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3 years ago

Which of the following statements is always true of similar polygons?

Mathematics

2 answers:

UkoKoshka [18]3 years ago

7 0

1. D

2. A

3. B

4. A

5. C

k0ka [10]3 years ago

6 0

The correct answer among all the other choices is D. both (a) and (b). The statements that are always true of similar polygons are corresponding angles of similar figures have the same measure and the lengths of corresponding sides form equivalent ratios. Thank you for posting your question. I hope this answer helped you. Let me know if you need more help.

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Solve quadratic equation 10x^2+x-2=0

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Answer:

$\{-\frac{1}{2} \} \cup \{\frac{2}{5}\}$

Step-by-step explanation:

$10x^2 + x - 2 = 0; \\ a = 10, b = 1, c = -2; \\ D = b^2 - 4ac = 1^2 - 4 * 10 * (-2) = 1 + 80 = 81 = 9^2, > 0; \\ x_{1, 2} = \frac{-b \pm \sqrt{D}}{2a} = \frac{-1 \pm \sqrt{9^2}}{2 * 10} = \frac{-1 \pm 9}{20} = \left \ [ {{\frac{-1 - 9}{20} = -\frac{10}{20} = -\frac{1}{2} } \atop {\frac{-1 + 9}{20} = \frac{8}{20} = \frac{2}{5}}} \right.$

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Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that th

jonny [76]

Answer:

The answer is $\sqrt{\frac{6}{5}}$

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

$\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy$

Solving:

$\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy$

$[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.$

Replacing the limits:

$6*\frac{1}{3} =2$

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ $\frac{1}{m}$

Solving the double integral with these new limits we have:

$\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy$

This part is a little bit tricky so let's solve the integral first for dy:

$\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx$

Replacing the limits:

$\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx$

Solving now for dx:

$[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.$

Replacing the limits:

$\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}$