All categories

2 years ago

Somebody help me please..

Mathematics

2 answers:

nadya68 [22]2 years ago

5 0

I believe the answer is B I pretty sure

Furkat [3]2 years ago

5 0

Answer: c) 2b - 3a + 7/2

Explanation:

(1 + b/2) - 3a + 3

(1 + b + 6/2) - 3a

= 2b - 3a + 7/2

You might be interested in

A water cup is shaped like a cone. It has a diameter of 3 inches and a slant height of 5 inches. About how many square inches of

Bezzdna [24]

The answer to that wil be A 24 sqaure inches

8 0

4 years ago

4n+3< 6n+8 -2n what is n

PolarNik [594]

Answer:

$8>3$

Step-by-step explanation:

STEP 1: Rewrite so n is on the left side of the inequality.

$6n+8-2n>4n+3$

STEP 2: Subtract 2 n from 6 n .

$4n+8>4n+3$

STEP 3: Move all terms containing n to the left side of the inequality.

$8>3$

6 0

3 years ago

(8x − 5)(2x2 − 5x − 6)

Elodia [21]

Answer:

(-5x - 2)(8x - 5)

I hope this helps!

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

What is the solution to 1/2n+3=6

Alona [7]

n in (-oo:+oo)

(1/2)*n-3 = 6 // - 6

(1/2)*n-6-3 = 0

1/2*n-9 = 0 // + 9

1/2*n = 9 // : 1/2

n = 9/1/2

n = 18

5 0

3 years ago

Read 2 more answers