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3 years ago

A number increased by 10

Mathematics

2 answers:

Natasha_Volkova [10]3 years ago

7 0

Answer:

Step-by-step explanation:

just got it right on plato

ra1l [238]3 years ago

7 0

Answer:

im pretty sure we need the previous value

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Which term is correct when simplifying √128? A.) 16 B.) 12 C.) 10 D.) 8√2

alina1380 [7]

When dealing with roots, we want to factor to see if it's possible

$128 = 2 \times 64$
$= 2 \times 2 \times 32$
$= 2 \times 2 \times 2 \times 16$
$= 2 \times 2 \times 2 \times 2 \times 8$
$= 2 \times 2 \times 2 \times 2 \times 2 \times 4$
$= 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

now since we're dealing with square roots, we circle every pair of numbers we see in the final factoring
so..
$= (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times 2$

now for every circled pair, we take only 1 of the 2 circled and bring it outside
anything left uncircled stays inside the root
so..
$= 2 \times 2 \times 2 \times \sqrt{2}$

now simply multiply
$8 \sqrt{2}$

3 0

3 years ago

Factor (16a2 - 49) completely

riadik2000 [5.3K]

Answer:

(4a+7)(4a-7)

Step-by-step explanation:

5 0

3 years ago

Consider the line integral Z C (sin x dx + cos y dy), where C consists of the top part of the circle x 2 + y 2 = 1 from (1, 0) t

pashok25 [27]

Direct computation:

Parameterize the top part of the circle $x^2+y^2=1$ by

$\vec r(t)=(x(t),y(t))=(\cos t,\sin t)$

with $0\le t\le\pi$ , and the line segment by

$\vec s(t)=(1-t)(-1,0)+t(2,-\pi)=(3t-1,-\pi t)$

with $0\le t\le1$ . Then

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)$

$=\displaystyle\int_0^\pi(-\sin t\sin(\cos t)+\cos t\cos(\sin t)\,\mathrm dt+\int_0^1(3\sin(3t-1)-\pi\cos(-\pi t))\,\mathrm dt$

$=0+(\cos1-\cos2)=\boxed{\cos1-\cos2}$

Using the fundamental theorem of calculus:

The integral can be written as

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=\int_C\underbrace{(\sin x,\cos y)}_{\vec F}\cdot\underbrace{(\mathrm dx,\mathrm dy)}_{\vec r}$

If there happens to be a scalar function $f$ such that $\vec F=\nabla f$ , then $\vec F$ is conservative and the integral is path-independent, so we only need to worry about the value of $f$ at the path's endpoints.