All categories

11 months ago

-√-12 I need help there’s a picture of my question

Mathematics

1 answer:

Sophie [7]11 months ago

3 0

Answer:

The rewritten expression would be " $-2i\sqrt{3}$ ".

You might be interested in

Solve the system of equations 2x + 3y = 40 and –2x + 2y = 20.

Flura [38]

2x + 3y = 40
+
-2x + 2y = 20
___________ Adding both equations

0x + 5y = 60
____________

5y = 60
y = 60/5 = 12

y = 12.

Since we have gotten y =12, and the only option with that is option D.

So we don't have to solve for x.

The answer is D.

7 0

3 years ago

Write the equation in standard form. Identity the center and radius. x² + y2 + 8x-4y-7=0

Masteriza [31]

Answer:

See below

Step-by-step explanation:

Given equation:

x² + y² + 8x - 4y - 7 = 0

Convert this to standard form by completing the square:

(x - h)² + (y - k)² = r², where (h, k) - center, r - radius
x² + 2*4x + 4² + y² - 2*2y + 2² - 16 - 4 - 7 = 0
(x + 4)² + (y - 2)² - 27 = 0
(x + 4)² + (y - 2)² = 27
(x + 4)² + (y - 2)² = (√27)²

The center is (- 4, 2) and the radius is √27

3 0

1 year ago

Question 10 of 10

o-na [289]

I hope this helps you

7 0

2 years ago

The height of a ball thrown into the air after t seconds have elapsed is h = −16t2 + 40t + 6 feet. What is the first time, t, wh

sp2606 [1]

<h3>The first time when the ball will reach a height of 20 feet is 0.42 seconds</h3>

Solution:

Given that,

The height of a ball thrown into the air after t seconds have elapsed is:

$h = -16t^2 + 40t + 6$

What is the first time, t, when the ball will reach a height of 20 feet?

Substitute h = 20

$20 = -16t^2 + 40t + 6\\\\-16t^2 + 40t + 6 -20 = 0\\\\-16t^2 + 40t -14 = 0\\\\16t^2 -40t + 14 = 0\\\\8t^2 -20t + 7=0$

Solve by quadractic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

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

$\mathrm{For\:}\quad a=8,\:b=-20,\:c=7$

$t = \frac{-\left(-20\right)\pm \sqrt{\left(-20\right)^2-4\cdot \:8\cdot \:7}}{2\cdot \:8}\\\\t = \frac{20 \pm \sqrt{176}}{16}\\\\t = \frac{20 \pm 4\sqrt{11}}{16}\\\\t = \frac{ 5 \pm \sqrt{11}}{4}\\\\We\ have\ two\ solutions\\\\ t=2.07915, \:t=0.42084$

Rounding off we get,

t = 2.08 , t = 0.42

Thus the first time when the ball will reach a height of 20 feet is 0.42 seconds

4 0

2 years ago

Evaluate the expression.

fgiga [73]

Answer:

-2.4

Step-by-step explanation:

numerator: 15.36

denominator: -6.4

15.36/-6.4 = -2.4

5 0

3 years ago

Read 2 more answers