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3 years ago

The value -1 is a solution for x^2 - 6x + 5 = 0 True or False?

Mathematics

1 answer:

solmaris [256]3 years ago

3 0

False -1 is not a solution.

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(2x^4 - 6x^2+8) + (-x^4+ 2x^2-12)

Neko [114]

(2x^4 - 6x^2+8) + (-x^4+ 2x^2-12)
=2x^4-6x^2-x^4+2x^2-12
=x^4-4x^2-12

Just watch your signs when opening up the brackets and then collect like terms.

3 0

4 years ago

What is the ordered pair for -x-y=6?

Stells [14]

(0,-6) (1,-7) (2,-8)

3 0

3 years ago

It allows us to express a very small or very large number in compact form, what is this.

lesya692 [45]

Scientific notation

Explanation:

…

4 0

2 years ago

The curves y = √x and y=(2-x) and the Cartesian axes form two distinct regions in the first quadrant. Find the volumes of rotati

makkiz [27]

Answer:

Step-by-step explanation:

If you graph there would be two different regions. The first one would be

$y = \sqrt{x} \,\,\,\,, 0\leq x \leq 1 \\$

And the second one would be

$y = 2-x \,\,\,\,\,, 1 \leq x \leq 2$ .

If you rotate the first region around the "y" axis you get that

$A_1 = 2\pi \int\limits_{0}^{1} x\sqrt{x} dx = \frac{4\pi}{5} = 2.51$

And if you rotate the second region around the "y" axis you get that

$A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188$

And the sum would be 2.51+4.188 = 6.698

If you revolve just the outer curve you get

If you rotate the first region around the x axis you get that

$A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708$

And if you rotate the second region around the x axis you get that

$A_2 = \pi \int\limits_{1}^{2} (2-x)^2 dx = \frac{\pi}{3} = 1.0472$

And the sum would be 1.5708+1.0472 = 2.618

7 0

4 years ago

If (x) =7-3x, determine the value of x when f(x)=16.

melomori [17]

Answer:

f=16/7.3x

Step-by-step explanation:

Lets solve for f:

f(7.3)=16

Step 1:Divide both sides by 7.3x

7.3f/7.3x=16/17.3x

f=16/7.3x

3 0

3 years ago