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Pani-rosa [81]
3 years ago
14

Solve for X and represent solution graphically 5x+2<x+7​

Mathematics
1 answer:
mel-nik [20]3 years ago
7 0

Answer:

x< 5/4

Step-by-step explanation:

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There are 42 female performers in a dance recital. The ratio of men to women is 2:3. How many men are in the dance recital
slava [35]

Answer:

28

Step-by-step explanation: This should be 28. 42 divided by 3 is 14. Why would you divide it by 3? Because the full ratio is 2:3. So 42 divided by 3 is 14. Then you add fourteen twice, for the two in the ratio. Fourteen plus fourteen is 28.

6 0
3 years ago
Just seeing how this works ​
SSSSS [86.1K]

Answer:

I do not know how to do this

4 0
3 years ago
A field is to be fertilized at a cost of $0.12 per square yard. The rectangular part of the field is 110 yards long and the diam
Lera25 [3.4K]

see the picture in the attached figure


step 1

find the area of rectangle

A=110*43

A=4730 yd²


step 2

find the area of semicircles

we know that two semicircles is equal to one circle

so

Area of the circle=pi*r²

r=43/2

r=21.5 yd

A=pi*21.5²

A=462.25 *pi yd²-------->1,451.465 yd²


total area=area rectangle + area of circle

total area=4,730+1,451.465---------> 6,181.465 yd²


step 3

find the cost

$0.12 per square yard

6,181.465*0.12=$741.78


therefore


the answer is

$741.78

4 0
3 years ago
The tables show the cost (in dollars) of colored pencils at two stores. Brendan says that the school store offers a better deal
LenaWriter [7]

Answer:

B

Step-by-step explanation:

I got it right on the test

8 0
2 years ago
A certain country has $10 billion in paper currency in circulation, and each day $50 million comes into the country's banks. The
mash [69]

Answer:

\bf x(t)=10(1-e^{-0.005*t})

Step-by-step explanation:

The differential equation

\bf \displaystyle\frac{dx}{dt}=0.005(10-x)

can be solved by separation of variables. Write the equation as

\bf \displaystyle\frac{dx}{10-x}=0.005dt

Integrate on both sides

\bf \int\displaystyle\frac{dx}{10-x}=\int0.005dt\Rightarrow -ln(10-x)=0.005t+C\Rightarrow\\\\\Rightarrow ln(10-x)^{-1}=0.005t+C\Rightarrow (10-x)^{-1}=e^{0.005t}e^C

where C is a constant.

\bf e^C is also a constant and we will keep calling it C, (there is no reason to change the letter). We have then

\bf (10-x)^{-1}=Ce^{0.005t}\Rightarrow \displaystyle\frac{1}{10-x}=Ce^{0.005t}\Rightarrow 10-x=\displaystyle\frac{1}{Ce^{0.005t}}\Rightarrow\\\\\Rightarrow x(t)=10-(1/C)e{-0.005t}

(1/C) is a constant, and for the same reason we will keep calling it C. So the general solution is

\bf x(t)=10-Ce^{-0.005t}

Now, we use the initial condition x(0)=0

\bf x(0)=10-Ce^{-0.005*0}=0\Rightarrow C=10

and the particular solution is

\bf x(t)=10-10e^{-0.005*t}=10(1-e^{-0.005*t})\\\\\boxed{x(t)=10(1-e^{-0.005*t})}

7 0
3 years ago
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