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Gelneren [198K]

2 years ago

11

A scuba diver ascends 5 3/4 feet to the top of a boat platform. The diver then descends 20 1/8 feet by jumping off the boat. Wri

te and evaluate on subtraction expression to show what lever under sea level the scuba currently is

1 answer:

Nana76 [90]2 years ago

8 0

-15 5/8 feet
5 6/8- 20 1/8

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The following table shows values from a linear function. What is the

Naddik [55]

Answer:

9

Step-by-step explanation:

The equation graphed from these points is y=3x+9, therefore the y-intercept is 9.

4 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>

Learn more about figures circumscribed by a circle here:

brainly.com/question/16478185

6 0

2 years ago

Which of the following is a fifth degree binomial?

Anna11 [10]

Answer:

C)

Step-by-step explanation:

$2xy^{2}-5y^{5}$

Binomial should have two terms. Five degree binomial means highest degree of the binomial should be 5

5 0

2 years ago

Bob sets the price of the plants according to their height in inches. A plant that is 9.5 inches high costs $2.28 in his store.

alekssr [168]

2.28 devide 9.5
=0.24

7 0

3 years ago

The CEO of a large corporation asks his Human Resource (HR) director to study absenteeism among its executive-level managers at

Fynjy0 [20]

Answer:

Following are the answer to this question:

Step-by-step explanation:

Given:

n = 30 is the sample size.

The mean $\bar X$ = 7.3 days.

The standard deviation = 6.2 days.

df = n-1

$= 30-1 \\ =29$

The importance level is $\alpha$ = 0.10

The table value is calculated with a function excel 2010:

$= tinv (\ probility, \ freedom \ level) \\= tinv (0.10,29) \\ =1.699127\\ = t_{al(2x-1)}= 1.699127$

The method for calculating the trust interval of 90 percent for the true population means is:

Formula:

$\bar X - t_{al 2,x-1} \frac{S}{\sqrt{n}} \leq \mu \leq \bar X+ t_{al 2,x-1} \frac{S}{\sqrt{n}}$

$=\bar X - t_{0.5, 29} \frac{6.2}{\sqrt{30}} \leq \mu \leq \bar X+ t_{0.5, 29} \frac{6.2}{\sqrt{30}}\\\\=7.3 -1.699127 \frac{6.2}{\sqrt{30}}\leq \mu \leq7.3 +1.699127 \frac{6.2}{\sqrt{30}}\\\\=7.3 -1.699127 (1.13196)\leq \mu \leq7.3 +1.699127 (1.13196) \\\\=5.37 \leq \mu \leq 9.22 \\$

It can rest assured that the true people needs that middle managers are unavailable from 5,37 to 9,23 during the years.

7 0

2 years ago