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3 years ago

A, b and c stand for three different numbers

1 answer:

5 0

To find the mean, you add up the numbers and divide by
how many numbers there are.

So if the mean of 'A' and 'B' is 40, then A + B = 80 .
And if the mean of 'B' and 'C' is 35 then B + C = 70.
==========================================
You said that (A + B) + C = 100 .

From something else that you said, I noticed that A + B = 80 .

So I can write (80) + C = 100

Subtract 80 from each side: C = 20
==========================================
I also noticed, from what you said, that B + C = 70 .

So I can write B + 20 = 70

Subtract 20 from each side: B = 50
==========================================
Now I know that C = 20 and B = 50.

Finally, you said that A + B + C = 100 ,

So I can write A + 50 + 20 = 100

Combine like terms on the left: A + 70 = 100

Subtract 70 from each side: A = 30
==========================================

  A = 30
 B = 50
 C = 20

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Ik this sounds dumb and it's very late so I don't want to do the math 600 Seconds is ___ hours what I'm asking is 600 seconds eq

fomenos

600 seconds would be 10 minutes, as there are 60 seconds in a minute.

Expressed as hours, that would be 1/6 of an hour.

6 0

3 years ago

Read 2 more answers

Help me please ..........................................................

ale4655 [162]

It would be the last one to add up to 180 witch is a triangle or supplementary

7 0

3 years ago

Beatrice calculated the slope between two pairs of points.

kumpel [21]

Answer:

The answer in the procedure

Step-by-step explanation:

we know that

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

step 1

Find the slope between (-3, -2) and (1, 0)

$m=\frac{0+2}{1+3}$

$m=\frac{2}{4}=\frac{1}{2}$

Find the equation of the line

$y-y1=m(x-x1)$

with m and the point (1,0)

substitute

$y-0=\frac{1}{2}(x-1)$

$y=\frac{1}{2}x-\frac{1}{2}$

step 2

Find the slope between (-2, -1) and (4, 2)

$m=\frac{2+1}{4+2}$

$m=\frac{3}{6}=\frac{1}{2}$

Find the equation of the line

$y-y1=m(x-x1)$

with m and the point (4,2)

substitute

$y-2=\frac{1}{2}(x-4)$

$y=\frac{1}{2}x-2+2$

$y=\frac{1}{2}x$

Compare the equation of the two lines

The two lines are parallel, because their slope is the same, but are different lines

therefore

Beatrice's conclusion is incorrect

All of these points are not on the same line, because are different parallel lines

The slope between (-2,-1) and (1,0) is equal to $\frac{1}{2}$

8 0

3 years ago

Read 2 more answers

The shadows of two vertical poles were measured at the same time to be 6m and 15m long. If the first pole is 8cm long, find the

son4ous [18]

Answer:

20 cm

Step-by-step explanation:

Shadow of vertical poles : 6m ; 15m

Length of poles : 8cm ; h

8cm long pole = shadow length, 6m

h cm long pole = shadow length, 15 m

Using cross multiplication :

8cm * 15 m = h cm * 6m

120 = 6h

h = 120 / 6

h = 20 cm

Height of second pole = 20cm

8 0

3 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)

(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

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

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

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

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

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

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

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

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

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

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

$x = \frac{2+3\sqrt{2}}{2}$ 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