|60| __ |60| a.= b.< c.>

Which statements about the system are true?

Kruka [31]

2 Answers:

B) The lines are parallel
C) The lines have the same slope.

Parallel lines always have equal slope, but different y intercepts.

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Explanation:

Let's solve the second equation for y

3y - x = -7

3y = -7+x

3y = x-7

y = (x-7)/3

y = x/3 - 7/3

y = (1/3)x - 7/3

The equation is in y = mx+b form with m = 1/3 as the slope and b = -7/3 as the y intercept. We see that the first equation, where y was already isolated, also has a slope of m = 1/3. The two equations of this system have the same slope. Choice C is one of the answers.

However, they don't have the same y intercept. The first equation has y intercept b = -4, while the second has b = -7/3. This means that they do not represent the same line. They need to have identical slopes, and identical y intercepts (though the slope can be different from the y intercept of course) in order to have identical lines. So we can rule out choice D and E because of this.

Because the two equations have the same slope, but different y intercepts, this means the lines are parallel. Choice B is the other answer.

Parallel lines never touch or intersect, which in turn means there is no solution point. A solution point is where the lines cross. We can rule out choice A.

I recommend using your graphing calculator, Desmos, GeoGebra, or any graphing tool (on your computer or online) to graph each equation given. You should see two parallel lines forming. I used GeoGebra to make the graph shown below.

5 0

3 years ago

You roll a fair six sided die twice. The first roll shows a six and the second row shows 3

vazorg [7]

Answer:

4/12 simplified to 1/3 of rolling 2 times.

Step-by-step explanation:

3 0

2 years ago

If the area of a rectangle 16s2t and the length is 8st2, what would the width of the rectangle, given that width is found by div

Dafna1 [17]

Area = Length x Width
Substituting the given values:
16S^2 t = 8St^2 x Width
Manipulating for Width:
Width = 16S^2 t / 8St^2
Rewriting the squared terms in simplified form :
Width = 16 x S x S x t / 8 x S x t x t
Cancelling the like terms
Width = 16 x S / 8 x t
Cancelling the numeric terms :

Width = 2S / t

6 0

3 years ago

A metal beam was brought from the outside cold into a machine shop where the temperature was held at 65degreesF. After 5 min, t

ivolga24 [154]

Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

$T(t)=T_a+(T_0-T_a)e^{-kt}$

where $T_a$ is the ambient temperature, $T_0$ is the initial temperature, $t$ is the time and $k$ is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say $T_0$

We know that the ambient temperature is $T_a=65$ , so

$T(t)=65+(T_0-65)e^{-kt}$

We also know that when $t=5 \:min$ the temperature is $T(5)=35$ and when $t=10 \:min$ the temperature is $T(10)=50$ which gives:

$T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}$

$T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}$

Rearranging,

$35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}$

$50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}$

If we divide these two equations we get

$\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}$

$\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}$

Now, that we know the value of $k$ we can use it to find the initial temperature of the beam,

$35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5$

so the beam started out at 5 °F.

6 0

3 years ago

For what interval is the value of (f – g)(x) negative?

mrs_skeptik [129]

Complete question:

Refer the attached diagram

Answer:

In reference to the attached figure, (-∞, 2) is the value where (f-g) (x) negative.

Step-by-step explanation:

From the attached figure, it shows that given data:

f (x) = x – 3

g (x) = - 0.5 x

To Find: At what interval the value of (f-g) (x) negative

So, first we need to calculate the (f-g) (x)

(f – g ) (x) = f (x) – g (x) = x-3 - (- 0.5 x)

⇒ (f - g) (x) =1.5 x - 3

Now we are supposed to find the interval for which (f-g) (x) is negative.

⇒ (f - g) (x) = x - 3+ 0.5 x = 1.5 x – 3 < 0

⇒ 1.5 x – 3 < 0

⇒ 1.5 x < 3

⇒ $x$

⇒ x < 2

Thus for (f - g) (x) negative x must be less than 2. Thereby, the interval is (-∞, 2). Function is negative when graph line lies below the x - axis.

7 0

2 years ago