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Vera_Pavlovna [14]
3 years ago
10

Whats the domain for this equation?

Mathematics
1 answer:
fgiga [73]3 years ago
7 0

Strictly speaking an equation does not have a domain. A function does. Though it may make sense to speak about a solution set to this equation over a domain.

The following could be helpful towards an answer: The equation has two sides. The left hand side is a non-linear function and the right hand side is a line with slope -1 and y-intercept of 1.

The function\log_5(6-5^x) has a domain over all Reals x for which it holds 6-5^x > 0 \\6 > 5^x\\\log_5 6 > x so the domain of the nonlinear function is -\infty.

The domain of the linear function 1-x is all Reals: -\infty

So, together you could say the equation's solutions are defined over the domain -\infty.

Lmk if you have questions.

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Could ΔABC be congruent to ΔADC by SSS? Explain.
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There are 5 ways to test if two figures are congruent, namely;
side-angle-side(SAS)
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Angle-angle-side(AAS)
Hypotenuse-leg (HL)
3 Sides (SSS)
here we shall focus on SSS. When the three corresponding sides of 2 figures say a triangle have the same length we will conclude that the triangles are congruent by SSS.
Therefore from our choices we can conclude that   triangles ABC and ADC are only congruent if the two other side have the same length and BC=DC.

The answer is yes;
B] Yes, but only if BC=DC
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Answer:

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Step-by-step explanation:

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The semicircle shown at left has center X and diameter W Z. The radius XY of the semicircle has length 2. The chord Y Z has leng
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{\bold{\red{\huge{\mathbb{QUESTION}}}}}

The semicircle shown at left has center X and diameter W Z. The radius XY of the semicircle has length 2. The chord Y Z has length 2. What is the area of the shaded sector formed by obtuse angle WXY?

\bold{ \red{\star{\blue{GIVEN }}}}

RADIUS = 2

CHORD = 2

RADIUS --> XY , XZ , WX

( BEZ THEY TOUCH CIRCUMFERENCE OF THE CIRCLES AFTER STARTING FROM CENTRE OF THE CIRCLE)

\bold{\blue{\star{\red{TO \:  \: FIND}}}}

THE AREA OF THE SHADED SECTOR FORMED BY OBTUSE ANGLE WXY.

\bold{  \green{ \star{ \orange{FORMULA \:  USED}}}}

AREA COVERED BY THE ANGLE IN A SEMI SPHERE

AREA = ANGLE   \: \: IN  \: \:  RADIAN  \times RADIUS

\huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}

Total Area Of The Semi Sphere:-

AREA =   \pi \times radius  \\  \\ AREA = \pi \times 2 = 2\pi

Area Under Unshaded Part .

Given a triangle with each side 2 units.

This proves that it's is a equilateral triangle which means it's all angles r of 60° or π/3 Radian

So AREA :-

AREA =  \frac{\pi}{3}  \times radius \\  \\ AREA =  \frac{\pi}{3}  \times 2 \\  \\ AREA =  \frac{2\pi}{3}

\green{Now:- } \\  \green{ \: Area  \: Under \:  Unshaded \:  Part }

Total Area - Area Under Unshaded Part

Area= 2\pi -  \frac{2\pi}{3}  \\ Area =  \frac{6\pi - 2\pi}{3}   \\ Area =  \frac{4\pi}{3}  \:  \: ans

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