TIMED URGENT HELP PLS

Solve the following system of equations:

Vladimir79 [104]

Answer:

(-1,1)

Step-by-step explanation:

We can solve this by either graphing and finding ther point the lines intersect, or mathematically, I'll do both.

Graphing:

Mathematically:

−2x + 4y = 6

y = 2x + 3

See the attached graph. The lines intersect at (-1,1)

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I'll rearrange the first equation (to make it easier for me):

−2x + 4y = 6

4y = 2x + 6

y = (1/2)x + 1.5

Now lets substitute the second equation into the first so that we can eliminate y:

y = 2x + 3

[(1/2)x + 1.5] = 2x + 3

- (3/2)x = (3/2)

x = -1

If x = -1:

y = 2(-1) + 3

y = 1

The solution is x = -1 and y = 1, or (-1,1)

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Both approaches give us (-1,1), the solution to the system of equations. It is the only point that satisfies both equations.

3 0

1 year ago

HELP ASAPP I WILL GIVE BRAINLIEST

JulijaS [17]

6.667 yards I think I got it of safari

8 0

2 years ago

Read 2 more answers

Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the corre

Scrat [10]

Answer:

4√(xy³)

Step-by-step explanation:

8√(x²y⁶)

The above expression can be simplified as follow:

8√(x²y⁶)

Recall:

m√a = a^1/m

Therefore,

8√(x²y⁶) = (x²y⁶)^1/8

Recall:

(aⁿ)^1/m = a^(n/m)

Therefore,

(x²y⁶)^1/8 = x^(2/8)•y^(6/8)

= x^1/4•y^3/4

= (xy³)^1/4

Recall :

a^1/m = m√a

Therefore,

(xy³)^1/4 = 4√(xy³)

Therefore,

8√(x²y⁶) = 4√(xy³)

6 0

2 years ago

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If N is the population of the colony and t is the time in days, express N as a function of t. Consider Upper N 0 is the origina

lesya [120]

Answer:

(a) $N(t)=Noe^{kt}$

(b)5,832 Mosquitoes

(c)5 days

Step-by-step explanation:

(a)Given an original amount $N_o$ at t=0. The population of the colony with a growth rate $k \neq 0$ , where k is a constant is given as:

$N(t)=Noe^{kt}$

(b)If $N_o=1000$ and the population after 1 day, N(1)=1800

Then, from our model:

N(1)=1800

$1800=1000e^{k}\\$Divide both sides by 1000\\e^{k}=1.8\\$Take the natural logarithm of both sides\\k=ln(1.8)$

Therefore, our model is:

$N(t)=1000e^{t*ln(1.8)}\\N(t)=1000\cdot1.8^t$

In 3 days time

$N(3)=1000\cdot1.8^3=5832$

The population of mosquitoes in 3 days time will be approximately 5832.

(c)If the population N(t)=20,000,we want to determine how many days it takes to attain that value.

From our model

$N(t)=1000\cdot1.8^t\\20000=1000\cdot1.8^t\\$Divide both sides by 1000\\20=1.8^t\\$Convert to logarithm form\\Log_{1.8}20=t\\\frac{Log 20}{Log 1.8}=t\\ t=5.097\approx 5\; days$

In approximately 5 days, the population of mosquitoes will be 20,000.

7 0

2 years ago

Read 2 more answers

Can you guys help me pls

Rama09 [41]

Answer

A

Step-by-step explanation:

So in the diagrams we see there is 2/3 shaded into a 12 cutted square. So we multiply 4x3 to get 12. So now we have to turn the 3 into 12.

so

$\frac{2}3}$ x $\frac{4}{4}$ = $\frac{8}{12}$

Since we multiplied 4 on the bottom we have to multiply 4 on the top.

sorry im not good at explaining but if you need me to re explain ill work on it. :)

5 0

2 years ago