What is the perimeter

Mathematics

1 answer:

Art [367]3 years ago

6 0

Answer:

989cm

Step-by-step explanation:

103, 131, 149, 111

Bottom length = 103 + 149 = 253

Left height = 131 + 111 = 242

103 + 131 + 149 + 111 + 253 + 242 = 989

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Given circle U with diameter ST and radius UQ. QR is tangent to U at Q. If

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Step-by-step explanation:

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bonufazy [111]

Answer:

option c is correct.

Step-by-step explanation:

$7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{16x}\right)-3\left(\sqrt[3]{8x}\right)$

WE need to simplify this equation.

Solve the parenthesis of each term.

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Now, We will find factors of the terms inside the square root

factors of 2: 2

factors of 16 : 2x2x2x2

factors of 8: 2x2x2

Putting these values in our equation: $=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2X2X2X2 x}\right)-3\left(\sqrt[3]{2X2X2 x}\right)\\=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2X2X2} \sqrt[3] {2 x}\right)-3\left(\sqrt[3]{2X2X2} \sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2^3} \sqrt[3] {2 x}\right)-3\left(\sqrt[3]{2^3} \sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2x}\right)-3*2\left(\sqrt[3] {2 x}\right)-3*2\left(\sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2}\sqrt[3]{x}\right)-6\left(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)$

Adding like terms we get:

$=7\left(\sqrt[3]{2}\sqrt[3]{x}\right)-6\left(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right\\=(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)\\$