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natima [27]
3 years ago
13

PLEASE HELP FAST...WILL MARK AS BRAINLIEST IF GIVEN GOOD ANSWER ​

Mathematics
1 answer:
EleoNora [17]3 years ago
3 0

Answer:

I think it B

Step-by-step explanation:

Cause if we are looking for an angle that is approximately = to the coise B has no angle

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Who can teel me all of the digits of pie i will give branliest
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Answer:

3.141592653589793......It will go on for infinity

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Maria's class was having an end of the school year party. The teacher ordered 6 large pizzas which were cut into 8 slices each.
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1. The height of a triangle is 6 m more than its base. The area of the triangle is 56 m². What is the length of the base? Enter
Elodia [21]
Answers:
1. 8 m 
2. 17 m
3. 7 cm
4. 2 s

Explanations:

1. Let x = length of the base
          x + 6 = height of the base

    Then, the area of the triangle is given by

    (Area) = (1/2)(base)(height)
       56 = (1/2)(x)(x + 6)
       56 = (1/2)(x²  + 6x) 
     
    Using the symmetric property of equations, we can interchange both sides      of equations so that 

    (1/2)(x²  + 6x) = 56
    
    Multiplying both sides by 2, we have
   
    x² + 6x = 112
    
    The right side should be 0. So, by subtracting both sides by 112, we have 

    x² + 6x - 112 = 112 - 112
    x² + 6x - 112 = 0

    By factoring, x² + 6x - 112 = (x - 8)(x + 14). So, the previous equation           becomes

    (x - 8)(x +14) = 0

   So, either 

    x - 8 = 0 or x + 14 = 0

   Thus, x = 8 or x = -14. However, since x represents the length of the base and the length is always positive, it cannot be negative. Hence, x = 8. Therefore, the length of the base is 8 cm.

2. Let x = length of increase in both length and width of the rectangular garden

Then,

14 + x = length of the new rectangular garden
12 + x = width of the new rectangular garden

So, 

(Area of the new garden) = (length of the new garden)(width of the new garden) 

255 = (14 + x)(12 + x) (1)

Note that 

(14 + x)(12 + x) = (x + 14)(x + 12)
                          = x(x + 14) + 12(x + 14)
                          = x² + 14x + 12x + 168 
                          = x² + 26x + 168

So, the equation (1) becomes

255 = x² + 26x + 168

By symmetric property of equations, we can interchange the side of the previous equation so that 

x² + 26x + 168 = 255

To make the right side becomes 0, we subtract both sides by 255:

x² + 26x + 168 - 255 = 255 - 255
x² + 26x - 87 = 0 

To solve the preceding equation, we use the quadratic formula.

First, we let

a = numerical coefficient of x² = 1

Note: if the numerical coefficient is hidden, it is automatically = 1.

b = numerical coefficient of x = 26
c = constant term = - 87

Then, using the quadratic formula 

x =  \frac{-b \pm  \sqrt{b^2 - 4ac} }{2a} =  \frac{-26 \pm  \sqrt{26^2 - 4(1)(-87)} }{2(1)}  
\newline x =  \frac{-26 \pm  \sqrt{1,024} }{2}
\newline
\newline x =  \frac{-26 \pm  32 }{2}

So, 

x = \frac{-26 + 32 }{2} \text{  or } x = \frac{-26 - 32 }{2}
\newline x = \frac{6 }{2} \text{  or } x = \frac{-58 }{2}
\newline \boxed{ x = 3 \text{  or } x = -29}

Since x represents the amount of increase, x should be positive.

Hence x = 3.

Therefore, the length of the new garden is given by 

14 + x = 14 + 3 = 17 m.

3. The area of the shaded region is given by

(Area of shaded region) = π(outer radius)² - π(inner radius)²
                                       = π(2x)² - π6²
                                       = π(4x² - 36)

Since the area of the shaded region is 160π square centimeters,

π(4x² - 36) = 160π

Dividing both sides by π, we have 

4x² - 36 = 160

Note that this equation involves only x² and constants. In these types of equation we get rid of the constant term so that one side of the equation involves only x² so that we can solve the equation by getting the square root of both sides of the equation.

Adding both sides of the equation by 36, we have

4x² - 36 + 36 = 160 + 36
4x² = 196 

Then, we divide both sides by 4 so that

x² = 49

Taking the square root of both sides, we have

x = \pm 7

Note: If we take the square root of both sides, we need to add the plus minus sign (\pm) because equations involving x² always have 2 solutions.

So, x = 7 or x = -7.

But, x cannot be -7 because 2x represents the length of the outer radius and so x should be positive.

Hence x = 7 cm

4. At time t, h(t) represents the height of the object when it hits the ground. When the object hits the ground, its height is 0. So,
 
h(t) = 0   (1)

Moreover, since v_0 = 27 and h_0 = 10, 

h(t) = -16t² + 27t + 10   (2)

Since the right side of the equations (1) and (2) are both equal to h(t), we can have

-16t² + 27t + 10 = 0

To solve this equation, we'll use the quadratic formula.

Note: If the right side of a quadratic equation is hard to factor into binomials, it is practical to solve the equation by quadratic formula. 

First, we let

a = numerical coefficient of t² = -16 
b = numerical coefficient of t = 27
c = constant term = 10

Then, using the quadratic formula 

t = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} = \frac{-27 \pm \sqrt{27^2 - 4(-16)(10)} }{2(-16)} \newline t = \frac{-27 \pm \sqrt{1,369} }{-32} \newline \newline t = \frac{-27 \pm 37 }{32}

So, 

t = \frac{-27 + 37 }{-32} \text{ or } t = \frac{-27 - 37 }{-32} \newline t = \frac{-10}{32}  \text{ or } t = \frac{-64 }{-32}   \newline \boxed{ t = -0.3125 \text{ or } t = 2}

Since t represents the amount of time, t should be positive. 

Hence t = 2. Therefore, it takes 2 seconds for the object to hit the ground.


 




 





3 0
4 years ago
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The addition property of equality states that a*b + c = a + b + c. True or False.
Anarel [89]

this is false.

the addition property of equality states that a + b = c, and b + a = c

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3 years ago
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<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7Bx-5%7D%7B2x%5E%7B2%7D-5x-3%20%7D" id="TexFormula1" title="f(x) = \f
Vedmedyk [2.9K]

i) The given function is

f(x)=\frac{x-5}{2x^2-5x-3}

The factored form is

f(x)=\frac{x-5}{(x-3)(2x+1)}

The domain are the values of  x for which the function is defined.

(x-3)(2x+1)\ne 0

(x-3)\ne0,(2x+1)\ne 0

x\ne3,x\ne-\frac{1}{2}

ii) To find the vertical asymptotes, equate the denominator to zero.

(x-3)(2x+1)=0

(x-3)=\ne0,(2x+1)=0

x=3,x=-\frac{1}{2}

iii) To find the roots, equate the numerator to zero.

x-5=0

The root is x=5

iv) To find the y-intercept, put x=0 into the function.

f(0)=\frac{0-5}{(0-3)(2(0)+1)}

f(0)=\frac{-5}{(-3)(1)}

f(0)=\frac{5}{3}

The y-intercept is \frac{5}{3}

v) The horizontal asymptote is given by;

lim_{x\to \infty}\frac{x-5}{2x^2-5x-3}=0

The horizontal asymptote is y=0

vi) The function is not reducible. There are no holes.

vii) The given function is a proper rational function.

Proper rational functions do not have oblique asymptotes.

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3 years ago
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