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3 years ago

10

What is a healthy breakfast choice when you are rushed for time in the morning? A. bagel and a yogurt drink B. celery sticks wit

h peanut butter and apple juice C. cereal with yogurt or milk and orange juice D. All of these answers are correct.

2 answers:

ikadub [295]3 years ago

6 0

D. All of these answers are correct

Ann [662]3 years ago

4 0

All of the above seem to be healthy breakfast choices

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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 <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

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What is the area of this figure?

sergiy2304 [10]

$1.5 \times 3 = 4.5 \: {(km)}^{2}$

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Please help me with these 3 (last one is optional if you want to do it) but please don’t give me link to anything or you will ge

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puedes o

Step-by-step explanation:

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3 years ago

Consider the graph of the quadratic function. Which interval on the x-axis has a negative rate of change?

Gre4nikov [31]

The average rate of change is defined as:

$AVR = \frac{f(x2)-f(x1)}{x2-x1}$

For AVR to be negative, it must comply with:

$x2 - x1> 0$

$f (x2)$

Therefore, we observe that the interval that fulfills these conditions is the whole interval to the right of the parabola.

Among the options given, this interval is:

1 to 2.5

Answer:

An interval on the x-axis that has a negative rate of change is:

D. 1 to 2.5

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B!!! Because they A equals -20, B equals 20, C equals 0, and D equals 0!

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