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3 years ago

PLEASE ANSWER ASAP WILL GOVE BRAINLYIEST

Mathematics

1 answer:

kirza4 [7]3 years ago

5 0

Answer:

Proved that GB ≅ AH.

Step-by-step explanation:

See the attached diagram.

Statement 1: ∠ GBH ≅ ∠ AHB

Reason 1: This is given.

Statement 2: ∠ GHB ≅ ∠ ABH

Reason 2: This is also given.

Statement 3: BH ≅ HB.

Reason 3: From the diagram. Reflexive property of congruence.

Statement 4: Δ GBH ≅ Δ AHB

Reason 4: By Angle-Side-Angle or ASA criteria of congruency.

Statement 5: GB ≅ AH

Reason 5: Corresponding sides of two congruent triangles. (Proved)

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Evaluate the definite integral. <br> 1 x4(1 + 2x5)5 dx.

Nata [24]

Answer:

Step-by-step explanation:

Given the definite integral $\int\limits {\dfrac{x^4}{(1-2x^5)^5} } \, dx$ , we to evaluate it. Using integration by substitution method.

Let u = 1-2x⁵ ...1

du/dx = -10x⁴

dx = du/-10x⁴.... 2

Substitute equation 1 and 2 into the integral function and evaluate the resulting integral as shown;

$= \int\limits {\dfrac{x^4}{u^5} } \, \dfrac{du}{-10x^4}$

$= \dfrac{-1}{10} \int\limits {\dfrac{du}{u^5} } \\\\= \dfrac{-1}{10} \int\limits {{u^{-5}du } \\= \dfrac{-1}{10} [{\frac{u^{-5+1}}{-5+1}] \\\\= \dfrac{-1}{10} ({\frac{u^{-4}}{-4})\\\\$

$= \dfrac{u^{-4}}{40} \\\\\\= \dfrac{1}{40u^4} +C$

substitute u = 1-2x⁵ into the result

$= \dfrac{1}{40(1-2x^5)^4} +C$

Hence $\int\limits {\dfrac{x^4}{(1-2x^5)^5} } \, dx$ $= \dfrac{1}{40(1-2x^5)^4} +C$

5 0

3 years ago

Find exact values for sin θ, cos θ and tan θ if csc θ = 3/2 and cos θ < 0.

kumpel [21]

Answer:

Part 1) $sin(\theta)=\frac{2}{3}$

Par 2) $cos(\theta)=-\frac{\sqrt{5}}{3}$

Part 3) $tan(\theta)=-\frac{2\sqrt{5}}{5}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we have

$csc(\theta)=\frac{3}{2}$

Remember that

$csc(\theta)=\frac{1}{sin(\theta)}$

therefore

$sin(\theta)=\frac{2}{3}$

step 2

Find the $cos(\theta)$

we know that

$sin^{2}(\theta) +cos^{2}(\theta)=1$

we have

$sin(\theta)=\frac{2}{3}$

substitute

$(\frac{2}{3})^{2} +cos^{2}(\theta)=1$

$\frac{4}{9} +cos^{2}(\theta)=1$

$cos^{2}(\theta)=1-\frac{4}{9}$

$cos^{2}(\theta)=\frac{5}{9}$

square root both sides

$cos(\theta)=\pm\frac{\sqrt{5}}{3}$

we have that

$cos(\theta) < 0$ ---> given problem

$cos(\theta)=-\frac{\sqrt{5}}{3}$

step 3

Find the $tan(\theta)$

we know that

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

we have

$sin(\theta)=\frac{2}{3}$

$cos(\theta)=-\frac{\sqrt{5}}{3}$

substitute

$tan(\theta)=\frac{2}{3}:-\frac{\sqrt{5}}{3}=-\frac{2}{\sqrt{5}}$

Simplify

$tan(\theta)=-\frac{2\sqrt{5}}{5}$

6 0

3 years ago

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V125BC [204]

Answer:

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3 years ago

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ad-work [718]

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Step-by-step explanation:ik

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3 years ago

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stepan [7]

Answer:

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Step-by-step explanation:

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3 years ago