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olya-2409 [2.1K]

3 years ago

13

How do you find out if the scale factor of a triangle is similar?

1 answer:

Norma-Jean [14]3 years ago

6 0

When two triangles<span> are </span>similar<span>, the reduced ratio of any two corresponding sides is called the </span>scale factor<span> of the </span>similar triangles. In Figure 1, Δ ABC∼ Δ DEF. Figure 1Similar triangles<span> whose </span>scale factor<span> is 2: 1. The ratios of corresponding sides are 6/3, 8/4, 10/5.

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The base of a triangle is 7 cm greater than the height. the area is 60 cm squared. find the height and base of the triangle.

tino4ka555 [31]

Base=x+7 Height=x Area=60. 60=((x+7)(x))/2----->60=(x^2+7x)/2

120=x^2+7x X^2+7X-120=0. QUDRATIC FORMULA: -7+/- SQRT(49+480)/2

(-7+/-23)/2... 8 and -15. -15 ISNT the answer because you cant have a negative length. X=8 Base=8+7---->15 Height=8

8 0

3 years ago

Read 2 more answers

3(2c-8) - 10c > 0 can some one explain how to solve this ?

Jet001 [13]

Answer:

c < - 6

Step-by-step explanation:

Given

3(2c - 8) - 10c > 0 ← distribute and simplify left side

6c - 24 - 10c > 0

- 4c - 24 > 0 ( add 24 to both sides )

- 4c > 24

Divide both sides by - 4, reversing the inequality symbol as a consequence of dividing by a negative quantity.

c < - 6

8 0

3 years ago

−4+8x)+3>x−4?<br><br> x>16<br><br> x<−1<br><br> x<−323<br><br> x>433

pentagon [3]

Answer:

x > 4/33

Step-by-step explanation:

(-4 +8x) +3> x/-4

8x - 1 > x/-4

-32x + 4 < x

4 < 33x

4/33 < x

6 0

2 years ago

What is a square root of (a-b)^2

Bess [88]

Hello,

$\sqrt{(a-b)^2}=|a-b| \\ 
= a-b\ if\ a \ \textgreater\ b \\
=b-a\ if\ a\ \textless \ b$

4 0

2 years ago

Calculate s f(x, y, z) ds for the given surface and function. g(r, θ) = (r cos θ, r sin θ, θ), 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π; f(x, y, z)

Triss [41]

$g(r,\theta)=(r\cos\theta,r\sin\theta,\theta)\implies\begin{cases}g_r=(\cos\theta,\sin\theta,0)\\g_\theta=(-r\sin\theta,r\cos\theta,1)\end{cases}$

The surface element is

$\mathrm dS=\|g_r\times g_\theta\|\,\mathrm dr\,\mathrm d\theta=\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

and the integral is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\int_0^{2\pi}\int_0^4((r\cos\theta)^2+(r\sin\theta)^2)\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

$=\displaystyle2\pi\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac\pi4(132\sqrt{17}-\sinh^{-1}4)$

###

To compute the last integral, you can integrate by parts:

$u=r\implies\mathrm du=\mathrm dr$

$\mathrm dv=r\sqrt{1+r^2}\,\mathrm dr\implies v=\dfrac13(1+r^2)^{3/2}$

$\displaystyle\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac r3(1+r^2)^{3/2}\bigg|_0^4-\frac13\int_0^4(1+r^2)^{3/2}\,\mathrm dr$

For this integral, consider a substitution of

$r=\sinh s\implies\mathrm dr=\cosh s\,\mathrm ds$

$\displaystyle\int_0^4(1+r^2)^{3/2}\,\mathrm dr=\int_0^{\sinh^{-1}4}(1+\sinh^2s)^{3/2}\cosh s\,\mathrm ds$

$\displaystyle=\int_0^{\sinh^{-1}4}\cosh^4s\,\mathrm ds$

$=\displaystyle\frac18\int_0^{\sinh^{-1}4}(3+4\cosh2s+\cosh4s)\,\mathrm ds$

and the result above follows.

4 0

3 years ago