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3 years ago

6

A square has a side length of 6x+2. a rectangle with width 3x+1 has the same area as the square. what is the length of the recta

ngle?

1 answer:

deff fn [24]3 years ago

3 0

Answer:

2

Step-by-step explanation:

(6x+2)/(3x+1) = 2

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Please help with 7a and 7b. The picture above is irrelevant to the problem. If possible explain or show your work. Thank you!!!!

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3 years ago

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Step-by-step explanation:

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3 years ago

The general form of an parabola is 2x2−12x−3y+12=0 .

Murljashka [212]

Answer:

The standard form of the parabola is $(x-3)^2=4*\frac{3}{8}(y+2)$

Step-by-step explanation:

The standard form of a parabola is

$(x-h)^2=4p(y-k)$ .

In order to convert $2x^2-12x-3y+12=0$ into the standard form, we first separate the variables:

$2x^2-12x+3y+12=0\\\\2x^2-12x+12=3y$

we now divided both sides by 2 to remove the coefficient from $2x^2$ and get:

$x^2-6x+6=\frac{3}{2}y$ .

We complete the square on the left side by adding 3 to both sides:

$x^2-6x+6+3=\frac{3}{2}y+3$

$x^2-6x+9=\frac{3}{2}y+3$

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

now we bring the right side into the form $4p(y-k)$ by first multiplying the equation by $\frac{2}{3}$ :

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

and then we multiplying both sides by $\frac{3}{2}$ to get

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

Here we see that

$4p=\frac{3}{2}$

$\therefore p=\frac{3}{8}$

Thus, finally we have the equation of the parabola in the standard form:

$\boxed{(x-3)^2=4*\frac{3}{8}(y+2)}$

5 0

3 years ago

Given the following trigonometric ratio, enumerate the meaning ratio

Likurg_2 [28]

Answer:

The trigonometric ratios are presented below:

$\sin \theta = \frac{AC}{\sqrt{AC^{2} + BC^{2}}}$

$\cos \theta = \frac{BC}{\sqrt{AC^{2} + BC^{2}}}$

$\cot \theta = \frac{BC}{AC}$

$\sec \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{BC}$

$\csc \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{AC}$

Step-by-step explanation:

From Trigonometry we know the following definitions for each trigonometric ratio:

Sine

$\sin \theta = \frac{y}{h}$ (1)

Cosine

$\cos \theta = \frac{x}{h}$ (2)

Tangent

$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{y}{x}$ (3)

Cotangent

$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{x}{y}$ (4)

Secant

$\sec \theta = \frac{1}{\cos \theta} = \frac{h}{x}$ (5)

Cosecant

$\csc \theta = \frac{1}{\sin \theta} = \frac{h}{y}$ (6)

Where:

$x$ - Adjacent leg.

$y$ - Opposite leg.

$h$ - Hypotenuse.

The length of the hypotenuse is determined by the Pythagorean Theorem:

$h = \sqrt{x^{2}+y^{2}}$

If $y = AC$ and $x = BC$ , then the trigonometric ratios are presented below:

$\sin \theta = \frac{AC}{\sqrt{AC^{2} + BC^{2}}}$

$\cos \theta = \frac{BC}{\sqrt{AC^{2} + BC^{2}}}$

$\cot \theta = \frac{BC}{AC}$

$\sec \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{BC}$

$\csc \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{AC}$

6 0

3 years ago