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Svetllana [295]
3 years ago
15

Why is it important to be able to solve for variables in equations?

Mathematics
1 answer:
stellarik [79]3 years ago
5 0
Solving for variables in equations is important for many reasons.  Of course, it's important to be able to solve for them so you can pass math tests.  But even more so, it's important for real-life applications.  There are many uses for solving equations.  They can provide answers for mathematicians, accountants, insurance companies- you name it!  There are way too many to list!  It may not seem important now, but just keep working at it!  I personally think solving equations is a ton of fun!
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Verify the identity. cotangent x equals StartFraction 1 plus cosine 2 x Over sine 2 x EndFractioncot x= 1+cos2x sin2x Use the ap
lutik1710 [3]

Answer with Step-by-step explanation:

We are given that

RHS

\frac{1+Cos2x}{Sin2x}

We have to verify the identity.

We know that

1+Cos2x=2Cos^2 x

Sin2x=2SinxCos x

Using the formula

\frac{2Cos^2x}{2SinxCosx}

\frac{Cosx}{Sinx}

Cot x

By using the formula

Cotx=\frac{Cosx}{Sinx}

LHS=RHS

Hence, verified

6 0
3 years ago
The product of x and the sum of 6 and 8 times the square of x
Vaselesa [24]

Answer:

x(6 + 8x²) or 6x + 8x³.

Step-by-step explanation:

"The square of x" can be represented by x² and 8 times that would be 8 * x² or 8x². The sum of 6 and 8x² can be represented by 6 + x² and the product of x and 6 + x² can be represented by x * (6 + 8x²) or x(6 + 8x²) which simplifies to 6x + 8x³.

7 0
3 years ago
Read 2 more answers
How do you do the shortcut method
Gala2k [10]
<span>Add both the result (1×3 + 1×3) = 6 and write down to the left of 9 (result ofstep 1).

</span>
8 0
3 years ago
PLZ HELP I WILL MARK AS BRAINLIEST
-Dominant- [34]

Answer:

  7 square units

Step-by-step explanation:

As with many geometry problems, there are several ways you can work this.

Label the lower left  and lower right vertices of the rectangle points W and E, respectively. You can subtract the areas of triangles WSR and EQR from the area of trapezoid WSQE to find the area of triangle QRS.

The applicable formulas are ...

  area of a trapezoid: A = (1/2)(b1 +b2)h

  area of a triangle: A = (1/2)bh

So, our areas are ...

  AQRS = AWSQE - AWSR - AEQR

  = (1/2)(WS +EQ)WE -(1/2)(WS)(WR) -(1/2)(EQ)(ER)

Factoring out 1/2, we have ...

  = (1/2)((2+5)·4 -2·2 -5·2)

  = (1/2)(28 -4 -10) = 7 . . . . square units

4 0
3 years ago
What is the answer for this math problem 4-4×7+3=
Varvara68 [4.7K]
4-4*7+3=4-28+3=-24+3=-21
8 0
3 years ago
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