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4 years ago

Which relation is also a function?

Mathematics

1 answer:

Mumz [18]4 years ago

8 0

Answer:

4th. as it had UNIQUE values for every x

unlike rest 3,

like 2nd has two values of 3, not a function

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8,264 is divisible by 3? True or false?

mars1129 [50]

$Number\ is\ divisible\ by\ 3\ if\ the\ sum\ of\ the\ digits\ is\ divisible\ by\ 3.\\\\8,264\\\\8+2+6+4=20-
is\ not\ divisible\ by\ 3.\\\\8,264\ is\ divisible\ by\ 3-FALSE$

7 0

3 years ago

Read 2 more answers

0.2(-4–2.5b–7c). write the product

Xelga [282]

Answer:

-0.8 - 0.5b - 1.4c

Step-by-step explanation:

Take the b and the c away

Then times the 0.2 by all the numbers but don't add them

Once your done you and the b and c back to you equation

4 0

3 years ago

Simplify. (-5.12) - (-4.7) + (7.021) 6.79 6.601 -2.799

liq [111]

The answer is 6.601, and therefore, the second option.

8 0

3 years ago

Read 2 more answers

I need help answering

jekas [21]

Put the values of the variables in place of the variables in the expression, then do the arithmetic.

$8p+3q-18\\\\=8\left(\dfrac{1}{2}\right)+3(7)-18=\dfrac{8}{2}+21-18\\\\=4+21-18\\\\=25-18\\\\=\bf{7}$

6 0

3 years ago

If x^2+1/x^2=3 find the value of x^2/(x^2+1)^2 Express answer as a common fraction. Thanks!!

timama [110]

$x^2+\dfrac1{x^2}=3\implies x^4+1=3x^2\implies x^4-3x^2+1=0$

By the quadratic formula,

$x^2=\dfrac{3\pm\sqrt5}2\implies x^2+1=\dfrac{5\pm\sqrt5}2$

Then

$(x^2+1)^2=\dfrac{25\pm10\sqrt5+5}4=\dfrac{15\pm5\sqrt5}2$

$\implies\dfrac{x^2}{(x^2+1)^2}=\dfrac{\frac{3\pm\sqrt5}2}{\frac{15\pm5\sqrt5}2}=\dfrac{3\pm\sqrt5}{15\pm5\sqrt5}$

Multiply numerator and denominator by the denominator's conjugate:

$\dfrac{3\pm\sqrt5}{15\pm5\sqrt5}\cdot\dfrac{15\mp5\sqrt5}{15\mp5\sqrt5}=\dfrac{45\pm15\sqrt5\mp15\sqrt5-25}{15^2-(5\sqrt5)^2}=\dfrac{20}{100}=\dfrac15$

3 0

3 years ago