21/5 as a whole number

Mathematics

2 answers:

Softa [21]3 years ago

4 0

4.2 is the answer of this question you are asking ✌️

mylen [45]3 years ago

3 0

4.2 is the answer. Hoped this helped a bit!

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Use the quadratic formula to solve the equation. -2x^2-x+7=0.

Furkat [3]

<span>-2x^2-x+7=0
Variable with the highest degree's (exponent) constant, -2 is a, next variable's constant, -1 is b, the constant or number without a variable, 7 is c

using substitution put the numbers into the formula
</span>(-b±√(b^(2)-4ac))/(a^(2))

(-(-1)±√((-1)^(2)-4(-2)(7))/((-2)^(2)) simplify

(1±√(1+56))/4

1±√(57)/4 is your answer

8 0

3 years ago

Please HELP! I DON'T KNOW HOW CAN I SOLVE THIS

Wewaii [24]

Answer:

Step-by-step explanation:

The Pythagorean Theorem:

$leg^2+leg^2=hypotenuse^2\to a^2+b^2=c^2$

$41.\\\\a=1,\ b=1,\ c=x\\\\x^2=1^2+1^2\\\\x^2=1+1\\\\x^2=2\to x=\sqrt2$

$42.\\a=1, b=2,\ c=x\\\\x^2=1^2+2^2\\\\x^2=1+4\\\\x^2=5\to x=\sqrt5$

$43.\\a=3,\ b=x,\ c=5\\\\3^2+x^2=5^2\\\\9+x^2=25\qquad\text{subtract 9 from both sides}\\\\x^2=16\to x=\sqrt{16}\to x=4$

$44.\\\\a=4,\ b=x,\ c=8\\\\4^2+x^2=8^2\\\\16+x^2=64\qquad\text{subtract 16 from both sides}\\\\x^2=48\to x=\sqrt{48}\\\\x=\sqrt{16\cdot3}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\x=\sqrt{16}\cdot\sqrt3\\\\x=4\sqrt3$