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3 years ago

If anyone is there, what would be the area to this? A= b1 b2 x h /2

Mathematics

1 answer:

GuDViN [60]3 years ago

6 0

Answer:

Area = 100 cm²

Step-by-step explanation:

A trapezium whose parallel sides are of length a and b, and the perpendicular distance between them is h has an area given by the formula

In this question, the figure actually comprises of two trapezium which

parallel sides are 18cm and 7cm and height = 4cm

So, area of this figure = 2(1/2 (a+b)h) = (a+b)h

= (18 + 7)*4

= 100 cm²

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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

2 years ago

What is an equation for the linear function whose graph contains the points (−1, −2) and (3, 10)

sladkih [1.3K]

The equation of a line starting from two points is:
$y-y_1=\frac{y_2-y_1}{x_2-x_1} \cdot (x-x_1)$

From the first point you get: x1 = -1, y1 = -2
From the second point you get: x2 = 3, y2 = 10

Replace x1, y1, x2, y2 in the equation of the line and you get:
$y+2=\frac{10+2}{3+1} \cdot (x+1)$
$y+2=\frac{12}{4} \cdot (x+1)$
$y+2=3 \cdot (x+1)$

From this you get the equation of your line:
$y(x)=3x+1$

4 0

3 years ago

Read 2 more answers

PLEASE HELP NEED THIS ASAP.....

horrorfan [7]

Answer: b would actually be the closest but the exact answer is 145.2

8 0

1 year ago

All geometric figures,shapes,and solids can be named using sets of

Naddika [18.5K]

Answer:

All geometric figures, shapes and solids can be named as a sets of surface regions and plane regions and they all lie in three dimensional spaces.

Step-by-step explanation:

7 0

3 years ago

How do I solve 1/2z=9 1/4?

Nastasia [14]

Answer:

z = 9/2

Step-by-step explanation:

1/2z = 9 1/4?

z = 9/2

Explanation:

Simplify

9 x 1/4 → 9/4

Multiply 2 to both sides to cancel out the 2

1z = 9/4

Note: 1z is just z

6 0

3 years ago