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3 years ago

Y=4x-1 y=2x-5 solve by substitution

Mathematics

1 answer:

Alenkasestr [34]3 years ago

8 0

Answer:

Step-by-step explanation:

4x-1= 2x-5

4x-2x= 1-5

2x=-4

x= -2

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What are the solution(s) of x2-4-0?

Elanso [62]

Answer:

For $x^2 - 4 = 0$ , x = 2, or x = - 2.

Step-by-step explanation:

Here, the given expression is :

$x^2 - 4 = 0$

Now, using the ALGEBRAIC IDENTITY:

$a^2 - b^2 = (a-b)(a+b)$

Comparing this with the above expression, we get

$x^2 - 4 = 0 = x^2 - (2)^2 = 0\\\implies (x-2)(x+2) = 0$

⇒Either (x-2) = 0 , or ( x + 2) = 0

So, if ( x- 2) = 0 ⇒ x = 2

and if ( x + 2) = 0 ⇒ x = -2

Hence, for $x^2 - 4 = 0$ , x = 2, or x = - 2.

4 0

4 years ago

Read 2 more answers

A viewfinder has a triangular lens. Some of the measurement of the length are shown below. Which of the following best represent

S_A_V [24]

Answer:

Step-by-step explanation:

First, we need to find the unknown angle of the triangle.

40 + 20 = 60

180 - 60 = 120

So the unknown angle is 120.

Then, we can use the law of sines to create a proportion.

sin(120)/8 = sin(40)/a

Then, we can simplify this.

a is about 5.9, or answer choice A.

5 0

3 years ago

Solve x2 = 12x – 15 by completing the square. Which is the solution set of the equation?

REY [17]

X ·2 = 12x - 15

2x = 12x - 15

2x + (-12x) = (12x - 15) + (-12x)

Result: x = 3/2

8 0

4 years ago

Read 2 more answers

Use implicit differentiation to find the points where the parabola defined by x2−2xy+y2+4x−8y+20=0 has horizontal and vertical t

Komok [63]

Answer:

The parabola has a horizontal tangent line at the point (2,4)

The parabola has a vertical tangent line at the point (1,5)

Step-by-step explanation:

Ir order to perform the implicit differentiation, you have to differentiate with respect to x. Then, you have to use the conditions for horizontal and vertical tangent lines.

-To obtain horizontal tangent lines, the condition is:

$\frac{dy}{dx}=0$ (The slope is zero)

--To obtain vertical tangent lines, the condition is:

$\frac{dy}{dx}=\frac{1}{0}$ (The slope is undefined, therefore the denominator is set to zero)

Derivating respect to x:

$\frac{d(x^{2}-2xy+y^{2}+4x-8y+20)}{dx} = \frac{d(x^{2})}{dx}-2\frac{d(xy)}{dx}+\frac{d(y^{2})}{dx}+4\frac{dx}{dx}-8\frac{dy}{dx}+\frac{d(20)}{dx}$ = $2x -2(y+x\frac{dy}{dx})+2y\frac{dy}{dx}+4-8\frac{dy}{dx}= 0$

Solving for dy/dx:

$\frac{dy}{dx}(-2x+2y-8)=-2x+2y-4\\\frac{dy}{dx}=\frac{2y-2x-4}{2y-2x-8}$

Applying the first conditon (slope is zero)

$\frac{2y-2x-4}{2y-2x-8}=0\\2y-2x-4=0$

Solving for y (Adding 2x+4, dividing by 2)

y=x+2 (I)

Replacing (I) in the given equation:

$x^{2}-2x(x+2)+(x+2)^{2}+4x-8(x+2)+20=0\\x^{2}-2x^{2}-4x+x^{2} +4x+4+4x-8x-16+20=0\\-4x+8=0\\x=2$

Replacing it in (I)

y=(2)+2

y=4

Therefore, the parabola has a horizontal tangent line at the point (2,4)

Applying the second condition (slope is undefined where denominator is zero)

2y-2x-8=0

Adding 2x+8 both sides and dividing by 2:

y=x+4(II)

Replacing (II) in the given equation:

$x^{2}-2x(x+4)+(x+4)^{2}+4x-8(x+4)+20=0\\x^{2}-2x^{2}-8x+x^{2}+8x+16+4x-8x-32+20=0\\-4x+4=0\\x=1$

Replacing it in (II)

y=1+4

y=5

The parabola has vertical tangent lines at the point (1,5)

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4 years ago

How do you answer the equation -4r+11=4(1-r)+7?

kow [346]

Answer:

Any value of

makes the equation true.

All real numbers

Interval Notation:

(-Any value of

makes the equation true.

All real numbers

Interval Notation:

( ∞ ,∞ )

Step-by-step explanation:

3 0

3 years ago