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3 years ago

Yukio says the scale from DEF to ABC is 3:1. Is Yukio Correct? Explain

Mathematics

1 answer:

Angelina_Jolie [31]3 years ago

6 0

Answer: Yes, Yukio is correct.

Step-by-step explanation:

Assuming that Triangle DEF and ABC have the same angles (they do because they are right-angled), we can take the length from the larger triangle (DEF) and divide it by the length of the smaller triangle (ABC).

Length of DEF = 6cm

Length of ABC = 2cm

= 6/2

= 3

Proves that scale of DEF to ABC is 3:1

You might be interested in

Simplify the rational expression:<br> 5y+5/2 ÷ 25y-20/40y²-32y

NNADVOKAT [17]

Step-by-step explanation:

Your problem → 5y+5/2 / 25y-20/40y^2-32y

5y+5/2÷25y-20/40y^2-32y

=5⋅y+5/2÷25×y-20/40×y^2-32×y

=5y+5/2÷25×y-20/40×y^2-32y

=5y+5/2×1/25×y-20/40×y^2-32y

=5y+y/10-20/40×y^2-32y

=5y+y/10-y^2/2-32y

=5y×10+y-y^2×5-32y×10 ÷ 10

=50y+y-5y2-320y ÷ 10

= -269y-5y^2 / 10

4 0

2 years ago

The sum of 45% of a number and 55% of the same number is 200. what is the number?

Ymorist [56]

.45x+.55x=200
Combine like terms
1x=200
So x is 100

Or you can do 45%+55%=100%
And 100% of 200 is 200

8 0

3 years ago

Which transformation produces another triangle that has the same area as the one shown?

vfiekz [6]

Answer:

A reflection in the line y = -2, then a translation 2 units right.

Step-by-step explanation:

Transformation is the movement of a point from its initial location to a new location. Types of transformation are rotation, reflection, translation and dilation.

Rigid transformation are transformation that preserves the shape and size of the object. Rotation, reflection, translation are rigid transformation whereas dilation is not a rigid transformation.

Any transformation regarding dilation changes the side length of the object and hence the area.

From the question, only one option is right. The rest options are not right because they involve dilation. The correct option is:

A reflection in the line y = -2, then a translation 2 units right.

7 0

2 years ago

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$