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3 years ago

15

13. Justify Conclusions Determine whether each statement is sometimes,

1 answer:

Helga [31]3 years ago

6 0

Answer:

False. It is not possible for a triangle to have two right angles.

Step-by-step explanation:

Every triangle has 3 sides and, therefore, 3 angles (hence its name "tri" (three) "angle" (angles). Always, the sum of the three angles of a triangle will have a value of 180º. Therefore If two of the angles of a triangle were right, that is, they measured 90º, the third angle would be 0º, that is, it would be non-existent.Therefore, it is impossible for two angles of a triangle to be right angles.

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What is the sum of (3+6)^2 and 3^3

Mademuasel [1]

(3+6)^2 + 3^3 =
9^2 + 3^3 =
81 + 27 =
108

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3 years ago

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The radius of a circle is 13 in. Find its area in terms of <br> π<br><br><br> Please help

Sunny_sXe [5.5K]

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Step-by-step explanation: 13x13xπ

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3 years ago

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3 years ago

Twice the difference of a number and four is less than the sum of the number and five.

Whitepunk [10]

Let the number be x.

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3 years ago

If f(x)=x squared-2x and g(x) = 6x+4 for which is a value of x does (f+g)(x) =0

ra1l [238]

Answer:

For $x = -2$

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}$

$\Delta = b^{2} - 4ac$

In this question:

$f(x) = x^2 - 2x$

$g(x) = 6x + 4$

$f(x) + g(x) = (f+g)(x) = x^2 - 2x + 6x + 4 = x^2 + 4x + 4$

Which is a value of x does (f+g)(x) =0

$x^2 + 4x + 4$

Quadratic equation with $a = 1, b = 4, c = 4$

So

$\Delta = 4^{2} - 4*1*4 = 0$

$x_{1} = \frac{-4 + \sqrt{0}}{2} = -2$

$x_{2} = \frac{-4 - \sqrt{0}}{2} = -2$

For $x = -2$

7 0

3 years ago