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3 years ago

14

Pedro has 5 jars of marbles. Each jar is 2/3 filled with marbles. How many full jars of marbles does Pedro have? Write an equati

on to represent the problem. ______________________ Solve the equation. 1

1 answer:

Fynjy0 [20]3 years ago

4 0

Answer:

3 1/3 is the answer mark brainliest

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Worth 50 points!!!!!!!!!!!!!!!!!!!!!!!!

erica [24]

Hi,

o) 357 - 486 = -129
f) 4572 + 609 = 5181
b) -12 - 31 = -43
n) - 15 + 8 = -7

Hope this helps.
r3t40

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3 years ago

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Answer!! Whoever has the best answer I'll give brainliest!!

RoseWind [281]

There are 2 ways to do this. I'm going to do it the way that I think is most simple.

You have to find their LCM which is 40. Now times 40 to 1/4 v and 7/10

The equation then becomes 10+40v=28

You simplify it: 40v=18

v= 9/20

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2 years ago

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I am an even,3 digit palindrome.(example:464). The product of the digits is 8. What number am I?

antiseptic1488 [7]

181 is the only palindrome that has digits that when multiplied together give you 8.

Hope that helps :)

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4 years ago

Can some one help me factor 6x^2+4x-10

Fudgin [204]

2(3x + 5) (x - 1)

This is done through factoring 2 out of the equation first...

2(3x^2 + 2x -5)

And then cross multiplying it

3 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