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3 years ago

5

Determine the solution to the system of equations

1 answer:

gizmo_the_mogwai [7]3 years ago

3 0

Answer:

no sé pero no se y no se y más no se

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What are the coordinates of the image of vertex G after a reflection across the line y = x?

Yuliya22 [10]

Answer:

The coordinates of the image of vertex G are (-5, 4)

Step-by-step explanation:

Let us revise some cases of reflection

If the point (x, y) reflected across the x-axis , then its image is (x, -y)

If the point (x, y) reflected across the y-axis , then its image is (-x, y)

If the point (x, y) reflected across the line y = x , then its image is (y, x)

If the point (x, y) reflected across the line y = -x , then its image is (-y, -x)

From the given figure

∵ The line of the reflection is y = x

→ That means we will switch the coordinates of the point to find its image

∵ The coordinates of vertex G are (4, -5)

∴ The x-coordinate = 4 and the y-coordinate = -5

→ Switch the two coordinates

∴ The coordinates of its image G' are (-5, 4)

∴ The coordinates of the image of vertex G are (-5, 4)

7 0

2 years ago

A botanist predicts that the height of a certain tree will increase by 2% every year. If the height of the tree is now 50 feet,

Neporo4naja [7]

To answer this kinda question, we gotta tackle it year by year.
By the end of the first year, the tree would grow by 2% = 102% of its original height.
Year 1: 102% x 50 = 51 feet
Year 2: 103% x 51 = 52.02 feet

Hope I helped!! xx

4 0

3 years ago

what is the common difference between succeseive terms in this sequence? 9 , 2.5 , -4 , -10.5 , -17 ....

OlgaM077 [116]

The difference is 6.5

3 0

3 years ago

56. How many tangent lines to the curve <img src="https://tex.z-dn.net/?f=y%3Dx%20%2F%28x%2B1%29" id="TexFormula1" title="y=x /(

PIT_PIT [208]

There are 2 tangent lines that pass through the point

$y=\frac{1}{(-1+\sqrt{3)^2} } (x-1)+2$

and

$y=\frac{1}{(-1-\sqrt{3)^2} } (x-1)+2$

Explanation:

Given:

$y=\frac{x}{x+1}$

The point-slope form of the equation of a line tells us that the form of the tangent lines must be:

$y=m(x-1)+2$ $[1]$

For the lines to be tangent to the curve, we must substitute the first derivative of the curve for $m$ :

$\frac{dy}{dx} =\frac{d(x)}{dx}(x+1)-x^\frac{d(x+1)}{dx} \\ \\$

$\frac{dy}{dx} =\frac{x+1-x}{(x+1)^2}$

$\frac{dy}{dx}= \frac{1}{(x+1)^2}$

$m=\frac{1}{(x+1)^2}$ $[2]$

Substitute equation [2] into equation [1]:

$y=\frac{x-1}{(x+1)^2}+2$ $[1.1]$

Because the line must touch the curve, we may substitute $y=\frac{x}{x+1}:$

$\frac{x}{x+1}=\frac{x-1}{(x+1)^2}+2$

Solve for x:

$x(x+1)=(x-1)+2(x+1)^2$

$x^2+x=x-1+2x^2+4x+2$

$x^2+4x+1$

$x\frac{-4±\sqrt{4^2-4(1)(1)} }{2(1)}$

$x=-2$ ± $\sqrt{3}$

$x=-2$ ± $\sqrt{3}$ <em> </em> $and$ $x=-2-\sqrt{3}$

There are 2 tangent lines.

$y=\frac{1}{(-1+\sqrt{3)^2} } (x-1)+2$

and

$y=\frac{1}{(-1-\sqrt{3)^2} } (x-1)+2$

8 0

2 years ago

A bulletin board measuring 28 dm long was divided among three subject areas in the ratio 2:2:3. How much space was alothed for e

Sphinxa [80]

Answer: Space was alloted for each subject is 8 dm, 8 dm, 12 dm.

Explanation:

Since we have given the length of bulletin board =28 dm

Let the share of three subject areas be 2x,2x,3x

According to question,

$2x+2x+3x=28\\7x=28\\x=\frac{28}{7}\\x=4dm$

so x=4 dm

Now,

Space was alloted for each subject is given as below:

First subject = $2x=2\times 4=8 dm$

Second subject = $2x=2\times 4=8 dm$

Third subject = $3x=3\times 4=12 dm$

4 0

2 years ago