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AURORKA [14]
3 years ago
11

Danielle has her finals coming up and she's worried she'll fail her class. Her past 3 test grades were a 72. an 85, and a 68. Sh

e needs a 70 to pass the class. What grade does she need to make on the final to get a 70?
Mathematics
1 answer:
natka813 [3]3 years ago
7 0

Answer:

55

Step-by-step explanation:

You might be interested in
Select the correct location on the coordinate plane. Ruhana owns a workshop where her team of technicians refurbishes TV sets an
yarga [219]

Answer:

Ruhana and her team should refurbish 15 tv sets and 20 dvd players each week to minimize the cost.

Step-by-step explanation:

For the given situation let the number of tv sets be x and number of dvd players be y.

Now we have to minimize the cost as it costs $75 to refurbish a tv set and $40 to refurbish a dvd player.

i.e. Minimize z=75x+40y

it gives subject to the constraints

4x+2y≥100......(1)(100 hours each week. It takes 4 hours to refurbish a tv set and 2 hours to refurbish a dvd player.)

x+y≥35.......(2)(weekly sales target is to refurbish at least 35 tv sets or dvd players.)

To plot equation (1) we need to find coordinates of points lying on line (1)

put x=0 gives 2y=100⇒y=50

put y=0 gives 4x=100⇒x=25

So we got points (0,50) and(25,0) for (1)..............(3)

Similarly for equation (2)

put x=0 gives y=35

put y=0 gives x=35

so we got points (0,35) and(35,0) for (2).................(4)

with the help of (3) and (4) we plot the following graph (assume x≥0 and y≥0)

The unbounded feasible region determined by constraints gives the corner points as A(0,50),B(15,20)and C(35,0).

from  we get the value of z is minimum at point B (15,20) .

hence we got our optimal solution at B (15,20), where x is the number of tv sets and y is the number of dvd players  .therefore Ruhana and her team should refurbish 15 tv sets and 20 dvd players each week to minimize the cost.

7 0
3 years ago
Please help me with my math
Natasha_Volkova [10]

Answer:

The answer is C

Step-by-step explanation:

the y-intercept is going through 0 therefore there would be no "b" in the y=mx+b formula

3 0
3 years ago
What are the possible names for 1 2/9
Harrizon [31]
A mixed number and one and two-ninths.
Hope this helps! :D
4 0
3 years ago
Read 2 more answers
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
The area of a playground is 266 yd. The width of the playground is 5 yd longer than its length. Find th
Mademuasel [1]

Answer:

C

Step-by-step explanation:

The width is 5 yds longer so we can mark out b and d if you multiply 19 and 24 is 456 so you are left with C

5 0
2 years ago
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