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3 years ago

Solve for z 10z − 7z = 3

Mathematics

2 answers:

Alina [70]3 years ago

6 0

10z − 7z = 3

The answer is Z = 1

valentinak56 [21]3 years ago

4 0

Answer:

z = 1

Step-by-step explanation:

10z - 7z = 3

collect like terms (10z and 7z)

10z-7z=3z

3z =3

divide for both sides

z = 1

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50 POINTS!!

Andre45 [30]

Answer:Triangle A: 38 degrees

Triangle B: Unknown (not enough information)

Triangle C: Unknown (not enough information)

Triangle D: 70 degrees

Triangle E: 40 degrees

Step-by-step explanation:

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2 years ago

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A total of 300 pine and maple trees will be planted in a park. There will be 2 pine trees planted for every 3 maple trees plante

mars1129 [50]

Answer:

120

Step-by-step explanation:

2 of every 5 trees are pine trees.

2/5 · 300 = 120

120 pine trees are going to be planted in the park.

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2 years ago

MAX POINTS!!! Write and solve a system of equations. Be sure to label the variables. The difference of two numbers is two. Two t

Cerrena [4.2K]

Answer:

5 and 3

Step-by-step explanation:

Let the numbers be x and y.

The equations are:

x - y = 2
2x + y = 13

Add up the equations to eliminate y and solve for x:

x - y + 2x + y = 2 + 13
3x = 15
x = 15/3
x = 5

Now find the value of y:

y = x - 2
y = 5 - 2
y = 3

3 0

2 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]

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

Please solve this quadratic equation: 3^2x -9 =0

Basile [38]

Answer:

${3}^{2} x - 9 = 0 \\ 9x = 9 \\ x = 1$

4 0

2 years ago