The number of books read by a mathematician 2 years after he turned 25 is given by the polynomial

Mathematics

1 answer:

stich3 [128]2 years ago

5 0

Evaluating a function means to plug the required value, substituting every x occurrence with that value.

So, you have

$f(x)=3x^2+80x \implies f(2) = 3\cdot 2^2+80\cdot 2 = 3\cdot 4 + 160 = 12+160=172$

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Pls Help! John wants to prove ABC = to DEF. He Knoes AB = DE and AC = DF. With Screenshot

matrenka [14]

Answer:

$\angle A \cong \angle D$

Step-by-step explanation:

The Side-Angle-Side method cana only be used when information given shows that an included angle which is between two sides of a ∆, as well as the two sides of the ∆ are congruent to the included side and two sides of the other ∆.

Thus, since John already knows that $\overline{AB} \cong \overline{DE}$ and $\overline{AC} \cong \overline{DF}$ , therefore, an additional information showing that the angle between $\overline{AB}$ and $\overline{DE}$ in ∆ABC is congruent to the angle between $\overline{AC}$ and $\overline{DF}$ in ∆DEF.

For John to prove that ∆ABC is congruent to ∆DEF using the Side-Angle-Side method, the additional information needed would be $\angle A \cong \angle D$ .

See attachment for the diagram that has been drawn with the necessary information needed for John to prove that ∆ABC is congruent to ∆DEF.

7 0

2 years ago

Help me out please!!!!

DedPeter [7]

Answer:

Answer (e)

Step-by-step explanation:

Note that this graph has a zero at x = 1 and a vertical asymptote at x = -4. Only Answer (e) corresponds to that. Try letting x = 1. y is zero, correct? Try letting x = -4. y is undefined, correct?

7 0

2 years ago

AD is a diameter of a circle and AB is a chord. If AD=34cm, AB=30cm, the distance of AB from the centre of the circle is :

e-lub [12.9K]

Answer:

The distance is 8 cm

Step-by-step explanation:

The chord and the diameter form one leg and the hypotenuse of a right triangle. The other leg, BD, has length ...

BD² +AB² = AD²

BD² = AD² -AB² = 34² -30² = 256

BD = √256 = 16

The segment from the center of the circle to the midpoint of the chord is the midline of triangle ABD, so is half the length of BD.