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pentagon [3]
3 years ago
9

Anyone good at composite functions??

Mathematics
1 answer:
sammy [17]3 years ago
6 0

\bf \begin{cases} ~\hfill f(x) = &-3x+4\\ ~\hfill g(x) = &x^2\\ (g \circ f)(0)=&g(~~f(0)~~) \end{cases} \\\\[-0.35em] ~\dotfill\\\\ f(0) = -3(0)+4\implies \boxed{f(0) = 4} \\\\\\ g(~~f(0)~~)\implies g\left( \boxed{4} \right) = (4)^2\implies \stackrel{(g \circ f)(0)}{g(4)} = 16

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Given: AB || DE , AD bisects BE.<br> Prove: ABC = DEC using the ASA postulate.
Ket [755]

Answer:

As per ASA postulate, the two triangles are congruent.

Step-by-step explanation:

We are given two triangles:

\triangle ABC and \triangle DEC.

AD bisects BE.

AB || DE.

Let us have a look at two properties.

1. When two lines are parallel and a line intersects both of them, then <em>alternate angles </em>are equal.

i.e. AB || ED and \angle B and \angle E are alternate angles \Rightarrow \angle B = \angle E.

2. When two lines are cutting each other, angles formed at the crossing of two, are known as <em>Vertically opposite angles </em>and they are are <em>equal</em>.

\Rightarrow \angle ACB = \angle DCE

Also, it is given that <em>AD bisects BE</em>.

i.e. EC = CB

1. \angle B = \angle E

2. EC = CB

3. \angle ACB = \angle DCE

So, we can in see that in \triangle ABC and \triangle DEC, two angles are equal and side between them is also equal to each other.

Hence, proved that \triangle ABC \cong \triangle DEC.

8 0
3 years ago
If sin x = 3/5, calculate tan x​
ZanzabumX [31]

Answer:

tan ( x )  =  ±  3 /4

Step-by-step explanation:

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3 years ago
Is 3.14512 a rational number
zvonat [6]
No it is not because you can't make it into a fraction.
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What whole number most closely estimates to <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B50%7D%20" id="TexFormula1" title=
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Answer:

7

Step-by-step explanation:

since \sqrt{50}>\sqrt{49}=7

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3 years ago
(5 pts) A quadratic function f is given by f(x) = ax2 + bx + c where a is not 0. Select all
Vsevolod [243]

Answers:

  • a) True. Plug in x = 0 and it leads to y = c. Therefore, the point (0,c) is on the parabola.
  • b) False. Plug in y = 0 and apply the quadratic formula. One x intercept may sometimes be x = c, but it could easily be other values as well. Or perhaps you may not get any real number solutions at all. See part d) below.
  • c) True. If a < 0, then the leading coefficient is negative. Overall, both endpoints will tend toward negative infinity to produce a parabola that opens downward.
  • d) False. The quadratic may have one x intercept or it may not have any x intercepts at all. It depends on what the discriminant d = b^2 - 4ac is equal to. If d < 0, then we have no x intercepts. If d = 0, then we have exactly 1 x intercept. If d > 0, then we have two different x intercepts.
  • e) True. The vertex's x coordinate is -b/(2a). If b = 0, then the x coordinate of the vertex is 0. The vertical line x = 0 is directly on top of the y axis.
8 0
3 years ago
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