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3 years ago

Pls help me with these problems it is due today and i need help:)

Mathematics

1 answer:

Solnce55 [7]3 years ago

7 0

Answer:

Pls be mentally strong and don't not take help if you want take help from your parents

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I need help finding missing angles

vagabundo [1.1K]

Ok. Could I have a picture of the question? :)

6 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

How do I graph g(x)=f(x)+5

sashaice [31]

Find the absolute value vertex. In this case, the vertex for y=|x−5|y=|x-5| is (5,0)(5,0).

(5,0)(5,0)

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

(−∞,∞)(-∞,∞)

{x|x∈R}{x|x∈ℝ}

For each xx value, there is one yy value. Select few xx values from the domain. It would be more useful to select the values so that they are around the xx value of the absolute valuevertex.

xy3241506172

8 0

3 years ago

Every evening Jenna empties her pockets and puts her change in a jar. At the end of the week she counts her money. One week she

Sergeeva-Olga [200]

A) .10 d + .25 q = 7.75
B) d + q = 40

Multiplying B) by -.10
B) -.10d -.10q = -4.0
Then adding this to A)
A) .10 d + .25 q = 7.75
.15q = 3.75

Quarters = 25
Therefore, dimes = 15
***************************************************
Double-Check
A) .10 d + .25 q = 7.75
A) .10 * 15 d + .25 * 25 = 7.75
A) 1.50 + 6.25 = 7.75

Correct!!

6 0

3 years ago

What type of stock receives an equal part of the profits on each share to be distributed after all other obligations of a compan

solong [7]

I think the answer is A. Common

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3 years ago

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