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3 years ago

So what is the answer

Mathematics

1 answer:

Dovator [93]3 years ago

3 0

We need the question in order to answer:(

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2 If = 9 and 3p = x, what is the value of 4p? A) 6 B) 12 C) 18 D) 24

ValentinkaMS [17]

Answer:

The question needs to rewritten

if=9 and 3p=x

assuming then 12 B

Step-by-step explanation:

7 0

3 years ago

How can we represent a time that came before a specific zero point

Oduvanchick [21]

Answer:

by stating what the time was before the zero point came.

Step-by-step explanation:

3 0

3 years ago

A large water tank holds 210 liters of water. About how many gallons of water does the same tank hold, if there are approximatel

shusha [124]

Answer:

Step-by-step explanation:

Set up a proportion

15 Liters / 4 gallons = 210 Liters / x Cross Multiply

15x = 4 * 210 Combine the right

15x = 840 Divide by 15

x = 840 / 15

x = 56 gallons

4 0

3 years ago

When will these two equations intersect y=-2x +5 y = 1/3 x - 2 2

max2010maxim [7]

-2x + 5 = 1/3x - 22
-7/3x = -27
-7x = -81
x = 81/7 or 11.57
They will intersect when x = 81/7 or 11.57

6 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