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2 years ago

Find the value of x. Round the answer to the nearest tenth, if needed.

Mathematics

2 answers:

wolverine [178]2 years ago

6 0

X/49 = 21/x
x^2 = 21(49) = 1029
x = √1029 = 32.1
x = 32.1

Paladinen [302]2 years ago

4 0

From the figure, we can set up three equations with three unknowns so that we can calculate the value x. The equations are as follows:

y^2 = x^2 + 49^2 --->
70^2 = z^2 + y^2
z^2 = 21^2 + x^2

Solving simultaneously, we obtain,
x = 7√21 = 32.1 <------- OPTION D

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Prove that if a and b are integers, then a^2-4b egal or non-egal 2

Lera25 [3.4K]

Answer:

tex]a^2 - 4b \neq 2[/tex]

Step-by-step explanation:

We are given that a and b are integers, then we need to show that $a^2 - 4b \neq 2$

Let $a^2 - 4b = 2$

If a is an even integer, then it can be written as $a = 2c$ , then,

$a^2 - 4b = 2\\(2c)^2 - 4b =2\\4(c^2 -b) = 2\\(c^2 -b) =\frac{1}{2}$

RHS is a fraction but LHS can never be a fraction, thus it is impossible.

If a is an odd integer, then it can be written as $a = 2c+1$ , then,

$a^2 - 4b = 2\\(2c+1)^2 - 4b =2\\4(c^2+c-b) = 2\\(c^2+c-b) =\frac{1}{4}$

RHS is a fraction but LHS can never be a fraction, thus it is impossible.

Thus, our assumption was wrong and $a^2 - 4b \neq 2$ .

7 0

3 years ago

69 points!!!

Amiraneli [1.4K]

Answer:

C) The solution for the given system of equations are A(0,-5) and B(-4,3)

Step-by-step explanation:

The given system of equation are : $x^{2} + y^{2} = 25\\2x + y = -5$

from equation 2, we get y = -5 - 2x .

Put the above value of y in the equation (1).

We get: $x^{2} + y^{2} = 25 \implies x^{2} + (-5-2x)^{2} = 25$

By ALGEBRAIC IDENTITY:

$(a+b)^{2} = a^{2} + b^{2} + 2ab\\ (-5-2x)^{2} = (-5)^{2} + (-2x)^{2} + 2(-5)(-2x)$

or, $x^{2} + (-5-2x)^{2} = 25 \implies x^{2} + (25 + 4x^{2} + 20x) = 25$

or, $5x^{2} + 20x = 0 \implies x(5x + 20) = 0$

⇒ x = 0 or, x = -20/5 = -4

So, the possible values for x are: x = 0 or x = -4

If x = 0, y = -5-2x = -5-2(0) = -5

and if x = -4, y = -5 -2(-4) = -5 + 8 = 3

Hence, the solution for the given system of equations are A(0,-5) and B(-4,3)

8 0

3 years ago

Q10 Q2.) Find the standard form of the equation of the parabola satisfying the given conditions.

Kipish [7]

Hello!!

Equation forms:

$y = \frac{x^2}{4} + \frac{x}{2} + \frac{13}{4}$

Therefore, $x^2 - 2x + 4y - 13 = 0$

And, this is how your graph would look

Good luck :)