A gallon is equal to 8 pints. How many pints are in 5 gallons?

Find x, round your answer to the nearest integer.

Makovka662 [10]

Answer:

B. 8

Step-by-step explanation:

cos37° = 4/5 = x/10

=> 4(10) = 5x

=> x = 40/5

=> x = 8

The correct option is B.

8 0

2 years ago

The manager at Tiffany's Furniture Store is trying to figure out how much to charge for a table that just arrived. If the table

Darina [25.2K]

$156.65
103.00 x .55 = 56.65
103.00 + 56..65 = $156.65

8 0

3 years ago

If f(x) = 3x 7, find f(a - 1) a. 3a 6 b. 3a 4 c. 3a 8 d. 3a 10

Rom4ik [11]

Hello,

Let's suppose f(x)=3x+7
f(a-1)=3(a-1)+7=3a-3+7=3x+4
Answer B with a sign "++++++++++++++++++++++++++++"

5 0

3 years ago

.What is the<br> percent increase from 1987 to today?

kozerog [31]

Answer:

In other words, $100 in 1987 is equivalent in purchasing power to about $227.09 in 2020, a difference of $127.09 over 33 years. The 1987 inflation rate was 3.65%. The current inflation rate (2019 to 2020) is now 2.49%

7 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

2 years ago