All categories

3 years ago

PLEASE HELP ME

Mathematics

1 answer:

Andru [333]3 years ago

7 0

8 people have finished the race in 30 minutes

You might be interested in

I'm giving a free brainliest to the first person who can answer my question. What is the fifth word of the national anthem?

anyanavicka [17]

Answer:

it's by "Say, can you see

<u>By</u>"

7 0

1 year ago

suppose you are in your last semester of college and have a 2.75 GPA after 105 credit hours if you are taking 15 credit hours in

Valentin [98]

Answer:

The overall CGPA would be 2.90 so it is not possible for hum to secure a CGPA of 3 for graduation.

Step-by-step explanation:

Given,

CGPA = 2.75

Credit hours = 105

Last semester GPA = 4

Last semester credit hours = 15

CGPA = $\frac{sum of (Credit hours of each subject * GPA of each subject)}{Total credit hours}$ credit hours * gpa

in our case, it would be:

CGPA = $\frac{(2.75*105)+(4*15}{105+15}$

=> $\frac{(288.75)+(60)}{120}$

=> $\frac{348.75}{120}$

=> 2.90

The overall CGPA would be 2.90 so it is not possible for hum to secure a CGPA of 3 for graduation.

7 0

3 years ago

I just need #21 and #23 answered please.

kykrilka [37]

#21: Even
#23: Neither

7 0

3 years ago

Read 2 more answers

(3x - 20°<br> (3x – 20)°<br> 4xº<br> 4xº<br> (3x – 20°<br> (3x – 20)<br> What is the value of x?

sergij07 [2.7K]

Step-by-step explanation:

<h3>this is given that the figure is a hexagon</h3><h3>let give a name to the given hexagon be ABCDEF</h3>

<h3>Angle a + B + C + D = 720 ( Ls sum pro. of hexagon)</h3><h3>3x-20+3x-20+4x+4x+3x-20+3x-20 = 720</h3><h3>3x+3x+3x+3x+4x+4x-20-20-20-20 = 720 (rearranging)</h3><h3>20x - 80 = 720</h3><h3>20x = 720 + 80 </h3><h3>x=800/20</h3>

<h2>x = 40 degrees </h2>

<h2>I HOPE THAT THIS ANSWER HELPS YOU</h2>

8 0

2 years ago

Someone please help me

allsm [11]

Answer:

Solutions: $x = \frac{-3}{ 4} + i \sqrt{39}$ , $x = \frac{-3}{4} - i \sqrt{39}$

Step-by-step explanation:

Given the quadratic equation, 2x² + 3x + 6 = 0, where a =2, b = 3, and c = 6:

Use the <u>quadratic equation</u> and substitute the values for a, b, and c to solve for the solutions:

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

$x = \frac{-3 +/- \sqrt{3^{2} - 4(2)(6)} }{2(2)}$

$x = \frac{-3 +/- \sqrt{9- 48} }{4}$

$x = \frac{-3 +/- \sqrt{-39} }{4}$

$x = \frac{-3 + i \sqrt{39} }{4}$ , $x = \frac{-3 - i \sqrt{39} }{4}$

Therefore, the solutions to the given quadratic equation are:

$x = -\frac{3}{ 4} + i \sqrt{39}$ , $x = -\frac{3}{4} - i \sqrt{39}$

4 0

2 years ago