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alekssr [168]
3 years ago
5

Just need to calculate the area of the shaded region, thanks

Mathematics
1 answer:
Elena-2011 [213]3 years ago
5 0

Answer:

Area of the shaded region= 82.2 cm²

Step-by-step explanation:

Please see attached picture for full solution.

You might be interested in
Is 1.73205 rational?
Assoli18 [71]

Answer:

No, it is not.

Step-by-step explanation:

It can not be turned into a fraction and isn't repeating with the same number.

6 0
3 years ago
Y=-x^2+20x-64 what is the max height
Inessa [10]

Answer:

36

Step-by-step explanation:

The maximum height is the y-coordinate of the vertex

given a quadratic in standard form : ax² + bx + c : a ≠ 0

then the x-coordinate of the vertex is

x_{vertex} = - \frac{b}{2a}

y = - x² + 20x - 64 is in standard form

with a = - 1, b = 20 and c = - 64, hence

x_{vertex} = - \frac{20}{-2} = 10

substitute x = 10 into the equation for y

y = - (10)² + 20(10) - 64 = 36 ← max height


8 0
3 years ago
A customer brings a check of 2,941. he wants 100 in cash, put 20% of the remaining into her savings account then the rest into a
Goryan [66]

Answer:

Amount theta she is putting in Checking account is 2272.80

Step-by-step explanation:

Given:

Amount on check = 2941

Amount that he want in cash = 100

Amount she put in saving account = 20% of remaining after getting cash

Remaining Amount she put in checking account.

To find: Amount in her Checking Account.

Amount left after taking cash = 2941 - 100 = 2841

Amount that she put in saving account = 20% of 2841 = \frac{20}{100}\times2841 =  568.20

Amount in her checking account = 2941 - 100 - 568.20  = 2272.8

Therefore, Amount theta she is putting in Checking account is 2272.80

8 0
3 years ago
in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)
Nimfa-mama [501]

Answer:

The component of \vec u orthogonal to \vec a is \vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right).

Step-by-step explanation:

Let \vec u and \vec a, from Linear Algebra we get that component of \vec u parallel to \vec a by using this formula:

\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a (Eq. 1)

Where \|\vec a\| is the norm of \vec a, which is equal to \|\vec a\| = \sqrt{\vec a\bullet \vec a}. (Eq. 2)

If we know that \vec u =(2,1,1,2) and \vec a=(4,-4,2,-2), then we get that vector component of \vec u parallel to \vec a is:

\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)

\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)

\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)

Lastly, we find the vector component of \vec u orthogonal to \vec a by applying this vector sum identity:

\vec  u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a} (Eq. 3)

If we get that \vec u =(2,1,1,2) and \vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right), the vector component of \vec u is:

\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10}    \right)

\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)

The component of \vec u orthogonal to \vec a is \vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right).

4 0
3 years ago
For what values of the variable are the following expressions defined?
Rom4ik [11]

Answer:

  • The Expression will be defined if the denominator is not equal to 0 since any number divided by 0 is undefined.
  • Therfore our main priority here is to check the denominator only because it's the only part that can make the expression undefined
  • the expression will be defined for all real numbers (a) BUT 3+a must not equal to 0 therfore (a) can be all real numbers but must never be equal to -3

Step-by-step explanation:

  • a \: is \: an \: element \: of \: all \: real \: numbers \: but \: not \: not \: equal \: to \:  -3
  • HOPE THIS HELPS!
3 0
2 years ago
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