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Daniel [21]
3 years ago
14

What is the slope of the line equation 7x-3y=21

Mathematics
1 answer:
kramer3 years ago
8 0
Equation: 7x - 3y = 21

3y = 7x - 21

Divide both sides by 3, 

y = 7/3x - 7

Compare it with principle equation, y = mx + c

Here, slope (m) = 7/3

In short, Your Answer would be: 7/3

Hope this helps!
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If ABC~DEF what is the length of BC?<br><br> If ABC~DEF what is the length of EF?
marysya [2.9K]

BC would equal the same as EF, they are similar. All you would have to do is figure out the pre image (BC) of EF.

3 0
3 years ago
Please help me for 30 points
myrzilka [38]

Answer: 1D, 2E, 3B, 4A, 5C

Step-by-step explanation:

First, eliminate choices that cannot go together.

The "greater than" and "less than" symbols, represented by > and <, are used to show that a number is greater than another set of numbers, but not equal to the set of numbers.

The "greater than or equal to" and "less than or equal to" symbols, represented by ≥ and ≤, are used to show that a number can also be equal to a set of numbers.

On the number line, open dots are used to represent > and <.

Closed dots are used to represent  ≥ and ≤.

Using this, C, D, and E must match with 1, 2, and 5.

Dealing with 3 now:

4m ≥ 32

Divide both sides by 4 to simplify m:

m ≥ 8 - This is used to show that m is greater than or equal to 8.

This lines up with choice B: the line starts at 8 and goes onwards to positive infinity.

This means that 4 matches to A, because it is the only other option with a closed dot.

Looking at 1 (x - 5 < 2), we can simplify it very easily by adding 5 to both sides of the inequality:

x < 7 - This is used to show that x is less than 7.

This lines up with choice D: The line starts at less than 7 and goes downwards to negative infinity.

Looking at 2 (m + 24 > 20), we can do this the same way. Subtract 24 from both sides to get:

m > -4 - This is used to show that m is greater than negative 4.

This lines up with choice E: The line starts at greater than 4 and goes to positive infinity.

The last choice, choice 5, must match with choice C.

6 0
3 years ago
The scale on a map says that 6 cm represents 15 km. What distance on the map in centimeters represents 20 kilometers?
ale4655 [162]

Answer:

8 centimeters.

Step-by-step explanation:

6cm\ on\ map\ -> 15km\\

Divide 3 both sides:

2cm\ on\ map\ -> 5km

Multiply 4 on both sides:

8cm\ on\ map\ -> 20km

8cm on the map represents 20km.

4 0
3 years ago
Find the distance from the origin to the graph of 7x+9y+11=0
Cerrena [4.2K]
One way to do it is with calculus. The distance between any point (x,y)=\left(x,-\dfrac{7x+11}9\right) on the line to the origin is given by

d(x)=\sqrt{x^2+\left(-\dfrac{7x+11}9\right)^2}=\dfrac{\sqrt{130x^2+154x+121}}9

Now, both d(x) and d(x)^2 attain their respective extrema at the same critical points, so we can work with the latter and apply the derivative test to that.

d(x)^2=\dfrac{130x^2+154x+121}{81}\implies\dfrac{\mathrm dd(x)^2}{\mathrm dx}=\dfrac{260}{81}x+\dfrac{154}{81}

Solving for (d(x)^2)'=0, you find a critical point of x=-\dfrac{77}{130}.

Next, check the concavity of the squared distance to verify that a minimum occurs at this value. If the second derivative is positive, then the critical point is the site of a minimum.

You have

\dfrac{\mathrm d^2d(x)^2}{\mathrm dx^2}=\dfrac{260}{81}>0

so indeed, a minimum occurs at x=-\dfrac{77}{130}.

The minimum distance is then

d\left(-\dfrac{77}{130}\right)=\dfrac{11}{\sqrt{130}}
4 0
3 years ago
Consider a rabbit population​ P(t) satisfying the logistic equation StartFraction dP Over dt EndFraction equals aP minus bP squa
maria [59]

Solution:

Given :

$\frac{dP}{dt}= aP-bP^2$         .............(1)

where, B = aP = birth rate

            D = $bP^2$  =  death rate

Now initial population at t = 0, we have

$P_0$ = 220 ,  $B_0$ = 9 ,  $D_0$ = 15

Now equation (1) can be written as :

$ \frac{dP}{dt}=P(a-bP)$

$\frac{dP}{dt}=bP(\frac{a}{b}-P)$    .................(2)

Now this equation is similar to the logistic differential equation which is ,

$\frac{dP}{dt}=kP(M-P)$

where M = limiting population / carrying capacity

This gives us M = a/b

Now we can find the value of a and b at t=0 and substitute for M

$a_0=\frac{B_0}{P_0}$    and     $b_0=\frac{D_0}{P_0^2}$

So, $M=\frac{B_0P_0}{D_0}$

          = $\frac{9 \times 220}{15}$

          = 132

Now from equation (2), we get the constants

k = b = $\frac{D_0}{P_0^2} = \frac{15}{220^2}$

        = $\frac{3}{9680}$

The population P(t) from logistic equation is calculated by :

$P(t)= \frac{MP_0}{P_0+(M-P_0)e^{-kMt}}$

$P(t)= \frac{132 \times 220}{220+(132-220)e^{-\frac{3}{9680} \times132t}}$

$P(t)= \frac{29040}{220-88e^{-\frac{396}{9680} t}}$

As per question, P(t) = 110% of M

$\frac{110}{100} \times 132= \frac{29040}{220-88e^{\frac{-396}{9680} t}}$

$ 220-88e^{\frac{-99}{2420} t}=200$

$ e^{\frac{-99}{2420} t}=\frac{5}{22}$

Now taking natural logs on both the sides we get

t = 36.216

Number of months = 36.216

8 0
3 years ago
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