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raketka [301]
4 years ago
6

Solve for x in the equation x^2 + 20x + 100 =36.

Mathematics
2 answers:
Stels [109]4 years ago
7 0

Answer:

The value of x for the equation x^{2}+20 x+100=36 is x=-16 or x=-4

<u>Step by Step Explanation:</u>

Given equation x^{2}+20 x+100=36

To find:

Value of x for the above equation

Solution:

We know that the given equation is x^{2}+20 x+100=36

From the options we shall evaluate the value of x by substituting in the above equation.

Option A: x=-16 or x=-4

Substitute these value of x in the given equation x^{2}+20 x+100=36 we get

(-16)^{2}+20(-16)+100=36

256-320+100=36

-64+100=36

36=36

Thus option A gives the correct value of x to the above equation.

Result:

Value of x=-16 or x=-4 is the correct answer for the above given equation.

arsen [322]4 years ago
5 0

Answer:

x = -16 or x = -4 ........

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Any triangle with one angle equal to 90⁰ produces a Pythagoras triangle and the Pythagoras equation can be applied in the triangle.

<h3>What is the Pythagoras Theorem?</h3>

The Pythagoras theorem states that if a triangle is right-angled (90 degrees), then the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In △ABD and △ACB,

∠A = ∠A (common)

∠ADB = ∠ABC (both are right angles)

Thus, △ABD ∼ △ACB (by AA similarity criterion)

Similarly, we can prove △BCD ∼ △ACB.

Thus △ABD ∼ △ACB,

Therefore, AD/AB = AB/AC.

We can say that AD × AC = AB².

Similarly, △BCD ∼ △ACB.

Therefore,

CD/BC = BC/AC.

We can also say that

CD × AC = BC².

Adding these 2 equations, we get

AB² + BC² = (AD × AC) + (CD × AC)

AB² + BC² =AC(AD +DC)

AB² + BC² = AC²

Hence proved

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Learn more about Pythagoras theorem from:

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2 years ago
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Answer:

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Step-by-step explanation:

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2x < 6

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The number line is indicated in the attached picture.

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we have been asked to find the sum of the series

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As we know that a geometric series has a constant ratio "r" and it is defined as

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The first term of the series is a_1=\left(\frac{1}{3}\right)^{1-1}=1

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S_n=a_1\frac{1-r^n}{1-r}

Plugin the values we get

S_5=1\cdot \frac{1-\left(\frac{1}{3}\right)^5}{1-\frac{1}{3}}

On simplification we get

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