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3 years ago

Help me please ! Thank you so so much :)

Mathematics

1 answer:

notka56 [123]3 years ago

5 0

#4 = B
Explanation= Because C&A are the same angle measurement, and the 90 angle is along a straight line making the other angle 90, and the triangles share the same side the postulate you can use is AAS because you can prove that the triangles share 2 common angles and 1 common side

#5 = B
Explanation= Because the triangles already have 1 congruent angle, and 1 congruent side the only thing left we can prove is that because the triangles mirror each other, they have congruent opposite angles (may be called vertical angles) as well so they have 2 angles and 1 side in common which tells you to use AAS

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10 months ago

What is one-fourth of 24

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Read 2 more answers

Suppose h(x)=3x-2 and j(x) = ax +b. Find a relationship between a and b such that h(j(x)) = j(h(x))

sergejj [24]

Answer:

$\displaystyle a = \frac{1}{3} \text{ and } b = \frac{2}{3}$

Step-by-step explanation:

We can use the definition of inverse functions. Recall that if two functions, f and g are inverses, then:

$\displaystyle f(g(x)) = g(f(x)) = x$

So, we can let j be the inverse function of h.

Function h is given by:

$\displaystyle h(x) = y = 3x-2$

Find its inverse. Flip variables:

$x = 3y - 2$

Solve for y. Add:

$\displaystyle x + 2 = 3y$

Hence:

$\displaystyle h^{-1}(x) = j(x) = \frac{x+2}{3} = \frac{1}{3} x + \frac{2}{3}$

Therefore, a = 1/3 and b = 2/3.

We can verify our solution:

$\displaystyle \begin{aligned} h(j(x)) &= h\left( \frac{1}{3} x + \frac{2}{3}\right) \\ \\ &= 3\left(\frac{1}{3}x + \frac{2}{3}\right) -2 \\ \\ &= (x + 2) -2 \\ \\ &= x \end{aligned}$

And:

$\displaystyle \begin{aligned} j(h(x)) &= j\left(3x-2\right) \\ \\ &= \frac{1}{3}\left( 3x-2\right)+\frac{2}{3} \\ \\ &=\left( x- \frac{2}{3}\right) + \frac{2}{3} \\ \\ &= x \stackrel{\checkmark}{=} x\end{aligned}$

3 0

2 years ago

If a quadratic equation has a real root, what do you know about the other roots of the equation? Explain.

notsponge [240]

Answer:

A Quadratic Equation can have upto 2 roots maximum. So,if one of the roots is a Real number, there are following two possibilities:

1) The other root is also a real number, but a different number

2) Its a repeated root, so the other root is the same number.

The other root cannot be a complex number as its not possible for one root to be real and other to be complex. Either no root will be complex or both will be complex roots.

Following are 3 possibilities for the roots of a quadratic equation:

2 Real and Distinct roots
2 Real and Equal roots
2 Complex roots

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3 years ago