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2 years ago

Y = 1/2x + 4

Mathematics

1 answer:

Mademuasel [1]2 years ago

6 0

no solution.

The two lines are parallel

(same slope but different y intercepts)

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Daniel [21]

$\huge \mathfrak \blue{Solution \: - }$

The formula for area of triangle is -

$\frac{1}{2} \times base \times height$

________________________________

So, the area of first triangle =

$\frac{1}{2} \times base \: \times height$

$\frac{1}{2} \times 4 \times 3$

$= \pink{6 \: units}$

________________________________

Area of second triangle =

$\frac{1}{2 } \times base \times height$

$\frac{1}{2} \times 4 \times 6$

$= \pink {12 \: units}$

________________________________

Area of second triangle - Area of first triangle

$= 12 - 6$

$= 6$

$\sf \underline\purple{The \: area \: increases \: by \: 6 \: square \: units.}$

3 0

3 years ago

Read 2 more answers

Maria earns $48 every three hours she works. How much will she earn by working 13 hours?

Valentin [98]

Answer:

$624

Step-by-step explanation:

48 x 13

5 0

3 years ago

Read 2 more answers

In triangle ABC, AB=10, BC=8, and angle B=30. In triangle KLJ, LJ=20, JK=16, and angle L=30. State whether the triangles are sim

Margaret [11]

Answer:

They are not similar.

Step-by-step explanation:

In triangle ABC, the angle we are given is in between the two defined sides(AB and BC with angle B). If these triangles were similar, we would see this with the SAS postulate.

In triangle KLJ, the angle is not in between the two sides, so they are not similar according to any of these two postulates.

6 0

1 year ago

Describe a sequence of transformations that will carry ABC onto DEF.

Mice21 [21]

Answer:

The sequence of transformations that maps ∆ABC to ∆A′B′C′ is a reflection across the line y = x followed by a translation 10 units right and 4 units up. Thus, the correct answer is (1) y = x and (2) 10 units to the right and 4 units up.

Step-by-step explanation:

7 0

3 years ago

find the coordinates of the point P on the parabola y=1-x^2 with domain 0≤x≤1 that minimize the area of the triangle enclosed by

coldgirl [10]

Let point P be with coordinates $(x_0,y_0).$ Find the equation of the tangent line.

1. If $y=1-x^2,$ then $y'=-2x.$

2. The equation of the tangent line at point P is

$y-y_0=-2x_0(x-x_0).$

Find x-intercept and y-intercept of this line:

when x=0, then $y=y_0+2x_0^2;$
when y=0, then $x=\dfrac{y_0}{2x_0}+x_0=\dfrac{y_0+2x_0^2}{2x_0}.$

The area of the triangle enclosed by the tangent line at P, the x-axis, and y-axis is

$A=\dfrac{1}{2}\cdot (2x_0^2+y_0)\cdot \left(\dfrac{y_0+2x_0^2}{2x_0}\right)=\dfrac{(y_0+2x_0^2)^2}{4x_0}.$

Since point P is on the parabola, then $y_0=1-x_0^2$ and

$A=\dfrac{(1-x_0^2+2x_0^2)^2}{4x_0}=\dfrac{(1+x_0^2)^2}{4x_0}.$

Find the derivative A':

$A'=\dfrac{2(1+x_0^2)\cdot 2x_0\cdot 4x_0-4(1+x_0^2)^2}{16x_0^2}=\dfrac{12x_0^4+8x_0^2-4}{16x_0^2}.$

Equate this derivative to 0, then

$12x_0^4+8x_0^2-4=0,\\ \\3x_0^4+2x_0^2-1=0,\\ \\D=2^2-4\cdot 3\cdot (-1)=16,\ \sqrt{D}=4,\\ \\x_0^2_{1,2}=\dfrac{-2\pm4}{6}=-1,\dfrac{1}{3},\\ \\x_0^2=\dfrac{1}{3}\Rightarrow x_0_{1,2}=\pm\dfrac{1}{\sqrt{3}}.$

And

$y_0=1-\left(\pm\dfrac{1}{\sqrt{3}}\right)^2=\dfrac{2}{3}.$

Answer: two points: $P_1\left(-\dfrac{1}{\sqrt{3}},\dfrac{2}{3}\right), P_2\left(\dfrac{1}{\sqrt{3}},\dfrac{2}{3}\right).$

6 0

3 years ago