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1 year ago

Using the numbers 1 through 9 only once in the boxes, what is the largest value of x you can get?

Mathematics

1 answer:

aliina [53]1 year ago

6 0

Using the numbers 1 through 9 only once in the boxes, the largest value of x you can get is <u>7</u>.

<h3>What are mathematical operations?</h3>

Mathematical operations refer to the calculation of values using operands and mathematical operators.

Mathematical operations are formed like functions to take zero or more input values (operands) to a well-defined output value.

Some of the mathematical operators perform addition, subtraction, multiplication, division, exponentiation, and modulus operations.

<h3>Data and Calculations:</h3>

Mx + y = z

Let m = 1

x = 7

y = 2

1 x 7 + 2 = 9

7 + 2 = 9

9 = 9

Thus, using the numbers 1 through 9 only once in the boxes, the largest value of x you can get is <u>7</u>.

Learn more about mathematical operations at brainly.com/question/20628271

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-1/3x+2x=3 3/4 what does the variable represent?

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Start off by combining like terms on the LHS (the terms with x in them).

So we get $\frac{-1}{3}x+2x= \frac{-1}{3}x+ \frac{6}{3}x= \frac{-5}{3}x$

Replacing this result with what we had before on the LHS, we get $\frac{-5}{3}x=3 \frac{3}{4}= \frac{9}{4}$

⇒Solve for x (divide both sides $\frac{-5}{3}$ )
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3 years ago

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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]

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

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AleksandrR [38]

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1. reduce brackets

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3. reduce -513 x -2 to 1026

4. subtract 1026 from both sides

5. reduce 823 - 1026 to -203

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