) of $a$ , then $a$ can only be one of the elements of this set that are multiples of 3, so

$a \in \{-12, -6, -3, 3, 6, 12\}$

But $a$ is positive, so there are 3 possible values $a$ can have.

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1 year ago

2(x-3)=5(x-3)+10 Need urgent help

ArbitrLikvidat [17]

Step-by-step explanation:

2(x-3)=5(x-3)+10

=> 2x - 6 = 5x - 15 + 10

=> -6 + 15 -10 = 5x - 2x

=> 5x -2x = 15 - 6 - 10

=> 3x = 15 - 16

=> 3x = -1

$= > x = \frac{ - 1}{3} (ans)$

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3 years ago

20 POINTS

Georgia [21]

Answer is in the photo, hope it helps. If you have any questions about the answer, please feel free to ask :)

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1 year 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:

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3 years ago

A bowl contains 15 granola bars and 10 fruit bars.

pishuonlain [190]

granola bars:fruit bars

15:10=3:2

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3 years ago

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