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Delvig [45]
4 years ago
5

Write the ratio 7/8 in two different forms

Mathematics
1 answer:
Rainbow [258]4 years ago
5 0
You can write it
7 to 8
7:8
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This set of points is on the graph of a function.
mars1129 [50]

Answers: (B) (2, -1) and (D) (5, 2)

<u>Step-by-step explanation:</u>

Inverse is when the x's and y's are swapped

Given: (-2, 5), (-1, 2), (0, 1), (2, 5)

Inverse: (5, -2), (2, -1) (1, 0), (5, 2)

6 0
3 years ago
A 18 foot ladder is leaning against a wall as shown
Inga [223]
I'm guessing the diagram shows a ladder leaning against a wall, making a right angle triangle with respect to the ground and the wall.

So, the wall's height is going to be the 'h', which will also be the 'opposite side' from the angle <span>ϴ which is made from the ladder and the ground.
</span>The ladder's length (18 foot) is going to be the 'hypotenuse' side and the other remaining side will be the 'adjacent'.

Now, once you've sorted out which side is which, we have to find the h (opp), and according to  SOH CAH TOA, we will choose Sin<span>ϴ = opp/hyp.
</span>so Sinϴ = h/18....now we gotta find h, so 'cross multiply' the equation to get  h = 18 x sin<span>ϴ.
</span>
To find angle ϴ, simply take the inverse of Sinϴ= h/18... and you'll get ϴ =  sin-1 (sin inverse) h/18

Hope this helps



6 0
3 years ago
a pen has an original price of $32 but is marked down by 15% which equation shows the new price of the pan
grigory [225]
31.85 because 32-0.15=31.85
7 0
4 years ago
In kick boxing, the force needed to break a
Molodets [167]

Answer: 20

Step-by-step explanation: for every 2 you have five, 8 is equal to 2x4 so do 4x5 and you get 20.

5 0
3 years ago
Suppose brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt.
Oliga [24]

Answer:

(a) 0.288 kg/liter

(b) 0.061408 kg/liter

Step-by-step explanation:

(a) The mass of salt entering the tank per minute, x = 0.2 kg/L × 5 L/minute = 1 kg/minute

The mass of salt exiting the tank per minute = 5 × (5 + x)/500

The increase per minute, Δ/dt, in the mass of salt in the tank is given as follows;

Δ/dt = x - 5 × (5 + x)/500

The increase, in mass, Δ, after an increase in time, dt, is therefore;

Δ = (x - 5 × (5 + x)/500)·dt

Integrating with a graphing calculator, with limits 0, 10, gives;

Δ = (99·x - 5)/10

Substituting x = 1 gives

(99 × 1 - 5)/10 = 9.4 kg

The concentration of the salt and water in the tank after 10 minutes = (Initial mass of salt in the tank + Increase in the mass of the salt in the tank)/(Volume of the tank)

∴ The concentration of the salt and water in the tank after 10 minutes =  (5 + 9.4)/500 = (14.4)/500 = 0.288

The concentration of the salt and water in the tank after 10 minutes = 0.288 kg/liter

(b) With the added leak, we now have;

Δ/dt = x - 6 × (14.4 + x)/500

Δ = x - 6 × (14.4 + x)/500·dt

Integrating with a graphing calculator, with limits 0, 20, gives;

Δ = 19.76·x -3.456 = 16.304

Where x = 1

The increase in mass after an increase in = 16.304 kg

The total mass = 16.304 + 14.4 = 30.704 kg

The concentration of the salt in the tank then becomes;

Concentration = 30.704/500 = 0.061408 kg/liter.

6 0
3 years ago
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