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3 years ago

Please help! 5x^2 - 3x - 2 = 0 Solve using the zero-factor property.

Mathematics

1 answer:

Alex3 years ago

7 0

Answer:

x = -2/5

x = 1

Step-by-step explanation:

5x² - 3x - 2 = 0

(5x + 2)(x - 1) = 0

⇒ 5x + 2 = 0

⇒ 5x = -2

⇒ x = -2/5

⇒ x - 1 = 0

⇒ x = 1

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2 years ago

The maximum height of a vehicle that can safely pass under a bridge is 12 feet 5 inches. A truck measures 162 inches in height.

Bumek [7]

<h2>Answer:</h2>

<em><u>The truck cannot pass safely under the bridge. The truck is 13 inches taller than the maximum height.</u></em>

<h2>Step-by-step explanation:</h2>

In the question,

The maximum height of the vehicle which is capable of passing under the bridge is 12 feet and 5 inches.

So,

Now we know that,

1 feet = 12 inches

So,

12 feet = 12 x 12 = 144 inches

So,

Total height of the vehicle which is permissible to pass under the bridge is,

12 feet 5 inches = 144 + 5 = 149 inches

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Height of the truck = 162 inches

Therefore, we can see that the permissible height is smaller than the height of the vehicle.

Height of vehicle which is more than permissible height is by,

162 - 149 = 13 inches

<em><u>Therefore, the truck cannot pass safely under the bridge. The truck is 13 inches taller than the maximum height.</u></em>

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3 years ago

Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?

cluponka [151]

Under the given transformation, the Jacobian and its determinant are

$\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2$

so that

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv$

where $D'$ is the region $D$ transformed into the $u$ - $v$ plane. The remaining integral is the twice the area of $D'$ .

Now, the integral over $D$ is

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y$

but through the given transformation, the boundary of $D'$ is the set of equations,

$\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}$

We require that $u>0$ , and the last equation tells us that we would also need $v>0$ . This means $1\le v\le\sqrt2$ and $0$ , so that the integral over $D'$ is

$\displaystyle2\iint_{D'}\mathrm du\,\mathrm dv=2\int_1^{\sqrt2}\int_0^{6v}\mathrm du\,\mathrm dv=\boxed6$

4 0

3 years ago