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scoundrel [369]
4 years ago
11

Hi, I'm struggling with these Areas & Perimeters.

Mathematics
1 answer:
bazaltina [42]4 years ago
8 0

1)

V=(3.14)(40^2)(1.3)

V=(3.14)(1600)(1.3)

V=5024(1.3)

V=6531.2

2)

Circumference of semicircle: (2*3.14*17)/2=53.38

CM to M: 4000/100=40, 700/100=7

40+10+7+34+53.38+33+68+10=255.35

Perimeter=255.35m

Area of semicircle: (3.14*17^2)/2=453.73

40*10=400m^2

68*33=2244m^2

400+2244+453.73=3097.73

Area=3097.73m^2

3)

100*75=7500

Area of semicircle: (3.14*25^2)/2=981.25

7500-981.25=6518.75

Area of car park not including the arena=6518.75m^2

4)

Meters to Centimeters: 0.4*100=40

45*40*50=90000

Centimeters to Meters: 90000/100=900

Volume=900m^3

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How do you express sin x + cos x in terms of sine only?
frez [133]

Answer:

\sin x + \sqrt{1-\sin^2x}

Step-by-step explanation:

Given: sin x + cos x

To change the given trigonometry expression in term of sine only.

Trigonometry identity:-

  • \sin^2x+\cos^2x=1
  • \cos x=\sqrt{1-\sin^2x}

Expression: \sin x+\cos x

We get rid of cos x from expression and write as sine form.

Expression: \sin x + \sqrt{1-\sin^2x}        \because \cos x=\sqrt{1-\sin^2x}

Hence, The final expression is only sine function.

4 0
4 years ago
I don't understand!What do i put?!
Serggg [28]
1 x 70
2 x 35<span> 
</span>5 x 14
7 x 10<span>.</span>
8 0
3 years ago
Someone help me with this
puteri [66]

Answer:

A) t = 1 second to reach the highest point

B) h = 96 feet

C) t = 3.449 seconds to hit the ground

Step-by-step explanation:

We notice that the equation that describes the position of the object in terms of the time, is a parabola (quadratic) with negative leading coefficient. therefore this is a parabola with arms pointing down, and with vertex at the top. It is at that top vertex that the maximum altitude of the object is reached.

We need therefore to find the position of the vertex (time "t" is the horizontal coordinate and height "h" is the vertical coordinate).

We know that the vertex (maximum of a parabola of this type - normally described by y=ax^2+bx+c) is given by x_{vertex} = -\frac{b}{2*a}. Therefore in our case, understanding that time "t" is equivalent to the variable "x", and eight "h" is equivalent to the variable "y", that "-16" is a, and "32" is b, we have that the maximum occurs for:

t_{max}=-\frac{32}{2*(-16)} = \frac{-32}{-32} = 1

which means at one second from the launching.

We then can find the answer to part B by replacing t with "1" second in the formula for the height "h":

h(1)=-16*(1)^2+32(1)+80=-16+32+80=96

That is : a height of 96 feet.

To find the answer for part C: the time to hit the ground, we solve for the variable "t" in the given expression for the height, by setting the height to zero (object touching the ground) and using the quadratic formula:

0=-16t^2+32t+80\\t=\frac{-32+/-\sqrt{32^2-4*(-16)*80} }{2*(-16)} \\ t=\frac{-32+/-\sqrt{6144} }{-32}

which gives us two solutions: t = -1.449 seconds, and t = 3.449 seconds

Since the negative time does not have physical meaning in our case (time before the object was launched), we adopt the second answer: t = 3.449 seconds

5 0
3 years ago
Solve the radical equation square root 25-40x=x-10
azamat

Answer:

15 or   -5


Step-by-step explanation:

√(25-40x)= x-10

Squaring both sides

25 - 40x = (x -10)²

25 - 40x = x² -20x + 100

x² - 20x + 40x + 100 - 25 =0

x² +20x = -75

x² + 20x + 10² = -75 + 100

(x + 10)² = 25

x - 10 = ±√25

x - 10 = ±5

x  = ±5 + 10

x = 15 or    x = -5


7 0
3 years ago
A triangle has a base of 3 2/5 inches and an area of 5 1/10 square inches. Find the height of the triangle.
allsm [11]

Answer:

Step-by-step explanation:

44

5 0
3 years ago
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