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3 years ago

-6x + 3x = -21 linear equations

Mathematics

2 answers:

snow_tiger [21]3 years ago

8 0

well combine like terms

-3x = -21

divide both sides by -3

x = 7

kompoz [17]3 years ago

5 0

Answer:

x = 7

Step-by-step explanation:

-6x + 3x = -21

Combine like terms

-3x = -21

Divide by -3

-3x/-3 = -21/-3

x = 7

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3 years ago

Convert 17/20 into percentages .

PilotLPTM [1.2K]

Answer:

85%

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3 years ago

Read 2 more answers

5. Write each comparison as a rate.

stepan [7]

Answer:

5, 0.55, 2.00, 3 cm.

Step-by-step explanation:

If you're asking about the unit rates, then..

a. For 1 day, there is 5 mm of rain. 15 / 3 = 5

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c. $14.00 / 7 = 2.00 Indu saves $2.00 for one day.

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I hope this helped :)

4 0

3 years ago

2. Tina uSUally brings her children every Sunday afternoon at the children's park in their barangay. The park is 350m Iong and 2

Verdich [7]

Answer: kindly check explanation

Step-by-step explanation:

Given the following :

Total area of parknand path way = 74,464 m2.

Dimension of park = 350m by 200m

Using the area of rectangle formula:

Area = Length * width

Area of park = 350m * 200m

Area of park = 70,000

Area of path = (74,464 - 70,000) = 4,464m2

Since path has uniform width, then width of the the path is sqrt(4464) = 66.8m

2.) the path of the park seems a bit narrow, hence, I wouldn't love children to bike and jog in such a narrow path

4 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