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3 years ago

Write the phrase as an expression. The quotient of 40 and the difference of a number y and 16.

Mathematics

1 answer:

klio [65]3 years ago

5 0

Answer:

40/y-16

Step-by-step explanation:

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7/8x=42<br><br> What is the value of x?

Artist 52 [7]

$\frac{7}{8}*x=42 \\ \\x=42: \frac{7}{8} \\ \\ x=42* \frac{8}{7} \\ \\ x= \frac{336}{7} \\ \\ \boxed{x=48}$

3 0

3 years ago

Whats the step by step answer to this equation i got -25 but its supposed to be positive not negative

andreev551 [17]

Answer:

Step-by-step explanation:

- 6/5x = -30 ⇒ multiply both sides by -1
6/5x = 30 ⇒ multiply both sides by 5
6x = 5*30
x = 150/6 ⇒ divide both sides by 6
x = 25 ⇒ answer

6 0

3 years ago

Read 2 more answers

Someone help if you can

Sav [38]

Answer:

sorry but the words are very blurry

3 0

3 years ago

Two professional baseball teams played a four game series. Attendance for the first three games was 126,503 people, what was the

ioda

Answer: 44,815 people; 126,503 + n = 171,318; n = 171,318 − 126,503

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