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2 years ago

WILL MARK AS BRAINLYIEST If x1 and x2 are solutions to ax2+bx+c=0 where a≠0, then x1+x2=

Mathematics

1 answer:

lesya692 [45]2 years ago

8 0

Answer:

x1 + x2 = -b/a

The answer is based on "Vieta's Formulas" for the sum and product of roots. In other words, in its simplest form, it states that:

for ax^2 + bx + c = 0

x1 + x2 = -b/a
x1 × x2 = c/a

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If 6^4x = 82, what is the value of x? Round your answer to three places.

neonofarm [45]

Answer:

$x=0.61485921$

Step-by-step explanation:

The given equation is in exponential form:

$6^{4x} =82$

To find $x$ we need to write it in logarithmic form:

$b^{a} =n$ ⇒ $log_{b} n=a$

$6^{4x} =82$ ⇒ $log_{ 6 } 82=4x$

$2.45943684 = 4x$

$\frac{2.45943684}{4} = x$

$0.61485921 =x$

To prove this, we can substitute the value into the given equation

$6^{4x} =82$

$6^{4(0.61485921)} =82$

$82=82$

Therefore, the answer is $x=0.61485921$

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Step-by-step explanation:

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Read 2 more answers

The figures below are made out of circles, semicircles, quarter circles, and a square. Find the area and the perimeter of each f

WITCHER [35]

Answer:

Part 1) $A=36(\pi-2)\ cm^2$

Part 2) $P=6(\pi+2\sqrt{2})\ cm$

Step-by-step explanation:

The picture of the question in the attached figure

Part 1) Find the area

we know that

The area of the shape is equal to the area of a quarter of circle minus the area of an isosceles right triangle

$A=\frac{1}{4}\pi r^{2}-\frac{1}{2}(b)(h)$

we have that the base and the height of triangle is equal to the radius of the circle

$r=12\ cm\\b=12\ cm\\h=12\ cm$

substitute

$A=\frac{1}{4}\pi (12)^{2}-\frac{1}{2}(12)(12)\\A=(36\pi-72)\ cm^2$

simplify

Factor 36

$A=36(\pi-2)\ cm^2$

Part 2) Find the perimeter

The perimeter of the figure is equal to the circumference of a quarter of circle plus the hypotenuse of the right triangle

The circumference of a quarter of circle is equal to

$C=\frac{1}{4}(2\pi r)=\frac{1}{2}\pi r$

substitute the given values

$C=\frac{1}{2}\pi (12)\\C=6\pi\ cm$

Applying the Pythagorean Theorem

The hypotenuse of right triangle is equal to

$AC=\sqrt{12^2+12^2}\\AC=\sqrt{288}\ cm$

simplify

$AC=12\sqrt{2}\ cm$

Find the perimeter

$P=(6\pi+12\sqrt{2})\ cm$

simplify

Factor 6

$P=6(\pi+2\sqrt{2})\ cm$

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