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2 years ago

Enter the values for the highlighted variables that show how to subtract the rational expressions correctly

Mathematics

2 answers:

I am Lyosha [343]2 years ago

8 0

Answer:

a = 6

b = 2

c = 6

d = 2

e = 6

f = 6

g=1

Step-by-step explanation:

a = 6

x^2 + 6x is equal to x(x+6)

b=2

Denominator and numerator of the first term are multiplied by x.

c=6

Second term is multiplied by (x-6)/(x-6)

d=2

Now that they have the same denominator, the two terms are combined. 2 is the coefficient of the first term

e=6

In the same way as d is carried over from b, e is carried over from c.

f = 6

2x - x + 6 = x + 6

g = 1

We factor out the (x+6) from the numerator and denominator

Lina20 [59]2 years ago

4 0

A = 6
x^2 + 6x is equal to x(x+6)
b=2
Denominator and numerator of the first term are multiplied by x.
c=6
Second term is multiplied by (x-6)/(x-6)
d=2
Now that they have the same denominator, the two terms are combined. 2 is the coefficient of the first term
e=6
In the same way as d is carried over from b, e is carried over from c.
f = 6
2x - x + 6 = x + 6
g = 1
We factor out the (x+6) from the numerator and denominator

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On a coordinate plane, 2 trapezoids are shown. The first trapezoid has points A (negative 1, negative 2), B (1, negative 2), C (

Liula [17]

Answer:

The true statements are:

The rule for the translation can be written as T-3, 1(x, y). ⇒ 1st

The rule for the translation can be written as (x, y) → (x - 3, y + 1) ⇒ 4th

Step-by-step explanation:

Let us revise the rule of the translation:

If the point (x , y) translated horizontally to the right by h units then its image is (x + h , y)
If the point (x , y) translated horizontally to the left by h units then its image is (x - h , y)
If the point (x , y) translated vertically up by k units then its image is (x , y + k)
If the point (x , y) translated vertically down by k units then its image is (x , y - k)

The vertices of trapezoid ABCD are:

A (-1 , -2) , B (1 , -2) , C (2 , -5) , D (-2 , -5)

The vertices Of trapezoid A'B'C'D' are:

A' (-4 , -1) , B' (-2 , -1) , C' (-1 , -4) , D' (-5 , -4)

Trapezoid ABCD is translated to form Trapezoid A'B'C'D'

By using the rules of translation above:

∵ A = (-1 , -2) and A' = (-4 , -1)

- The rule of translation is (x , y) → (x + h , y + k)

∵ x = -1 and x + h = -4

∴ -1 + h = -4

- Add 1 to both sides

∴ h = -3

∵ y = -2 and y + k = -1

∴ -2 + k = -1

- Add 2 to both sides

∴ k = 1

The rule of translation is (x , y) → (x - 3 , y + 1)

The trapezoid ABCD has been translated 3 units to the left and 1 unit up

You can check your answer by doing the same steps with the other 3 vertices you will have the same values of h and k

The true statements are:

The rule for the translation can be written as T-3, 1(x, y).

The rule for the translation can be written as (x, y) → (x - 3, y + 1)

9 0

2 years ago

Read 2 more answers

6x4=8xN<br> how do i solve this equation

kow [346]

Answer:

whats equivelent to 6x4?

Step-by-step explanation:

6 x 4 = 24

8 x 2 = 16 so 8 x 3 = 24 because 8 x 4 would be

8 x 2 = 16

16 x 2 =

6 + 6 = 12

carry the one.

1 + 1 + 1 (We now have three ones from carrying that 1) and our answer is

32.

So N = 3

5 0

2 years ago

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Write the number 1.8 in the form a/b, using integers, to show that it is a rational number.

Pepsi [2]

1. Note that the number 1.8 consists of two parts 1 and 0.8.

2. The decimal 0.8 can be written as the fraction $\dfrac{8}{10}.$

3. Add these two numbers:

$1+\dfrac{8}{10}=\dfrac{10}{10}+\dfrac{8}{10}=\dfrac{10+8}{10}=\dfrac{18}{10}.$

4. Since 18=2·3·3 and 10=2·5, the fraction $\dfrac{18}{10}$ can be simplified to the rational number

$\dfrac{18}{10}=\dfrac{2\cdot 3\cdot 3}{2\cdot 5}=\dfrac{3\cdot 3}{5}=\dfrac{9}{5}.$

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2 years ago

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The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 -57, where t is the time

cupoosta [38]

A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?

Answer:

$1\leq t\geq \sqrt{2}$

Step-by-step explanation:

Given:

A stuntman jumping off a 20-m-high building is modeled by the equation

$h =20-5t^{2}$ -----------(1)

A high-speed camera is ready to making film between 15 m and 10 m above the ground

when the stuntman is 15m above the ground.

height $h = 15m$

Put height value in equation 1

$15 =20-5t^{2}$

$5t^{2} =20-15$

$5t^{2} =5$

$t^{2} =1$

$t =\pm1$

We know that the time is always positive, therefore $t=1$

when the stuntman is 10m above the ground.

height $h = 10m$

Put height value in equation 1

$10 =20-5t^{2}$

$5t^{2} =20-10$

$5t^{2} =10$

$t^{2} =\frac{10}{5}$

$t^{2} =2$

$t=\pm\sqrt{2}$

$t=\sqrt{2}$

Therefore ,time interval of camera film him is $1\leq t\geq \sqrt{2}$

7 0

2 years ago

Byron has 1 7/10 kilograms of black pepper. He uses 7/8 of the black pepper and splits between 7 pepper shakers. How much pepper

IrinaVladis [17]

Amount of black pepper = 1 7/10 kg = 17/10 kg.
Amount used is
(7/8)*(17/10) = 119/80 kg

This amount is split between 7 shakers. Each shaker has
(119/80)/7 = 17/80 kg = 0.2125 kg

Answer: 17/80 kg (or 0.2125 kg)

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2 years ago