4 years ago

Solve the system y=x^2-1 y=x-1

Mathematics

1 answer:

Vikki [24]4 years ago

3 0

That is the answer
Have a Great day.

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Please please need help on this one

damaskus [11]

Answer:

20%

Step-by-step explanation:

890+400+650+420+590= 2950

590/2950= 20%

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3 years ago

on a scale drawing of a park, the sand box is 2 inches wide. The actual sand box is 8 feet wide. What is the scale of the drawin

cricket20 [7]

2 inches : 8 feet

simplified scale:

1 inch : 4 feet

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3 years ago

16 = d - 12 / 14 <br> what is the answer?

Aleonysh [2.5K]

Answer:

16=d-12/14

We simplify the equation to the form, which is simple to understand

16=d-12/14

Simplifying:

16=d-0.857142857143

We move all terms containing d to the left and all other terms to the right.

-1d=-0.857142857143-16

We simplify left and right side of the equation.

-1d=-16.8571428571

We divide both sides of the equation by -01 to get d.

d=16.8571428571

Step-by-step explanation:

3 0

3 years ago

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Find dx/dt when y=2 and dy/dt=1, given that x^4=8y^5-240<br><br> dx/dt=

katrin2010 [14]

Answer:

The value of $\frac{dx}{dt}$ is $\frac{160}{x^3}$ .

Step-by-step explanation:

The given equation is

$x^4=8y^5-240$

We need to find the value of $\frac{dx}{dt}$ .

Differentiate with respect to t.

$4x^3\frac{dx}{dt}=8(5y^4)\frac{dy}{dt}-0$ $[\because \frac{d}{dx}x^n=nx^{n-1},\frac{d}{dx}C=0]$

$4x^3\frac{dx}{dt}=40y^4\frac{dy}{dt}$

It is given that y=2 and dy/dt=1, substitute these values in the above equation.

$4x^3\frac{dx}{dt}=40(2)^4(1)$

$4x^3\frac{dx}{dt}=40(16)(1)$

$4x^3\frac{dx}{dt}=640$

Divide both sides by 4x³.

$\frac{dx}{dt}=\frac{640}{4x^3}$

$\frac{dx}{dt}=\frac{160}{x^3}$

Therefore the value of $\frac{dx}{dt}$ is $\frac{160}{x^3}$ .

4 0

3 years ago

Is it true that all multiples of 5 end with 5 or 0?

N76 [4]

Answer:

yes

Step-by-step explanation:

5 0

3 years ago

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