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3 years ago

Austin makes $5.50 per hour. Which expression represents his pay?

Mathematics

1 answer:

Naddika [18.5K]3 years ago

7 0

The answer is 5.50h because h equals the hours and every hour u are receiving 5.50 so you will be multiplying

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What is the constant of variation if y varies inversely as x and y = 3 when x = 6? PLZ HELP ME I DONT UNDERSTAND

mojhsa [17]

Answer:

k = 18

Step-by-step explanation:

Given that y varies inversely as x then the equation relating them is

y = $\frac{k}{x}$ ← k is the constant of variation

To find k use the condition y = 3 when x = 6, that is

3 = $\frac{k}{6}$ ( multiply both sides by 6 )

18 = k

The constant of variation is k = 18

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3 years ago

Solve the system of linear equations below.

Olegator [25]

Answer:

B. x = 2, y = 6

Step-by-step explanation:

$2x + 5y = 34...(1) \\ \\ x + 2y = 14 \\ \therefore \: x = 14 - 2y...(2) \\ \\ from \: equations \: (1) \: and \: (2) \\ \\ 2(14 - 2y) + 5y = 34 \\ \therefore \: 28 - 4y + 5y = 34 \\ \therefore \: 28 + y = 34 \\ \therefore \: y = 34 - 28 \\ \huge \red{ \boxed{\therefore \: y = 6}} \\ substituting \: y = 6 \: in \: equation \: (2) \\ x = 14 - 2 \times 6 \\ \therefore \: x =14 - 12 \\ \huge \purple{ \boxed{\therefore \: x =2}} \\ \\ \huge \orange{ \boxed{ \therefore \: x = 2, \: \: y = 6}} \\$

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iragen [17]

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1110

Step-by-step explanation:

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

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3 years ago