I need help on question 9 please!! Thank you!

PLZZ HELPP!! Find the perimeter of the following polygon.

hodyreva [135]

Answer:

43in

Step-by-step explanation:

Perimeter = sum of the sides

Perimeter = 11 + 10 + 10 + 12 = 43

4 0

3 years ago

Read 2 more answers

I need help 2(x + 1) = 2x + 2.

Ksju [112]

$2(x + 1) = 2x + 2\\
2x+2=2x+2\\
0=0\\
x\in\mathbb{R}$

7 0

3 years ago

Si camilo tiene 6 años mas que juan, juan tiene 7 años menos que andrés y andrés tiene 35 años

lidiya [134]

Answer:

Part a) Camilo is 34 years old and Juan is 28 years old

Part b) The difference between Camilo's age and Juan's age is 6 years

Part c) The sum of the ages of Camilo and Juan not coincide with the age of Andres' grandmother

Step-by-step explanation:

The question in English is

If Camilo is 6 years older than Juan, Juan is 7 years younger than Andrés and Andrés is 35 years old .

Part a) How old is Camilo and how old is Juan? raises the equation.

Part b) What's the difference between Camilo's age and Juan's age?

Part c) If Andres's grandmother is 65 years old, does the sum of the ages of Camilo and Juan coincide with the age of Andres' grandmother?

Part a) How old is Camilo and how old is Juan? raises the equation.

Let

x -----> Camilo's age

y -----> Juan's age

z -----> Andres's age

we know that

$x=y+6$ -----> equation A

$y=z-7$ -----> equation B

$z=35$ -----> equation C

Find the value of y

substitute the value of z in the equation B

$y=35-7=28$

Find the value of x

substitute the value of y in the equation A

$x=28+6=34$

therefore

Camilo is 34 years old and Juan is 28 years old

Part b) What's the difference between Camilo's age and Juan's age?

Subtract Juan's age from Camilo's age

so

$x-y=34-28=6\ years$

Part c) If Andres's grandmother is 65 years old, does the sum of the ages of Camilo and Juan coincide with the age of Andres' grandmother?

Adds Camilo's age and Juan's age

$x+y=34+28=62\ years$

$62\ years\neq 65\ years$

therefore

The sum of the ages of Camilo and Juan not coincide with the age of Andres' grandmother

3 0

3 years ago

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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"A supplier of digital memory cards claims that no more than 1% of the cards are defective. In a random sample of 600 memory car

Nady [450]

Answer:

Both the values fall in the critical region so we reject H0: p ≤ 0.01 and conclude that the results are significantly different from chance identification and it is not a sample fluctuation.

Step-by-step explanation:

1) Formulate the hypothesis

H0: p ≤ 0.01 against the claim Ha: p > 0.01

2) Choose the significance level ∝= 0.01

3) The test statistics under H0 is

z= x -np /√npq ( without continuity correction)

z= x ± 1/2 -np /√npq ( with continuity correction)

4) the critical region is z> ±2.33 because the alternate hypothesis is stated on greater than basis.

5)Computations:

Z= x -np /√npq ( without continuity correction)

Z= 6- 600( 0.03)/√ 600( 0.03)(0.97)

z= - 2.82 ( without continuity correction)

z= x ± 1/2 -np /√npq ( with continuity correction)

Z= (6-0.5)- 600( 0.03)/√ 600( 0.03)(0.97)

z= -12.5/4.1785

z= -0.29915 ( with continuity correction)

6) Both the values fall in the critical region so we reject H0: p ≤ 0.01 and conclude that the results are significantly different from chance identification and it is not a sample fluctuation.

8 0

3 years ago