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3 years ago

How many times further is 4,500,000,000and 58,000,000

Mathematics

2 answers:

sukhopar [10]3 years ago

6 0

Answer: 4442000000

Step-by-step explanation:

Oksana_A [137]3 years ago

3 0

Answer:

I know the answer but ill give you a hint all you do is subtract

Step-by-step explanation:

because it says how farthere

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What is the center and radius of the equation (x-8)^2 +(y+6)^2=13

tatyana61 [14]

Answer:

see explanation

Step-by-step explanation:

the equation of a circle in standard form is

(x - a)² + (y - b)² = r²

where (a, b) are the coordinates of the centre and r is the radius

(x - 8)² + (y + 6)² = 13 is in this form

with centre = (8, - 6) and r = $\sqrt{13}$

6 0

3 years ago

Is 4/16 equivalent to 0.25 and explain

egoroff_w [7]

Yes because if you do 4÷16 it is .25.
Hope it helps. If it doesn't comment on the bottom. I'll try to explain it more

7 0

3 years ago

Does anyone know the answer

murzikaleks [220]

Answer:

20,160

Step-by-step explanation:

6 0

3 years ago

you currently have 24 credit hours and a 2.8 gpa you need a 3.0 gpa to get into the college. if you are taking a 16 credit hours

Juliette [100K]

Answer:

$\sum_{i=1}^n w_i *X_i = 2.8*24 = 67.2$

And for this case we want a gpa of 3.0 taking in count that in this semester he/ she is going to take 16 credits so then the new mean would be given by:

$\bar X_f = \frac{\sum_{i=1}^n w_i *X_i+w_f *X_f }{24+16} = 3.0$

And we can solve for $\sum_{i=1}^n w_f *X_f$ and solving we got:

$3.0 *(24+16) =\sum_{i=1}^n w_i *X_i +\sum_{i=1}^n w_f *X_f$

And from the previous result we got:

$3.0 *(24+16) =67.2 +\sum_{i=1}^n w_f *X_f$

And solving we got:

$\sum_{i=1}^n w_f *X_f =120 -67.2= 52.8$

And then we can find the mean with this formula:

$\bar X_2 = \frac{\sum_{i=1}^n w_f *X_f}{16}= \frac{52.8}{16}=16=3.3$

So then we need a 3.3 on this semester in order to get a cumulate gpa of 3.0

Step-by-step explanation:

For this case we know that the currently mean is 2.8 and is given by:

$\bar X = \frac{\sum_{i=1}^n w_i *X_i }{24} = 2.8$

Where $w_i$ represent the number of credits and $X_i$ the grade for each subject. From this case we can find the following sum:

$\sum_{i=1}^n w_i *X_i = 2.8*24 = 67.2$

And for this case we want a gpa of 3.0 taking in count that in this semester he/ she is going to take 16 credits so then the new mean would be given by:

$\bar X_f = \frac{\sum_{i=1}^n w_i *X_i+w_f *X_f }{24+16} = 3.0$

And we can solve for $\sum_{i=1}^n w_f *X_f$ and solving we got:

$3.0 *(24+16) =\sum_{i=1}^n w_i *X_i +\sum_{i=1}^n w_f *X_f$

And from the previous result we got:

$3.0 *(24+16) =67.2 +\sum_{i=1}^n w_f *X_f$

And solving we got:

$\sum_{i=1}^n w_f *X_f =120 -67.2= 52.8$

And then we can find the mean with this formula:

$\bar X_2 = \frac{\sum_{i=1}^n w_f *X_f}{16}= \frac{52.8}{16}=16=3.3$

So then we need a 3.3 on this semester in order to get a cumulate gpa of 3.0

6 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Kitty [74]

Answer: (-2, 5) and (2, -3)

<u>Step-by-step explanation:</u>

Graph the line y = -2x + 1 (which is in y = mx + b format) by plotting the y-intercept (b = 1) on the y-axis and then using the slope (m = -2) to plot the second point by going down 2 and right 1 unit from the first point:

y - intercept = (0, 1) 2nd point = ( -1, 1).

Graph the parabola y = x² - 2x - 3 by first plotting the vertex and then plotting the y-intercept (or some other point):

$y = x^2-2x-3\quad \rightarrow \quad a=1,\ b=-2,\ c=-3\\\\\text{axis of symmetry:}\ x = \dfrac{-b}{2a}\ \longrightarrow \ x=\dfrac{-(-2)}{2(1)}=\dfrac{2}{2}=1\\\\\text{y-value of vertex:}\ f(1) = (1)^2-2(1)-3\quad \longrightarrow \quad y = 1 - 2 - 3=-4\\\\\text{y-intercept:}\ f(0)= (0)^2-2(0)-3\ \longrightarrow \ y=0 - 0 - 3 = -3 \\$

vertex = (1, -4) 2nd point (y-intercept) = (0, -3)

<em>see attached</em> - the graphs intersect at two points: (-2, 5) and (2, -3)

8 0

3 years ago