All categories

2 years ago

Question 5 IM IN A HURRY Find The slope A 1/2 B -2 C -1/2 D 2

Mathematics

2 answers:

Over [174]2 years ago

6 0

Answer:

C is the answer its -1/2

Levart [38]2 years ago

3 0

The answer is c !! -1/2

You might be interested in

In 2011, the population of mail was about 1.584 x 10^3 people. What is this number written in standard notation

frutty [35]

Answer:

1584

Step-by-step explanation:

1.584 x 1000 = 1584

8 0

3 years ago

From 145 pounds to 132 pounds

Mariana [72]

From 145 pounds to 132 pounds

145 - 132 = 13

13 pounds

Check:-
132 + 13 = 145
Correct!

13 pounds

4 0

3 years ago

1 . a sequence in which a fixed amount is added on to get the next term arithmetic sequence 2 . an individual quantity or number

Dmitry [639]

$T_n = a_1 + d(n-1)$

If a = 4 and d = 3,

$T_n = 4+ 3(n-1)$

$T_n = 4+ 3n-3$

$T_n = 3n + 1$

---------------------------------------------------------------------------------------
Answer: The general equation for the nth term is 3n + 1.
---------------------------------------------------------------------------------------

6 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 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:

brainly.com/question/16478185

6 0

2 years ago

What is the approximate area of a sector given O= 56 degrees with a diameter of 12m?

jeyben [28]

Area of sector is 17.584 meters

Solution:

Given that we have to find the approximate area of a sector given O= 56 degrees with a diameter of 12m

Diameter = 12 m

Radius = Diameter / 2 = 6 m

An angle of 56 degrees is the fraction $\frac{56}{360}$ of the whole rotation

A sector of a circle with a sector angle of 56 degrees is therefore also the fraction $\frac{56}{360}$ of the circle

The area of the sector will therefore also be $\frac{56}{360}$ of the area

$\text{ sector area } = \frac{56}{360} \times \pi r^2\\\\\text{ sector area } = \frac{56}{360} \times 3.14 \times 6^2\\\\\text{ sector area } = 17.584$

Thus area of sector is 17.584 meters

3 0

3 years ago