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3 years ago

Hey can you please help me posted picture of question

Mathematics

2 answers:

ruslelena [56]3 years ago

8 0

Yes he’s right it’s positive infinity so it’s C

xenn [34]3 years ago

4 0

The discriminant can be determined from the number of roots of the graph.

If Disc> 0, the function has two distinct roots
If Disc = 0, the function has a repeated root
If Disc < 0, the function has no real root.

Since, the given function has two distinct roots i.e. it cross x-axes at two different points, its discriminant will be positive.

So, the answer to this question is option C

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How do you know a radical expression is in simplest form?

Diano4ka-milaya [45]

Answer: To know whether a radical expression is in simplest form or not you should put the numbers and letters inside the radical in terms of prime factors. Then, the radical expression is in the simplest form if all the numbers and letters inside the radical are prime factors with a power less than the index of the radical

Explanation:

Any prime factor raised to a power greater than the index of the root can be simplified and any factor raised to a power less than the index of the root cannot be simplified

For example simplify the following radical in its simplest form:

$\sqrt[5]{3645 a^8b^7c^3}$

1) Factor 3645 in its prime factors: 3645 = 3^6 * 5

2) Since the powr of 3 is 6, and 6 can be divided by the index of the root, 5, you can simplify in this way:

- 6 ÷ 5 = 1 with reminder 1, so 3^1 leaves the radical and 3^1 stays in the radical

3) since the factor 5 has power 1 it can not leave the radical

4) the power of a is 8, then:

8 ÷ 5 = 1 with reminder 3 => a^1 leaves the radical and a^3 stays inside the radical.

5) the power of b is 7, then:

7 ÷ 5 = 1 with reminder 2 => b^1 leaves the radical and b^2 stays inside the radical

6) the power of c is 3. Since 3 is less than 5 (the index of the radical) c^3 stays inside the radical.

7) the expression simplified to its simplest form is

$3ab \sqrt[5]{3.5.a^3b^2c^3}$

And you know it cannot be further simplified because all the numbers and letters inside the radical are prime factors with a power less than the index of the radical.

7 0

3 years ago

Read 2 more answers

Carly has a coupon for 15% off the

wel

15% = 0.15

234.00 x 0.15 = 35.10

234.00 - 35.10 = 198.90

The camera will cost $198.90 after the discount.

6 0

3 years ago

Read 2 more answers

Help quick please ________

Kruka [31]

Hey

Area of first rectangle = 18 x 21 = 378 in^2

Area of second rectangle = 17 x 9 = 153 in^2

Total area = 378+153 = 531 in^2

Option 1

Hope this helps :)

5 0

2 years ago

Read 2 more answers

Which of the following is the smallest volume?

Lyrx [107]

B. 250 cm3 hope this helps

5 0

3 years ago

Read 2 more answers

The equation giving a family of ellipsoids is u = (x^2)/(a^2) + (y^2)/(b^2) + (z^2)/(c^2) . Find the unit vector normal to each

Fynjy0 [20]

Answer:

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Step-by-step explanation:

Given equation of ellipsoids,

$u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}$

The vector normal to the given equation of ellipsoid will be given by

$\vec{n}\ =\textrm{gradient of u}$

$=\bigtriangledown u$

$=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})$

$=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}$

$=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}$

Hence, the unit normal vector can be given by,

$\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}$

$=\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}$

$=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Hence, the unit vector normal to each point of the given ellipsoid surface is

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

3 0

3 years ago