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Ostrovityanka [42]

2 years ago

10

What is the equation of the line shown in this graph?

1 answer:

navik [9.2K]2 years ago

7 0

Answer:

x=-2

Step-by-step explanation:

Since it's a vertical line the answer is going to be x= which ever number the line is on. In this case x=-2

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Please answer due today (one question) Algebra and WILL GIVE BRAINLEST!!!

chubhunter [2.5K]

180(6 - 2) = 720

4x + 180 = 720

4x = 630

x = 157.5

S = 720

X = 157.5

5 0

3 years ago

Express the statement "A rectangular field with an area of 120m' has it length 12mn longer than the width" as quadratic equation

navik [9.2K]

Answer:

The quadratic equation is x^2 -12x -120 = 0

where x represents the length of the field

Step-by-step explanation:

We want to make an expression in terms of a quadratic equation;

Let the length of the field be x mm , then the width will be (x-12) mm

Mathematically , the area of the field will be L * B

Thus;

x * (x-12) = 120

That gives;

x^2 -12x = 120

x^2 -12x -120 = 0

6 0

3 years ago

Is 6,000 g bigger than 6kg

Helen [10]

They are equal in mass.

<u>1 kilogram is 1,000 grams.</u>

Meaning, 6,000 grams is 6 kilograms.

5 0

3 years ago

Read 2 more answers

Find the point (,) on the curve =8 that is closest to the point (3,0). [To do this, first find the distance function between (,)

ELEN [110]

Question:

Find the point (,) on the curve $y = \sqrt x$ that is closest to the point (3,0).

[To do this, first find the distance function between (,) and (3,0) and minimize it.]

Answer:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

Step-by-step explanation:

$y = \sqrt x$ can be represented as: $(x,y)$

Substitute $\sqrt x$ for $y$

$(x,y) = (x,\sqrt x)$

So, next:

Calculate the distance between $(x,\sqrt x)$ and $(3,0)$

Distance is calculated as:

$d = \sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2}$

So:

$d = \sqrt{(x-3)^2 + (\sqrt x - 0)^2}$

$d = \sqrt{(x-3)^2 + (\sqrt x)^2}$

Evaluate all exponents

$d = \sqrt{x^2 - 6x +9 + x}$

Rewrite as:

$d = \sqrt{x^2 + x- 6x +9 }$

$d = \sqrt{x^2 - 5x +9 }$

Differentiate using chain rule:

Let

$u = x^2 - 5x +9$

$\frac{du}{dx} = 2x - 5$

So:

$d = \sqrt u$

$d = u^\frac{1}{2}$

$\frac{dd}{du} = \frac{1}{2}u^{-\frac{1}{2}}$

Chain Rule:

$d' = \frac{du}{dx} * \frac{dd}{du}$

$d' = (2x-5) * \frac{1}{2}u^{-\frac{1}{2}}$

$d' = (2x - 5) * \frac{1}{2u^{\frac{1}{2}}}$

$d' = \frac{2x - 5}{2\sqrt u}$

Substitute: $u = x^2 - 5x +9$

$d' = \frac{2x - 5}{2\sqrt{x^2 - 5x + 9}}$

Next, is to minimize (by equating d' to 0)

$\frac{2x - 5}{2\sqrt{x^2 - 5x + 9}} = 0$

Cross Multiply

$2x - 5 = 0$

Solve for x

$2x =5$

$x = \frac{5}{2}$

Substitute $x = \frac{5}{2}$ in $y = \sqrt x$

$y = \sqrt{\frac{5}{2}}$

Split

$y = \frac{\sqrt 5}{\sqrt 2}$

Rationalize

$y = \frac{\sqrt 5}{\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$y = \frac{\sqrt {10}}{\sqrt 4}$

$y = \frac{\sqrt {10}}{2}$

Hence:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

3 0

3 years ago

Let $G$ denote the centroid of triangle $ABC$. If triangle $ABG$ is equilateral with side length 2, then determine the perimeter

belka [17]

Hey there!

You are trying to find the perimeter of triangle ABC. Each side is 2. Since there are 3 sides on a triangle, multiply 2 by 3 to get 6. You do not need the centroid for this problem.

I hope this helps!

5 0

3 years ago