describe how to find the value of 3 the the ninth power using the fact that three to the eighth power equals 6561

Mathematics

1 answer:

9966 [12]3 years ago

5 0

Hello,

$3^9=3*3^8=3*6561=19683$

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Efectuati calculele:

NNADVOKAT [17]

$\sqrt{(5 - 1)^{2}} + \sqrt{20}$

$\sqrt{4^{2}} + \sqrt{4 * 5}$

$\sqrt{16} + \sqrt{4} \sqrt{5} 
$

$4 + 2 \sqrt{5} 
$

$4\sqrt{5} + 2 \sqrt{5} 
$

$6 \sqrt{5}$

$\sqrt{(10 - 3)^{2}} + \sqrt{20 * 18}$

$\sqrt{7^{2}} + \sqrt{4 * 9 * 5 * 2}$

$\sqrt{49} + \sqrt{36 * 10}$

$7 + \sqrt{36}\sqrt{10}$

$7 + 6\sqrt{10}$

$7\sqrt{10} + 6\sqrt{10}$

$13\sqrt{10}$

$\sqrt[3]{(3 + 1)^{2}} - \sqrt{(3 - 2)x^{2}$

$\sqrt[3]{4^{2}} - \sqrt{x^{2}$

$\sqrt[3]{16} - x[tex] \\[tex] \sqrt[3]{8 * 2} - x$

$\sqrt[3]{8} \sqrt[3]{2} - x
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$2\sqrt[3]{2} - x$

8 0

3 years ago

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.