Write each sum as a product of two factors 7g+ 14h + 21i

Mathematics

1 answer:

frutty [35]3 years ago

5 0

Answer:

7 (g+2h+3i)

I hope this was the right answer:)

You might be interested in

Find the measurement indicated. Round to the nearest tenth.

fiasKO [112]

9514 1404 393

Answer:

22.0
15.0
30.0°
137.0°

Step-by-step explanation:

These are all Law of Cosine problems. A generic expression for the length of side 'c' opposite angle C, which is defined by sides 'a' and 'b' is ...

c² = a² +b² -2ab·cos(C)

The square root of this gives the side length:

c = √(a² +b² -2ab·cos(C))

Rearranging the equation, we can obtain an expression for the angle C.

C = arccos((a² +b² -c²)/(2ab))

These two formulas are used to solve the offered problems.

1) AC = √(13² +14² -2·13·14·cos(109°)) ≈ √483.506

AC ≈ 22.0

2) BC = √(7² +10² -2·7·10·cos(123°)) ≈ √225.249

BC ≈ 15.0

3) ∠B = arccos((24² +28² -14²)/(2·24·28)) = arccos(1164/1344)

∠B ≈ 30.0°

4) ∠B = arccos((6² +9² -14²)/(2·6·9)) = arccos(-79/108)

∠B ≈ 137.0°

4 0

3 years ago

Help pls !!!! Math is fun

kumpel [21]

Answer:

Step-by-step explanation:

The ratio fo the lengths of the sides of a 30-60-90 triangle is

short leg : long leg : hypotenuse

1 : sqrt(3) : 2

In a 30-60-90 triangle, the length of the short leg is the length of the long leg divided by sqrt(3).

The length of the short leg is 8 cm.

The length of the hypotenuse is twice the length of the short leg.

g = 2 * 8 cm = 16 cm

4 0

3 years ago

Find the sum of series 4/(4n-3) (4n+1)

ANTONII [103]

$\dfrac4{(4n-3)(4n+1)}=\dfrac1{4n-3}-\dfrac1{4n-1}$

Assuming the sum starts at $n=1$ , the $N$ th partial sum is

$\displaystyle\sum_{n=1}^N\left(\frac1{4n-3}-\frac1{4n-1}\right)=\left(1-\frac15\right)+\left(\frac15-\frac19\right)+\cdots+\left(\frac1{4N-7}-\frac1{4N-3}\right)+\left(\frac1{4N-3}-\frac1{4N+1}\right)$
$\displaystyle\sum_{n=1}^N\left(\frac1{4n-3}-\frac1{4n-1}\right)=1-\frac1{4N-1}$

As $N\to\infty$ , you're left with simply 1.