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2 years ago

Solve the system of equations. 3x+2y+z=16 4x−y=−5 y+z=11

Mathematics

1 answer:

labwork [276]2 years ago

6 0

Answer:

I have no clue j need more free answer

Step-by-step explanation:

3x+2y+z=16 4x-y=-5 y+x=11

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The function p(x) = _2(x _ 9)2 + 100 is used to determine the profit on T-shirts sold for x dollars. What would the profit from

Valentin [98]

Answer:

Option (b) is correct.

The profit from sales if the price of the T-shirts were $15 a pieces is $28.

Step-by-step explanation:

Given : The function $p(x)=-2(x-9)^2+100$ is used to determine the profit on T-shirts sold for x dollars.

We have to find the profit from sales if the price of the T-shirts were $15 a pieces.

Consider the given profit function $p(x)=-2(x-9)^2+100$

For the profit, if the price of a t-shirt is $ 15

Put x = 15, we have,

$p(15)=-2(15-9)^2+100$

Simplify, we have,

$p(15)=-2(6)^2+100=-2\cdot 36+100$

Also, $p(15)=-72+100=28$

Thus, we get, The profit from sales if the price of the T-shirts were $15 a pieces is $28.

5 0

3 years ago

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What the vertex of the equation: y = x^2 - 6x + 11

pychu [463]

Answer:

the vertex is of the equation is (3,2)

7 0

3 years ago

8(5x–3)–11(3x+7)=102

enyata [817]

40x - 24 - 33x - 77 = 102
40x - 33x = 102 + 24 + 77
7x = 203
x = 29

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3 years ago

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If Nancy uses 1 74 tablespoons of coffee to make 10 cups of coffee, how

Anna [14]

Answer:

if you mean 17/4 tablespoons then it would take .425 tablespoons to make one cup of coffee, because 17/4 is 4.25 as a decimal.

Step-by-step explanation:

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2 years ago

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Consider a series circuit consisting of a resistor of R ohms, an inductor of L henries, and variable voltage source of V(t) volt

ser-zykov [4K]

Answer:

$I(t)=\frac{1}{3}(1-e^{-30t})$

Step-by-step explanation:

We are given that

$\frac{dI}{dt}+\frac{R}{L}I=\frac{V(t)}{L}$

R=150 ohm

L=5 H

V(t)=10 V

$P=\frac{R}{L}=\frac{150}{5}=30$

$I.F=e^{\int Pdt}=e^{\int 30 dt}=e^{30 t}$

$I(t)\times I.F=\int e^{30 t}\times 10 dt+C$

$I(t)\times e^{30 t}=\frac{10}{30}e^{30 t}+C$

$I(t)=\frac{1}{3}+Ce^{-30 t}$

I(0)=0

Substitute t=0

$0=\frac{1}{3}+C$

$C=-\frac{1}{3}$

Substitute the values

$I(t)=\frac{1}{3}-\frac{1}{3}e^{-30 t}$

$I(t)=\frac{1}{3}(1-e^{-30t})$

5 0

3 years ago