Solve forj. j + 20 > 17 orj + 17

99.00000000000 x 9/11

julia-pushkina [17]

Step-by-step explanation:

=99*9/11

- -

_____

=891/11

=81

3 0

2 years ago

Find all real solutions to the equation (x² − 6x +3)(2x² − 4x − 7) = 0.

Jet001 [13]

Answer:

x = 3 + √6 ; x = 3 - √6 ; $x = \frac{2+3\sqrt{2}}{2}$ ; $x = \frac{2-(3)\sqrt{2}}{2}$

Step-by-step explanation:

Relation given in the question:

(x² − 6x +3)(2x² − 4x − 7) = 0

Now,

for the above relation to be true the following condition must be followed:

Either (x² − 6x +3) = 0 ............(1)

or

(2x² − 4x − 7) = 0 ..........(2)

now considering the equation (1)

(x² − 6x +3) = 0

the roots can be found out as:

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

for the equation ax² + bx + c = 0

thus,

the roots are

$x = \frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(3)}}{2\times(1)}$

or

$x = \frac{6\pm\sqrt{36-12}}{2}$

or

$x = \frac{6+\sqrt{24}}{2}$ and, x = $x = \frac{6-\sqrt{24}}{2}$

or

$x = \frac{6+2\sqrt{6}}{2}$ and, x = $x = \frac{6-2\sqrt{6}}{2}$

or

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

$x = \frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-7)}}{2\times(2)}$

or

$x = \frac{4\pm\sqrt{16+56}}{4}$

or

$x = \frac{4+\sqrt{72}}{4}$ and, x = $x = \frac{4-\sqrt{72}}{4}$

or

$x = \frac{4+\sqrt{2^2\times3^2\times2}}{2}$ and, x = $x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}$

or

$x = \frac{4+(2\times3)\sqrt{2}}{2}$ and, x = $x = \frac{4-(2\times3)\sqrt{2}}{4}$

or

$x = \frac{2+3\sqrt{2}}{2}$ and, $x = \frac{2-(3)\sqrt{2}}{2}$

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ; $x = \frac{2+3\sqrt{2}}{2}$ ; $x = \frac{2-(3)\sqrt{2}}{2}$

7 0

2 years ago

Helppp i will give brainlist

Mice21 [21]

Answer:

5 units right, 2 units up

3 0

3 years ago

Which of the following points are more than 5 units from the point P(−2, −2)? Select all that apply. A A (2, 1) B B (4, −1) C C

netineya [11]

The distance between any 2 points P(a,b) and Q(c,d) in the coordinate plane, is given by the formula:

 $|PQ|= \sqrt{ (a-c)^{2} + (b-d)^{2}}$

Using this formula we calculate the distances |PA|, |PB|, |PC|, |PD| and |PE| and compare to 5.

$|PA|= \sqrt{ (-2-2)^{2} + (-2-1)^{2}}= \sqrt{16+9}= \sqrt{25}=5$

$|PB|= \sqrt{ (-2-4)^{2} + (-2+1)^{2}}= \sqrt{36+1}= \sqrt{37} \approx 6$

$|PC|= \sqrt{ (-2-2)^{2} + (-2+3)^{2}}= \sqrt{16+1}= \sqrt{17}\approx4$

$|PD|= \sqrt{ (-2+6)^{2} + (-2+6)^{2}}= \sqrt{16+16}= \sqrt{32}\ \textgreater \ \sqrt{25}=5$

$|PE|= \sqrt{ (-2+5)^{2} + (-2-1)^{2}}= \sqrt{9+9}= \sqrt{16}=4$

Answer: B and D

3 0

3 years ago

Need help can anyone help me???

Paladinen [302]

Answer:

it is greater than 45°

Step-by-step explanation:

From the relationship of angles and secants/tangents, we have ...

m∠GSO = (long arc GO -arc GT)/2

Solving for (long arc GO), we have ...

2(m∠GSO) +arc GT = (long arc GO)

We know that (long arc GO) > 180°, so we can write ...

2(m∠GSO) +90° > 180° . . . . arc GT = 90°

2(m∠GSO) > 90° . . . . . . subtract 90°

m∠GSO > 45° . . . . . . . . divide by 2

_____

Alternate solution

Inscribed ∠GOT has half the measure of arc GT, so is 45°.

You know that if angle G were 90°, then the right triangle would be isosceles, and angle S would also be 45°. In this triangle, arc GTO is less than 180°, so angle G is less than 90°.

When angle G gets smaller, the sum of angles remains the same, so angle S must be larger than 45°.

This reasoning is written more formally in the math above.

8 0

3 years ago

Solve forj. j + 20 > 17 orj + 17 < 5