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3 years ago

Find the equation of the function in the graph

Mathematics

2 answers:

andrezito [222]3 years ago

5 0

Answer:

Step-by-step explanation:

it is a graph of $x^{2}$

marshall27 [118]3 years ago

4 0

Answer:

D.) y=x2

Step-by-step explanation:

Look in pic above, on EDGE NUITY I always used to graph to find the answer

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How are the volume formulas of prisms and cylinders similar? How are they<br> different?

AysviL [449]

Answer:

They are both similar in the way that the formula is base multiplied by height. They are different because of the type of shape on the base.

Step-by-step explanation:

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3 years ago

What is the answer to: -|4| - 9 =??

Delvig [45]

$-|4| - 9 =\\
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-13$

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3 years ago

What is the explicit formula for this sequence?<br> -9, -3, 3, 9, 15, ...

jok3333 [9.3K]

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8 0

2 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

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Sever21 [200]

Answer:

$900

Step-by-step explanation:

Use the formula for simple interest: i = p r t, where i is the interest earned, p is the principal amount, r is the interest rate as a decimal fraction, and t is the time in years.

Here, i = $27 = p(0.015)(2), or

$27 = 0.03p

Dividing both sides by 0.03, we get:

p = $27 / 0.03 = $900

The principal, in this situation, was $900.

6 0

3 years ago