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3 years ago

Angle Relationship Equations ... pose help I don’t do this but my friend has to so help

Mathematics

2 answers:

Naddika [18.5K]3 years ago

4 0

26 + ( 3x + 1 ) = 180

27 + 3x = 180

3x = 180 - 27

3x = 153

x = 153/3

x = 51

___________________

( 4x + 1 ) + ( 3x + 4 ) = 180

7x + 5 = 180

7x = 180 - 5

7x = 175

x = 175/5

x = 25

eduard3 years ago

4 0

Answer:

3) $x=51$

4) $x=25$

Step-by-step explanation:

$3) ~ 3x+1+26=180\\ \\ ~ (linear ~pair)$

$3x+27=180\\~ ~ -27 ~ ~-27$

$subtract ~ -27 ~ from ~ both ~ sides$

$\frac{3x}{3} =\frac{153}{3}$

$x=51$

--------------------------

$4)~ 4x+1+4+3x=180$

$7x=180-5$

$x=\frac{175}{7}$

$x=25$

----------------------------

hope it helps...

have a great day!!

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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
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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

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

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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
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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

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

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
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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

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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

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

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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.

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goldenfox [79]

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3 years ago

￼Angle Relationship Equations ... pose help I don’t do this but my friend has to so help

Angle Relationship Equations ... pose help I don’t do this but my friend has to so help