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4 years ago

Estimate the product by rounding 2 x 6,254

Mathematics

1 answer:

Pavlova-9 [17]4 years ago

6 0

5 × 6,254= 31,300 is the awnser :)

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−4(w+1)=−24 what is t? plz help

Blizzard [7]

- 4 w - 4 = - 24
- 4w = - 20
w = 5

5 0

4 years ago

Solve. Write your answer in interval notation.

lina2011 [118]

Answer: x < 3 or x ≥ 11

(-∞, 3) or [11, ∞)

Step-by-step explanation:

subtract 4 from each side

2x < 6 . or . 3x ≥ 33

x < 3 or x ≥ 11

make sure you have an OPEN DOT at 3 pointing to negative infinity and a CLOSED DOT at 11 pointing to positive infinity.

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3 years ago

What is the reference angle for 120°?

Drupady [299]

Since the angle 240° 240 ° is in the third quadrant, the reference angle formula is Ar=Ac−180°<span> A r = A c </span>- 180 °<span> .</span>

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3 years ago

Read 2 more answers

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

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

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3 years ago

Which of the following graphs represents a function

Nata [24]

Where are the graphs? a function is when every input is paired with exactly one output(every x value must have exactly 1 y value). the vertical line test helps determine a function, a vertical line can’t intersect a graph at more then one point, if it does intersect in more then 1 point then it’s not a function

7 0

3 years ago