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tester [92]
2 years ago
7

Does the give value make the inequality true?

Mathematics
2 answers:
Rudiy272 years ago
6 0

Answer:

It's true.

This is correct I know for a fact.

Advocard [28]2 years ago
4 0

Answer: false

Step-by-step explanation:

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6(X - 2) = -48<br><br> Find x
Darya [45]

Answer:

6x-12=-48

6x=-36

x=-6

Step-by-step explanation:

3 0
3 years ago
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mart [117]

Answer:

G. 17.3

solution:

5 0
3 years ago
f(x)=(2x−1)(3x+5)(x+1) has zeros at x= -5/3, x= -1 and x= 1/2 What is the sign of f on the interval -5/3
Crazy boy [7]

The sign of f on the interval -5/3 is negative

<h3>How to determine the sign?</h3>

The function is given as:

f(x)=(2x−1)(3x+5)(x+1)

The zeros of the function are

x= -5/3, x= -1 and x= 1/2

Next, we plot the graph of the function

At x = -5/3, the function approaches negative infinity

This means that the sign of f on the interval -5/3 is negative

Read more about function intervals at:

brainly.com/question/27831985

#SPJ1

6 0
2 years ago
Will give 50 points!!! ASAP!!!!!!!!!!!
Nostrana [21]

Answer:

\frac{1}{4}

Step-by-step explanation:

Since there are a total of 4 colors, and there are 5 of each colors, the total marbles in the bad would be;

5 x 4 = 20

If a color is selected at random the probability would be

\frac{part}{whole}

our part is 5

our whole is 20

so the probability would be;

\frac{5}{20} or \frac{1}{4} when reduced

Hope this helps!

5 0
2 years ago
Read 2 more answers
The Pacific halibut fishery has been modeled by the differential equation dy dt = ky 1 − y K where y(t) is the biomass (the tota
Dafna1 [17]

Answer:

a. The biomass weighs 2.30 * 10^7 kg after a year

b. It'll take 2.56 years to get to 4*10^7kg

Step-by-step explanation:

a.

k = 0.78,K = 6E7 kg

Given

dy/dt = ky(1- y/K)

Make ky dt the subject of formula

ky dt = dy/(1-y/K) --- make k dt the subject of formula

k dt = dy/(y(1-y/K))

k dt = K dy / y(K-y)

k dt = ((1/y) + (1/(K-y)))dy ---- integrate both sides

kt + c = ln(y/(K-y))

Ce^(kt) = y/(K-y)

Substitute the values of k and K

Ce^(0.78t) = y/(6*10^7 - y) ----- (1)

Given that y(0) = 2 * 10^7kg

(1) becomes

Ce^(0.78*0) = (2 * 10^7)/(6*10^7 - 2*10^7)

Ce° = (2*10^7)/(4*10^7

C = 2/7

Substitute 2/7 for C in (1)

2/7e^0.78t = y/(6*10^7 - y) ---(2)

We're to find the biomass a year later

So, t = 1

2/7e^0.78 = y/(6*10^7 - y)

0.62 = y/(6*10^7 - y)

y = 0.62(6*10^7 - y)

y = 0.62*6*10^7 - 0.62y

y + 0.62y = 0.62*6*10^7

1.62y = 0.62*6*10^7

1.62y = 3.72 * 10^7

y = 2.30 * 10^7kg.

Hence, the biomass weighs 2.30 * 10^7 kg after a year

b.

Here, we're to calculate the time it'll take the biomass to get to 4*10^7 kg

Substitute 4*10^7 for y in (2)

2/7e^0.78t = 4*10^7/(6*10^7 - 4*10^7)

2/7e^0.78t = 4*10^7/2*10^7

2/7e^0.78t = 2

e^0.78t = (2*7)/2

e^0.78t = 2

t = 2 * 1/0.78

t = 2.56 years

Hence, it'll take 2.56 years to get to 4*10^7kg

8 0
3 years ago
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