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3 years ago

Solve for the variable indicated. 2p=kx-t, for x

Mathematics

2 answers:

algol [13]3 years ago

7 0

hello there

the answer is

Let's solve for f.

2p=kx−tforx

Step 1: Flip the equation.

−fortx+kx=2p

Step 2: Add -kx to both sides.

−fortx+kx+−kx=2p+−kx

−fortx=−kx+2p

Step 3: Divide both sides by -ortx.

−fortx

−ortx

−kx+2p

−ortx

2kx−4p

2ortx

Answer:

2kx−4p

2ortx

thank you

oksian1 [2.3K]3 years ago

4 0

2p = kx - t

Add t to both sides.

2p + t = kx

Divide both sides by k.

2p/k + t/k = x

Reorder.

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Answer:

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Step-by-step explanation:

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3 years ago

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Step-by-step explanation:

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3 years ago

How do you solve this

marusya05 [52]

Hope this helps! Good luck!

3 0

3 years ago

An equilateral triangle has has an era of 43.5 square feet the height is 8.5 feet what is the Perimeter of the triangle

enot [183]

Answer:

30.7 ft

Step-by-step explanation:

An equilateral triangle with area 43.5 has a height of 8.68 not 8.5

I will do the problem as written even though it is not a possible equilateral triangle

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a = (1/2)bh

43.5 = (1/2)b(8.5)

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8 0

2 years ago

Find an upper limit for the zeroes 2x^4 -7x^3 + 4x^2 + 7x - 6 = 0

erik [133]

Answer-

2 is the upper limit for the zeros.

Solution-

The given function f(x) is,

$2x^4 -7x^3 + 4x^2 + 7x - 6 = 0$

For calculating the zeros,

$\Rightarrow f(x)=0$

$\Rightarrow 2x^4 -7x^3 + 4x^2 + 7x - 6 = 0$

$\Rightarrow 2x^4-4x^3-3x^3+6x^2-2x^2+ 4x+3x-6=0$

$\Rightarrow 2x^3(x-2)-3x^2(x-2)-2x(x-2)+3(x-2)=0$

$\Rightarrow (x-2)(2x^3-3x^2-2x+3)=0$

$\Rightarrow (x-2)(x^2(2x-3)-1(2x-3))=0$

$\Rightarrow (x-2)(x^2-1)(2x-3)=0$

$\Rightarrow (x-2)(x+1)(x-1)(2x-3)=0$

$\Rightarrow x-2=0,\ x+1=0,\ x-1=0,\ 2x-3=0$

$\Rightarrow x=2,\ x=-1,\ x=1,\ x=\frac{3}{2}$

From all the 4 roots, it can be obtained that 2 is the greatest zero.

7 0

3 years ago