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3 years ago

How do you solve 1 1/2 × ( − 6/15 ) =?

Mathematics

1 answer:

Andreas93 [3]3 years ago

8 0

Answer:

1.5 x -0.4 = -0.6 or -6/10 or -3/5

Step-by-step explanation:

1 1/2= 3/2x(-6/15)

3/2 = 1.5 -6/15 = 0.4

1.5 x -0.4 = -0.6 or -6/10 or -3/5

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A triangle has a base of x 1/2 m and a height of x 3/4 m. If the area of te triangle is 16m to the power of 2, what are the base

Olenka [21]

Answer:

The base of triangle is $\frac{8}{\sqrt{3}} \ m$ and the height of triangle is $\frac{12}{\sqrt{3}} \ m.$

Step-by-step explanation:

Given:

A triangle has a base of x 1/2 m and a height of x 3/4 m. If the area of the triangle is 16m to the power of 2.

Now, to find the base and height of the triangle.

The base of triangle = $x\times\frac{1}{2} =\frac{x}{2} \ m.$

The height of triangle = $x\times \frac{3}{4} =\frac{3x}{4}\ m.$

The area of triangle = $16\ m^2.$

Now, we put the formula of area to solve:

$Area=\frac{1}{2} \times base\times height$

$16=\frac{1}{2} \times \frac{x}{2} \times \frac{3x}{4}$

$16=\frac{3x^2}{16}$

Multiplying both sides by 16 we get:

$256=3x^2$

Dividing both sides by 3 we get:

$\frac{256}{3} =x^2$

Using square root on both sides we get:

$\frac{16}{\sqrt{3}}=x$

$x=\frac{16}{\sqrt{3}}$

Now, by substituting the value of $x$ to get the base and height:

$Base=\frac{x}{2}\\\\Base=\frac{\frac{16}{\sqrt{3}}}{2} \\\\Base=\frac{8}{\sqrt{3}} \ m.$

So, the base of triangle = $\frac{8}{\sqrt{3}} \ m.$

$Height=x\times\frac{3}{4} \\\\Height=\frac{16}{\sqrt{3}}\times \frac{3}{4} \\\\Height=\frac{12}{\sqrt{3}} \ m.$

Thus, the height of triangle = $\frac{12}{\sqrt{3}} \ m.$

Therefore, the base of triangle is $\frac{8}{\sqrt{3}} \ m$ and the height of triangle is $\frac{12}{\sqrt{3}} \ m.$

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