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4 years ago

What is -4 3/4% of 98.95

1 answer:

5 0

-4.75% of 98.95 is a simple multiplication problem.

first convert the percentage into a decimal. To do this mulitply -4.75 times 100 to get -0.0475.

Then multiply -0.0475 times 98.95 and that will give you the percentage of 98.95. The answer should be -4.700. I hope these are the right numbers because usually you don't deal with negative percentages.

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Please help your sis out!!!!!!!!!!

spayn [35]

You would input 10 as m into the equation
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t=0.5+60
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7 0

3 years ago

Read 2 more answers

Haleigh split 4/9 of a gallon of juice into 8 cups. How much juice was in each cup?

Brut [27]

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▹ Answer

<em>1/18 of a gallon of juice in each cup.</em>

▹ Step-by-Step Explanation

Since we have 4/9 of a gallon of juice for 8 cups, we will divide to find the amount of juice in each cup.

$\frac{4}{9} / 8$

Since this operation is division, we have to change 8 to it's reciprocal which is 1/8:

$\frac{4}{9}/ \frac{1}{8} = \frac{4}{72} = \frac{1}{18}$

Our final answer will be 1/18 of a gallon of juice in each cup.

Hope this helps!

CloutAnswers ❁

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5 0

3 years ago

Yall are doing great thanks sm for the help

blondinia [14]

Answer:

That is an obtuse isosceles

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

(9x + 10) - (2x + 4) in subtract linear expressions

ale4655 [162]

Answer:

Step-by-step explanation:

5x+14

4 0

3 years ago