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3 years ago

When using the substitution method for solving a system of linear equations in two variables, what should be the first step?

1 answer:

8 0

The first step would be to isolate a variable (solve for a variable)....so u will have something to sub into the other equation

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Solve the inequality 2(4+2x)25x+5 0 X5-2 O x2-2 O x3 O X23

Sveta_85 [38]

If you’re simplifying it’s 100x^2+200x+5

3 0

3 years ago

Which statement is true about the extreme value of the given quadratic equation? y = -3x2 + 12x − 33

Zina [86]

Answer:

The equation has a maximum value with a y-coordinate of -21.

Step-by-step explanation:

Given

$y =-3x^2 + 12x - 33$

Required

The true statement about the extreme value

First, write out the leading coefficient

$Leading = -3$

$-3 < 0$ means that the function would be a downward parabola;

Downward parabola always have their vertex on top of the parabola and as such, the function has a maximum value.

The maximum value is:

$x = -\frac{b}{2a}$

Where:

$a= -3; b =12; c =-33$

So, we have:

$x = -\frac{12}{2 * -3}$

$x = -\frac{12}{-6}$

$x =2$

Substitute $x =2$ in $y =-3x^2 + 12x - 33$

$y = -3*2^2 + 12 * 2 - 33$

$y = -21$

Hence, the maximum is -21.

8 0

2 years ago

What is the sum of 4 3/10 and (-6.2)

erastovalidia [21]

Answer:

-1.9

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

A solution initially contains 200 bacteria. 1. Assuming the number y increases at a rate proportional to the number present, wri

GuDViN [60]

Answer:

1. $\frac{dy}{dt}=ky$

2.543.6

Step-by-step explanation:

We are given that

y(0)=200

Let y be the number of bacteria at any time

$\frac{dy}{dt}$ =Number of bacteria per unit time

$\frac{dy}{dt}\proportional y$

$\frac{dy}{dt}=ky$

Where k=Proportionality constant

2. $\frac{dy}{y}=kdt$ ,y'(0)=100

Integrating on both sides then, we get

$lny=kt+C$

We have y(0)=200

Substitute the values then , we get

$ln 200=k(0)+C$

$C=ln 200$

Substitute the value of C then we get

$ln y=kt+ln 200$

$ln y-ln200=kt$

$ln\frac{y}{200}=kt$

$\frac{y}{200}=e^{kt}$

$y=200e^{kt}$

Differentiate w.r.t

$y'=200ke^{kt}$

Substitute the given condition then, we get

$100=200ke^{0}=200 \;because \;e^0=1$

$k=\frac{100}{200}=\frac{1}{2}$

$y=200e^{\frac{t}{2}}$

Substitute t=2

Then, we get $y=200e^{\frac{2}{2}}=200e$

$y=200(2.718)=543.6=543.6$

e=2.718

Hence, the number of bacteria after 2 hours=543.6

4 0

3 years ago

Marcie can swim 20 laps in an hour. How many hours will it take her to swim 100 laps

Margarita [4]

Answer:

5 hours

Step-by-step explanation:

We can write a ratio

20 laps 100 laps

-------------- = --------------

1 hour x hours

Using cross products

20x = 100

Divide by 20

20x/20 = 100/20

x = 5 hours

4 0

2 years ago

Read 2 more answers