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Digiron [165]
3 years ago
5

Consider the two triangles.

Mathematics
1 answer:
GuDViN [60]3 years ago
4 0

Answer:

(A)Show that the ratios StartFraction U V Over X Y EndFraction , StartFraction W U Over Z X EndFraction , and StartFraction W V Over Z Y EndFraction are equivalent.

\dfrac{UW}{XZ}=\dfrac{WV}{ZY}=\dfrac{UV}{XY}

Step-by-step explanation:

In Triangles WUV and XZY:

\angle VUW$ and \angle YXZ$ are congruent. \\\angle U W V$ and \angle X Z Y$ are congruent.\\ \angle U V W$ and \angle Z Y X$ are congruent.

Therefore:

\triangle UWV \cong  \triangle XZY

To show that the triangles are similar by the SSS similarity theorem, we have:

\dfrac{UW}{XZ}=\dfrac{WV}{ZY}=\dfrac{UV}{XY}

As a check:

\dfrac{UW}{XZ}=\dfrac{40}{32}=1.25\\\\\dfrac{WV}{ZY}=\dfrac{60}{48}=1.25\\\\\dfrac{UV}{XY}=\dfrac{50}{40}=1.25

The correct option is A.

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8 0
3 years ago
What is the true solution to the equation below?
Marina CMI [18]

Answer:

The solution is:

  • x=4

Step-by-step explanation:

Considering the expression

lne^{lnx}+lne^{lnx}^{^2}=2ln8

\ln \left(e^{\ln \left(x\right)}\right)+\ln \left(e^{\ln \left(x\right)\cdot \:2}\right)=2\ln \left(8\right)

\mathrm{Apply\:log\:rule}:\quad \:log_a\left(a^b\right)=b

\ln \left(e^{\ln \left(x\right)}\right)=\ln \left(x\right),\:\space\ln \left(e^{\ln \left(x\right)2}\right)=\ln \left(x\right)2

\ln \left(x\right)+\ln \left(x\right)\cdot \:2=2\ln \left(8\right)

\mathrm{Add\:similar\:elements:}\:\ln \left(x\right)+2\ln \left(x\right)=3\ln \left(x\right)

3\ln \left(x\right)=2\ln \left(8\right)

\mathrm{Divide\:both\:sides\:by\:}3

\frac{3\ln \left(x\right)}{3}=\frac{2\ln \left(8\right)}{3}

\ln \left(x\right)=\frac{2\ln \left(8\right)}{3}.....A

Solving the right side of the equation A.

\frac{2\ln \left(8\right)}{3}

As

\ln \left(8\right):\quad 3\ln \left(2\right)

Because

\ln \left(8\right)

\mathrm{Rewrite\:}8\mathrm{\:in\:power-base\:form:}\quad 8=2^3

⇒ \ln \left(2^3\right)

\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)

\ln \left(2^3\right)=3\ln \left(2\right)

So

\frac{2\ln \left(8\right)}{3}=\frac{2\cdot \:3\ln \left(2\right)}{3}

\mathrm{Multiply\:the\:numbers:}\:2\cdot \:3=6

          =\frac{6\ln \left(2\right)}{3}

\mathrm{Divide\:the\:numbers:}\:\frac{6}{3}=2

          =2\ln \left(2\right)

So, equation A becomes

\ln \left(x\right)=2\ln \left(2\right)

\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)

         =\ln \left(2^2\right)

         =\ln \left(4\right)

\ln \left(x\right)=\ln \left(4\right)

\mathrm{Apply\:log\:rule:\:\:If}\:\log _b\left(f\left(x\right)\right)=\log _b\left(g\left(x\right)\right)\:\mathrm{then}\:f\left(x\right)=g\left(x\right)          

x=4

Therefore, the solution is

  • x=4
6 0
3 years ago
Read 2 more answers
Q7Which statement is incorrect for the graph of thefunction y = -5 cosGcos( 3 (x – 3)) +62The period is 4.The amplitude is 3.The
guajiro [1.7K]

The given function is,

y=-5\cos (\frac{\pi}{2}(x-3))+6

The graph can be drawn as,

The period will be distance between two consequetive maximas,

\text{Period}=1-(-3)=4

Thus, period is correct.

The amplitude can be determined as,

\begin{gathered} Amplitude=\frac{\max -\min }{2} \\ =\frac{11-1}{2} \\ =5 \end{gathered}

Thus, amplitue is incorrect.

The range is,

y\in\lbrack1,11\rbrack

Thus, the range is correct.

The midline is,

\begin{gathered} \text{midline}=\frac{\max +\min }{2} \\ =\frac{1+11}{2} \\ =6 \end{gathered}

Thus, the midline is correct.

Thus, option (b) is the solution.

5 0
1 year ago
Is x = −8 a solution of the equation 3(x + 1) = 3x + 3?
n200080 [17]

Answer:

Yes

Step-by-step explanation:

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5 0
2 years ago
Read 2 more answers
What is 0.800 rounded to nearest tenth
arsen [322]
0.800 rounded to the nearest tenth is0.8
5 0
3 years ago
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