Area of triangle: A=1/2bh; Solve for b.

2 answers:

4 0

As given

$A = \frac{1}{2} * b* h$

Firstly we will multiply both sides by 2.

$2*A = 2* \frac{1}{2} * b * h$

$2A = bh$

Now dividing both sides of equation from h.

$\frac{2A}{h} = \frac{bh}{h}$

$\frac{2A}{h} = b$

So b = $\frac{2A}{h}$

BARSIC [14]2 years ago

4 0

Answer: B=2a/h

Step-by-step explanation:

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Fraction calculator 66 1/2 - 43=

zhuklara [117]

Answer:

$\frac{47}{2}$

Step-by-step explanation:

66 $\frac{1}{2}$ - 43 = 47/2

6 0

2 years ago

Read 2 more answers

(ASAP PICTURE ADDED) What is the simplified form of the following expression?

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.

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

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:

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