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2 years ago

Solve for x: -2(x + 3) = -4x + 20

Mathematics

1 answer:

Phantasy [73]2 years ago

4 0

Answer:

Step-by-step explanation:

Lets start with 13

-2(13 + 3) = -4(13) + 20

-32 = -52 + 20

-32 = -32

x = 1 basically

now 19

-2(19 + 3) = -4(19) +20

-44 = -76 + 20

-44 = -56

so divide those

-2(31 + 3) = -4(31) + 20

-68 = -124 + 20

-68 = -104

divide them

-2(37 + 3) = -4(37) + 20

-80 = -148 + 20

-80 = -128

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allochka39001 [22]

Answer:

1) Yes 2)No

Step-by-step explanation:

1) Yes because the two shorter lines combined are longer than the base and so those two lines are able to make a triangle.

2) No because the two lines on top will be equal to base line and so the two shorter lines will not be able to make a third angle because it will be level with the base.

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3 years ago

The amount Cami raised during last year’s charity walk, $45.50, is StartFraction 7 over 10 EndFraction of the amount she raised

barxatty [35]

Answer:

The equation is given by:

$n = \frac{7}{10} \times 45.50$

Step-by-step explanation:

During last years's charity walk, she raised $45.50.

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$n = \frac{7}{10} \times 45.50$

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2 years ago

Monique has 2 hours to complete 3 homework assignments.She wants to spend the same amount of time on each assignment.How meany m

slamgirl [31]

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2 years ago

A large tank is filled to capacity with 300 gallons of pure water. brine containing 5 pounds of salt per gallon is pumped into t

Licemer1 [7]

If $A(t)$ is the amount of salt in the tank at time $t$ , then the rate at which this amount changes over time is given by the ODE

$A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{5\text{ lb}}{1\text{ gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$

$A'(t)+\dfrac1{100}A(t)=15$

We're told that the tank initially starts with no salt in the water, so $A(0)=0$ .

Multiply both sides by an integrating factor, $e^{t/100}$ :

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=15e^{t/100}$
$\left(e^{t/100}A(t)\right)'=15e^{t/100}$
$e^{t/100}A(t)=1500e^{t/100}+C$
$A(t)=1500+Ce^{-t/100}$

Since $A(0)=0$ , we have

$0=1500+C\implies C=-1500$

so that the amount of salt in the tank over time is given by

$A(t)=1500(1-e^{-t/100})$

After 10 minutes, the amount of salt in the tank is

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3 years ago

Which value completes the equation?

Ivahew [28]

Step-by-step explanation:

56,000÷7=8000

56,000÷70=800

56,000÷700=80

56,000÷7000=8

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2 years ago