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3 years ago

Plz answer and thank you

Mathematics

2 answers:

MAXImum [283]3 years ago

7 0

It haz a greater value than 8!

icang [17]3 years ago

4 0

A . It has a value GREATER than 8

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Question 9 of 10 Which of these groups of values plugged into the TVM Solver of a graphing calculator will return the same value

hodyreva [135]

The TVM solver is a tool found in graphing calculators, that solve Time Value of Money problems.

The group of values that will return the same value as the given expression is;

D. N = 24; I% = 3.6; PV =; PMT = -415; FV = 0; P/Y = 12; C/Y = 12; PMT :END

<h3>What is Present Value?</h3>

Present value (PV) formula finds application in finance to calculate the present day value of an amount that is received at a future date.

In the TVM solver, we have;

I = The annual percentage rate

N = n × t

t = The number of years

PV = Present value

PMT = Payment

P/Y = Number of payments per year = n

C/Y = Number of compounding periods per year = n

The formula for monthly payment is presented as follows;

$P = \frac{M[(1+r/n)^{nt} - 1]}{(r/n)(1+r/n)^{nt}}$

M = $\frac{P.(r/n)(1+ r/n)^{nt} }{(1+ r/n)^{nt} - 1}$

Therefore, we get;

Where;

M = PMT = -415

P = PV

r = I

P/Y = n = 12

Therefore;

0.003 = I/12

I = 0.003 X 12 = 3.6%

N = n X t = 24

The value of the equation is the present value, PV = ?

When payment are made based on the PV, we have FV = 0

The group of values the same value as the expression

P = $\frac{(415)[(1+0.003)^{24}-1]}{(0.003)(1+0.003)^{24}}$

when plugged into the TVM solver of a calculator is;

D. N = 24; I% = 3.6; PV =; PMT = -415; FV = 0; P/Y = 12; C/Y = 12; PMT :END

Learn more about Present Value Solver from:

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7 0

2 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>

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6 0

3 years ago

Which of the following inequality should be graphed with a dashed line? Select all that apply.

AURORKA [14]

Answer:

B, D

Step-by-step explanation:

Any inequality that does not include the "or equal to" case will be graphed with a dashed line.

<h3>Application</h3>

The "or equal to" case is included when the inequality symbol is one of ≤ or ≥. When symbols > or < are used, the boundary line of the solution space is dashed (or the end point on the number line is an open circle).

The inequalities that will be graphed with a dashed line are ...