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3 years ago

X^2 + y^2=40 and y=x-4 solve this simultaneous equation

1 answer:

3 0

Write x-4 instead of y.
x²+(x-4)²=40
x²+x²-8x+16=40
2x²-8x-24=0
x²-4x-12=0
(x-6)(x+2)=0
x=6 or x=-2

for x=6, y=6-4 = 2
for x=-2, y=-2-4 = -6

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Factor x^3 -7x^2 -5x+35 by grouping. what is the resulting expression

Anni [7]

Answer:

(x-7) (x^2-5)

Step-by-step explanation:

x^3 -7x^2 -5x+35

Make 2 groups

x^3 -7x^2 -5x+35

Factor x^2 from the first group and -5 from the second group

x^2 (x-7) -5(x-7)

Now factor (x-7) out

(x-7) (x^2-5)

8 0

3 years ago

Which of the following statements accurately describe the dilation shown? Select all that apply.

Anon25 [30]

Answer:

A, D, and F I think. Hope this helps

8 0

2 years ago

Find the difference 1-(5)

bazaltina [42]

Answer:

-4

Step-by-step explanation:

Subtract 1 - 5 = -4...

Since 1 is less than 5, we know that the difference is negative, so we can rewrite this expression as -(5-1).

5 - 1 = 4, so - (5-1) = -4.

6 0

3 years ago

Read 2 more answers

A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per

Tems11 [23]

Answer:

The pressure is changing at $\frac{dP}{dt}=3.68$

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ and we want to find at what rate is the pressure changing.

The equation that model this situation is

$PV^{1.4}=k$

Differentiate both sides with respect to time t.

$\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\$

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

$\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)$

Apply this rule to our expression we get

$V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0$

Solve for $\frac{dP}{dt}$

$V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}$

when P = 23 kg/cm2, V = 35 cm3, and $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ this becomes

$\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68$

The pressure is changing at $\frac{dP}{dt}=3.68$ .

7 0

3 years ago

Which term best completes the state all _______ are rectangles A. Parallelograms B.squares C.rhombi D. None of theses

liq [111]

Answer:

The answer is B. squares.

Step-by-step explanation:

All squares are considered rectangles. Not all parallelograms are rectangles, and rhombi are not rectangles.

I hope this helped! :-)

8 0

3 years ago