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KengaRu [80]
3 years ago
13

Simplify the expression: (9x+10) - (3x+7)

Mathematics
2 answers:
mote1985 [20]3 years ago
6 0
(9x+10) - (3x+7) = 6x+3

Step-by-Step:

In this equation, our main goal is to combine our ‘like terms’ or terms that share the same variable or power. In this case, the ( ) of both equations holds one term that has an X and one that is a lone number. We are now subtracting our two sets of like terms from one another.

9x-3x = 6x and 10-7= 3

After this, we just combine our answers from largest variable/exponent to the smallest, in this case, our answer ends up being 6x+3
vovikov84 [41]3 years ago
3 0
19x-10x would be the answer
You might be interested in
Match the hyperbolas represented by the equations to their foci.
Arte-miy333 [17]

Answer:

1) (1 , -22) and (1 , 12) ⇔ (y + 5)²/15² - (x - 1)²/8² = 1

2) (-7 , 5) and (3 , 5) ⇔ (x + 2)²/3² - (y - 5)²/4² = 1

3) (-6 , -2) and (14 , -2) ⇔ (x - 4)²/8² - (y + 2)²/6² = 1

4) (-7 , -10) and (-7 , 16) ⇔ (y - 3)²/5² - (x + 7)²/12² = 1

Step-by-step explanation:

* Lets study the equation of the hyperbola

- The standard form of the equation of a hyperbola with

  center (h , k) and transverse axis parallel to the x-axis is

  (x - h)²/a² - (y - k)²/b² = 1

- the coordinates of the foci are (h ± c , k), where c² = a² + b²

- The standard form of the equation of a hyperbola with

  center (h , k) and transverse axis parallel to the y-axis is

  (y - k)²/a² - (x - h)²/b² = 1

- the coordinates of the foci are (h , k ± c), where c² = a² + b²

* Lets look to the problem

1) The foci are (1 , -22) and (1 , 12)

- Compare the point with the previous rules

∵ h = 1 and k ± c = -22 ,12

∴ The form of the equation will be (y - k)²/a² - (x - h)²/b² = 1

∵ k + c = -22 ⇒ (1)

∵ k - c = 12 ⇒ (2)

* Add (1) and(2)

∴ 2k = -10 ⇒ ÷2

∴ k = -5

* substitute the value of k in (1)

∴ -5 + c = -22 ⇒ add 5 to both sides

∴ c = -17

∴ c² = (-17)² = 289

∵ c² = a² + b²

∴ a² + b² = 289

* Now lets check which answer has (h , k) = (1 , -5)

  and a² + b² = 289 in the form (y - k)²/a² - (x - h)²/b² = 1

∵ 15² + 8² = 289

∵ (h , k) = (1 , -5)

∴ The answer is (y + 5)²/15² - (x - 1)²/8² = 1

* (1 , -22) and (1 , 12) ⇔ (y + 5)²/15² - (x - 1)²/8² = 1

2) The foci are (-7 , 5) and (3 , 5)

- Compare the point with the previous rules

∵ k = 5 and h ± c = -7 ,3

∴ The form of the equation will be (x - h)²/a² - (y - k)²/b² = 1

∵ h + c = -7 ⇒ (1)

∵ h - c = 3 ⇒ (2)

* Add (1) and(2)

∴ 2h = -4 ⇒ ÷2

∴ h = -2

* substitute the value of h in (1)

∴ -2 + c = -7 ⇒ add 2 to both sides

∴ c = -5

∴ c² = (-5)² = 25

∵ c² = a² + b²

∴ a² + b² = 25

* Now lets check which answer has (h , k) = (-2 , 5)

  and a² + b² = 25 in the form (x - h)²/a² - (y - k)²/b² = 1

∵ 3² + 4² = 25

∵ (h , k) = (-2 , 5)

∴ The answer is (x + 2)²/3² - (y - 5)²/4² = 1

* (-7 , 5) and (3 , 5) ⇔ (x + 2)²/3² - (y - 5)²/4² = 1

3) The foci are (-6 , -2) and (14 , -2)

- Compare the point with the previous rules

∵ k = -2 and h ± c = -6 ,14

∴ The form of the equation will be (x - h)²/a² - (y - k)²/b² = 1

∵ h + c = -6 ⇒ (1)

∵ h - c = 14 ⇒ (2)

* Add (1) and(2)

∴ 2h = 8 ⇒ ÷2

∴ h = 4

* substitute the value of h in (1)

∴ 4 + c = -6 ⇒ subtract 4 from both sides

∴ c = -10

∴ c² = (-10)² = 100

∵ c² = a² + b²

∴ a² + b² = 100

* Now lets check which answer has (h , k) = (4 , -2)

  and a² + b² = 100 in the form (x - h)²/a² - (y - k)²/b² = 1

∵ 8² + 6² = 100

∵ (h , k) = (4 , -2)

∴ The answer is (x - 4)²/8² - (y + 2)²/6² = 1

* (-6 , -2) and (14 , -2) ⇔ (x - 4)²/8² - (y + 2)²/6² = 1

4) The foci are (-7 , -10) and (-7 , 16)

- Compare the point with the previous rules

∵ h = -7 and k ± c = -10 , 16

∴ The form of the equation will be (y - k)²/a² - (x - h)²/b² = 1

∵ k + c = -10 ⇒ (1)

∵ k - c = 16 ⇒ (2)

* Add (1) and(2)

∴ 2k = 6 ⇒ ÷2

∴ k = 3

* substitute the value of k in (1)

∴ 3 + c = -10 ⇒ subtract 3 from both sides

∴ c = -13

∴ c² = (-13)² = 169

∵ c² = a² + b²

∴ a² + b² = 169

* Now lets check which answer has (h , k) = (-7 , 3)

  and a² + b² = 169 in the form (y - k)²/a² - (x - h)²/b² = 1

∵ 5² + 12² = 169

∵ (h , k) = (-7 , 3)

∴ The answer is (y - 3)²/5² - (x + 7)²/12² = 1

* (-7 , -10) and (-7 , 16) ⇔ (y - 3)²/5² - (x + 7)²/12² = 1

7 0
3 years ago
Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 60
Umnica [9.8K]

Sample mean : \overline{x}=10.6x=10.6

Standard deviation : s=1.7s=1.7

Significance level : \alpha:1-0.95=0.05α:1−0.95=0.05

Critical value : z_{\alpha/2}=1.96

Hence the 95% confidence interval for the number of chocolate chips per cookie for big chip cookies= (10.1989,\ 11.0011)(10.1989, 11.0011)

4 0
2 years ago
A beach ball a radius 15 centimeters what is the volume of the beach ball?
sladkih [1.3K]

Answer:

volume = 4/3 (pi x r^3)

Volume = 4/3 (pi x 15^3)

Volume = 4/3 (3.14 x 3375)

Volume = 14130 cubic cm

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
You are choosing between two health clubs. Club A offers membership for a fee of $ 28 plus a monthly fee of $ 22. Club B offers
Mazyrski [523]

Answer: 2 months, Total cost= $72

Step-by-step explanation:

Let x denote the number of months then

For Club A fee= 28+22x

For Club B fee= 16+28x

Equating both we get

28 + 22x= 16+28x

28-16= 28-22x

12= 6x

x= 2

So after 2 months the fee will be same i.e. 28+22(2)= $72

8 0
3 years ago
This math I just can't figure out
Elena-2011 [213]

Step-by-step explanation:

Here's the first question and how to solve the second.

So the first business start with 250, that's your b in slope.

y = mx + b

y = mx + 250

Now, they charge 14 an hour so that's your m.

y = 14x + 250

x is the number of hours it takes.

The second business is straight up 22 per hour, so it's just y = mx, where b is 0.

y = 22x VS y = 14x + 250

Try plugging in different numbers and seeing how much it take to level out!

4 0
3 years ago
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