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2 years ago

A number decreased by 12 is equal to 9 what is the number.

SAT

1 answer:

777dan777 [17]2 years ago

4 0

Answer:

21 is the correct answer

Explanation:

9+12=21

21-12=9

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2 years ago

Complete the square to rewrite the following equation. identify the center and radius of the circle. you must show all work and

insens350 [35]

The center of the circle is the point (-2, 3) and the radius is 3 units.

<h3>How to identify the center and radius?</h3>

Here we have the equation:

$x^2 + 4x + y^2 -6y = -4$

We can rewrite it as:

$x^2 + 2*2x + y^2 - 2*3y = -4$

Now, we can add and subtract 4 and add and subtract 9, so we get:

$x^2 + 2*2x + y^2 - 2*3y +4 - 4 + 9 - 9= -4\\\\(x^2 + 2*2*x + 4) + (y^2 - 2*3*y + 9) - 4 - 9 = -4\\\\(x + 2)^2 + (y- 3)^2 = 9 = 3^2$

So the center of the circle is the point (-2, 3) and the radius is 3 units.

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2 years ago

Which of the following BEST describes the relationship of Desktop Analytics to Windows Analytics

Mazyrski [523]

The answer choice that BEST describes the relationship of Desktop Analytics to Windows Analytics is It is a technique that uses data stored in data warehouses to create systems that allow it to perform information processing.

<h3>What is Desktop Analytics?</h3>

This refers to the cloud-based service that is used to give detailed insights to help the user to make better-informed decisions about Windows clients.

Hence, we can see that based on the relationship that exists between Desktop Analytics to Windows Analytics is the use of stored data in the data warehouses to help in information processing.

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2 years ago

A point charge q1 = -4. 00 nc is at the point x = 0. 60 m, y = 0. 80 m , and a second point charge q2 = +6. 00 nc is at the poin

Alekssandra [29.7K]

The net electric field is the vector sum of the components of the electric

field produced by the two charges.

The values of the magnitude and direction of the net electric field at the origin (approximate values) are;

131.6 N/C

12.6 ° above the negative x–axis

<h3>How are the net electric field magnitude and direction calculated?</h3>

The possible questions based on a similar question posted online are;

(a) The net electric field at the origin.

The electric field due to charge q₁ is given as follows;

$\vec E_{1x} = \mathbf{ \dfrac{1}{4 \cdot \pi \cdot \epsilon_0} \cdot \dfrac{q_1}{\vec{r}^2_2}}$

Which gives;

$\vec{E}_{1x} =\mathbf{ \dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(-4 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2}} \cdot cos\left(arctan\left(\dfrac{0.8}{0.6} \right) \right) =-21.6 \, N/C$

$\vec{E}_{1y} = \mathbf{\dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(-4 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2}} \cdot sin\left(arctan\left(\dfrac{0.8}{0.6} \right) \right) = 28.8 \, N/C$

Which gives;

$\vec{E}_1 = \mathbf{21.6 \, N/C \cdot \hat x + 28.8 \, N/C \hat y}$

$\vec{E}_{2x} = \dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(6.00 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2} = 150 \, N/C$