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3 years ago

Easy question: what is 1 plus 1 (Hint: not 2)

Mathematics

2 answers:

sveticcg [70]3 years ago

6 0

Answer:

1 + 1 = 2

Step-by-step explanation:

1 - (-(-1) = 1- 1 = 0

Gekata [30.6K]3 years ago

4 0

The answer is -2. You multiply 1 X (Not 2), that would be the negation of 2. That equals -2. Everyone knows that a negative number times itself is just itself. So there is your answer.

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What is 43% of 600 gallons of water

Leokris [45]

43% of 600 gallons of water is 258

Convert your percentage into a decimal:
$\frac{43}{100} = 0.43$

Multiply your decimal and 600:
$0.43 \times 600 = 258$

3 0

3 years ago

Read 2 more answers

M∠1=28°, m∠6=65°, m∠5=65°. Find m∠MAX

kirill115 [55]

Answer:

<MAX = 93

Step-by-step explanation:

Since <MAX is technically <1 + <5, we know that <1 is 28 and <5 is 65. We can add both of these angles up to solve for <MAX, which is 28 + 65 = 93.

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4 years ago

A girls hair is 23 inches long how long and grows about half an inch every month how long will her hair be in 2 years

jasenka [17]

Answer:

Step-by-step explanation:

1/2 multiplied by (12 multiplied by 2)

1/2 (24)

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= 35 inches long

4 0

4 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

How long in days is a mega second? Hi, I'm helping my little brother with a maths assignment at the moment and I couldn't quite

Marat540 [252]

I've heard of milliseconds, microseconds and nanoseconds, but never a megasecond. By the way:
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1 second = 1e+009 nanoseconds

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3 years ago