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1 year ago

6

A three-dimensional representation of an object or structure having all parts in the same proportion of their true size is calle

d a(n) _______________.

1 answer:

Rasek [7]1 year ago

5 0

Answer:

Is called an Isometric projection

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Find the length of X in simplest radical form with a rational denominator

Darya [45]

Answer:

$x = \sqrt 6$

Step-by-step explanation:

Given

The attached triangle

Required

Find x

Considering angle 30 degrees,

$Opposite = \sqrt{2$

$Adjacent = x$

Using the tangent formula:

$tan\ 30 = \frac{\sqrt 2}{x}$

Make x the subject

$x = \frac{\sqrt 2}{tan\ 30}$

$x = \frac{\sqrt 2}{1/\sqrt 3}$

$x = \sqrt 2 * \sqrt 3$

$x = \sqrt 6$

6 0

2 years ago

scZoUnD [109]

Answer:

ok so just get the v12{{9-))0} and =14

Step-by-step explanation:

4 0

2 years ago

If you know that the exact volume of a cylinder is 500 π cm3, how can you find the approximate volume, without knowing the heigh

grandymaker [24]

Answer:

The approximate volume is $1,570\ cm^{3}$

Step-by-step explanation:

we know that

The exact volume of the cylinder is

$V=500\pi \ cm^{3}$

so

To find the approximate volume of the cylinder, substitute the term $\pi$ by the approximate value of $3.14$ and then multiply

so

$V=500(3.14)=1,570\ cm^{3}$

6 0

3 years ago

The simplified expression

Romashka-Z-Leto [24]

Answer:

$5x^2 y^2$

Step-by-step explanation:

We need to use the properties shown below to solve this:

1. $\sqrt[n]{x^a} =x^{\frac{a}{n}}$

2. $\sqrt{x}\sqrt{x} =x$

3. $\sqrt{x} \sqrt{y}=\sqrt{x*y}$

Area of a triangle is given by 1/2 * base * height, so we do that and simplify:

$A=\frac{1}{2}(\sqrt{5x^3} )(2\sqrt{5xy^4} )\\A=\frac{1}{2}(5x^3)^{\frac{1}{2}}*2*(5xy^4)^{\frac{1}{2}}\\A=\sqrt{5}x^{\frac{3}{2}}*\sqrt{5}\sqrt{x} } y^2\\A=\sqrt{5} \sqrt{5}x^{\frac{3}{2}} x^{\frac{1}{2}}y^2\\A=5*x^2y^2\\A=5x^2 y^2$

6 0

3 years ago

Geometry Question Help if you can please

docker41 [41]

Answer:

≈ 78.55 cm

Step-by-step explanation:

Using the ratio of arc (AB) to circumference (C) is equal to the ratio of angle subtended at centre by arc AB to 360°, that is

$\frac{AB}{C}$ = $\frac{110}{360}$ , that is

$\frac{24}{C}$ = $\frac{110}{360}$ ( cross- multiply )

110C = 8640 ( divide both sides by 110 )

C = $\frac{8640}{110}$ ≈ 78.55 cm ( to 2 dec. places )

5 0

2 years ago