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2 years ago

A wire of length 7x is bent into the shape of a square. Express the area A of the square as a function of z.

Mathematics

1 answer:

Nostrana [21]2 years ago

3 0

7×

4+4+4+4

4+4=8

8+8=16

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ROOTS AND CUBES Can someone please help me answer all 4 following?

Alex Ar [27]

NOTES:

squared (²) means multiply that number by itself 2 times
cubed (³) means multiply that number by itself 3 times
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Answer: (C) 41

Step-by-step explanation:

$\quad 6^2+\sqrt[3]{125} \\= 6 \cdot 6+\sqrt[3]{5\cdot 5 \cdot 5}\\= 36 + 5\\= 41$

***************************************************************************************

Answer: (C) 10

Step-by-step explanation:

$\bigg(\dfrac{7}{3}\times \sqrt[3]{27}-2\bigg)\times \dfrac{1}{5} + \sqrt{81}$

$=\bigg(\dfrac{7}{3}\times \sqrt[3]{3\cdot 3 \cdot 3}-2\bigg)\times \dfrac{1}{5} + \sqrt{9\cdot 9}$

$=\bigg(\dfrac{7}{3}\times3-2\bigg)\times \dfrac{1}{5}+9$

$=(7 - 2)\times \dfrac{1}{5}+9$

$=5 \times \dfrac{1}{5}+ 9$

$= 1 + 9$

$= 10$

***************************************************************************************

8² = 8 · 8 = 64 11² = 11 · 11 = 121

5³ = 5 · 5 · 5 = 125 3³ = 3 · 3 · 3 = 27

$\sqrt{1600}=\sqrt{40\cdot 40}=40$ $\sqrt[3]{64}=\sqrt[3]{4\cdot 4\cdot 4}=4$

$\sqrt{144}=\sqrt{12\cdot 12}=12$ $\sqrt[3]{8000}=\sqrt[3]{20\cdot 20\cdot 20}=20$

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2 years ago

CAN I GET SOMEONES HELP ASAP!!

PIT_PIT [208]

The answer would be <QRZ

Since you are looking for an angle congruent to <UQR using the alternate interior angles theorem, interior suggests that the angle must be inside the parallel lines, se we can get rid of options <WRT and <TRZ since they are exterior angles

Furthermore, in the alternate interior angles theorem, the two angles must be alternate or opposite of each other which would show that the only possible answer would be <QRZ

hope this helped!

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2 years ago