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ivann1987 [24]
3 years ago
14

So who's going to do this aka the queen / King Master ​

Mathematics
1 answer:
attashe74 [19]3 years ago
5 0
590 is your answer mark me brainist please
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F(x) = -4x2 + 10<br> Find f(-2)
Veronika [31]

Answer:

f(-2) = -6

Step-by-step explanation:

In the expression, to get f(-2);

we simply substitute the value of x with -2

We have this as;

f(-2) = -4(-2)^2 + 10

f(-2) = -4(4) + 10

f(-2) = -16 + 10

f(-2) = -6

3 0
3 years ago
Bryan and Jadyn had barbeque potato chips and soda at a football party. Bryan ate 3 oz of chips and drank 2 cups of soda for a t
Diano4ka-milaya [45]

Answer:

200 mg sodium is in 1 oz of chips and 50 mg sodium is in 1 cup of soda.

Step-by-step explanation:

Let x mg sodium is in 1 oz of chips and and y mg is in 1 cup of soda.

∵ Bryan ate 3 oz of chips and drank 2 cups of soda for a total of 700 mg of sodium.

i.e. 3x + 2y = 700 --------(1),

Jadyn ate 1 oz of chips and drank 3 cups of soda for a total of 350 mg of sodium.

i.e. x + 3y = 350 ---------(2),

Equation (1) - 3 × equation (2),

We get,

2y - 9y = 700 - 1050

-7y = -350

\implies y = \frac{-350}{-7}= 50

From equation (1),

3x + 2(50) = 700

3x + 100 = 700

3x = 700 - 100

3x = 600

\implies x = \frac{600}{3}=200

Hence, 200 mg sodium is in 1 oz of chips and 50 mg sodium is in 1 cup of soda.

7 0
3 years ago
Find each sum simplify if possible
padilas [110]
3/5
This will help you
8 0
3 years ago
Read 2 more answers
Let f(x) = x - 3 and g(x) = x + 11 find f(x) times g(x)<br><br> What is the answer?
notsponge [240]

f(x) *g(x) = (x-3)(x+11) = x^2 +11x -3x -33 = x^2 +8x -33

Sorry if its wrong :(

6 0
3 years ago
A tank with a capacity of 1000 L is full of a mixture of water and chlorine with a concentration of 0.02 g of chlorine per liter
faltersainse [42]

At the start, the tank contains

(0.02 g/L) * (1000 L) = 20 g

of chlorine. Let <em>c</em> (<em>t</em> ) denote the amount of chlorine (in grams) in the tank at time <em>t </em>.

Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of

(<em>c</em> (<em>t</em> )/(1000 + (10 - 25)<em>t</em> ) g/L) * (25 L/s) = 5<em>c</em> (<em>t</em> ) /(200 - 3<em>t</em> ) g/s

In case it's unclear why this is the case:

The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after <em>t</em> s there will be (1000 + 10<em>t</em> ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time <em>t</em> is (1000 - 15<em>t </em>) L. Then the concentration of chlorine per unit volume is <em>c</em> (<em>t</em> ) divided by this volume.

So the amount of chlorine in the tank changes according to

\dfrac{\mathrm dc(t)}{\mathrm dt}=-\dfrac{5c(t)}{200-3t}

which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5<em>t</em> )^(5/3), then integrate both sides to solve for <em>c</em> (<em>t</em> ):

\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{200-3t}=0

\dfrac1{(200-3t)^{5/3}}\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{(200-3t)^{8/3}}=0

\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0

\dfrac{c(t)}{(200-3t)^{5/3}}=C

c(t)=C(200-3t)^{5/3}

There are 20 g of chlorine at the start, so <em>c</em> (0) = 20. Use this to solve for <em>C</em> :

20=C(200)^{5/3}\implies C=\dfrac1{200\cdot5^{1/3}}

\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}

7 0
3 years ago
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