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2 years ago

Please help -45 < 6s + 1

Mathematics

2 answers:

Mashutka [201]2 years ago

8 0

I think the answer should be -23/3

Alekssandra [29.7K]2 years ago

7 0

Answer:

s> -23/3

Step-by-step explanation:

ISOLATE THE VARIABLE S

-6s-45<1 --- - 6s< 1 + 45 = 46

ADD LIKE TERMS

DIVIDE

46/-6 = -23/3

-6s/-6 = s

Since you divide with a negative, switch the sign the opposite way then originally!

s>-23/3

I really hope this helped! :)

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AleksAgata [21]

Answer:

x=7

Step-by-step explanation:

Add 10 to both sides to isolate x.

4x=28

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7 0

2 years ago

Read 2 more answers

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 $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

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2 years ago

Can someone answer these please !!!!!!

katen-ka-za [31]

A. m^2 - 36 = 0
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b. 3x^2 + 15 = 0
3x^2 = -15
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x = √(-5) not possible unless complex numbers is applied
-1 = i^2
x = √(-5) = √(5i^2) = √5i

c. 4d^2 + 16 = 16
4d^2 = 0
d^2 = 0
d = 0

Hope it helped!

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Answer:

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2 years ago

Please help -45 < 6s + 1 ​

Please help -45 < 6s + 1