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4 years ago

X^2-8x+5=0 Solve by completing the square.

Mathematics

1 answer:

erastovalidia [21]4 years ago

8 0

$x^2-8x+5=0 \\
x^2-8x+16-11=0\\
(x-4)^2=11\\
x-4=-\sqrt{11} \vee x-4=\sqrt{11}\\
x=4-\sqrt{11} \vee x=4+\sqrt{11}\\$

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For geometry:(<br><br> i’ll give a brainist!!!

kvasek [131]

<h3>Answer: Angle Q = 133 degrees</h3>

========================================================

Work Shown:

Recall that for any triangle, the three angles always add to 180

P+Q+R = 180

(x+13) + (10x+13) + (2x-2) = 180

(x+10x+2x) + (13+13-2) = 180

13x+24 = 180

13x = 180-24

13x = 156

x = 156/13

x = 12

Now that we know x, we can find the angle measures

angle P = x+13 = 12+13 = 25 degrees
angle Q = 10x+13 = 10*12+13 = 120+13 = 133 degrees
angle R = 2x-2 = 2*12-2 = 24-2 = 22 degrees

As a way to check if we have the right answer or not, we see that,

P+Q+R = 25+133+22 = 180

So the answer is confirmed.

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3 years ago

Which is a correct first step in solving the inequality -4(2x - 1) > 5 - 3x?

Genrish500 [490]

Answer:

Distribute -4 to get -8x + 4 > 5 - 3x.

Step-by-step explanation:

Order of operations

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3 years ago

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How my calories are in 48 oz if tea

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Answer:

2 calories

Step-by-step explanation:

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4 years ago

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All three sides of a triangle are initially 4 m in length. One of the triangle's sides is oriented horizontally. The triangle is

Arlecino [84]

Answer:

the new height of the triangle = 2.449

Step-by-step explanation:

Given that:

the sides of the triangle are 4m in length i.e they are equal. It shows that the triangle is known to be an equilateral triangle.

Let say the triangle is a triangle IJK

Let the length of the side to be i = 4

Definitely

IJ = JK = IK = i = 4

If a midline is drawn and cuts the equilateral triangle in two equal halves of a right-angle triangle. Then, suppose the midline is L

Then ;

$JL = \dfrac{1}{2} JK$

$JL = \dfrac{1}{2} i$

Let consider triangle IJL

(IL)² = (IJ)² - (JL)²

$(IL)^2 = i^2 - \dfrac{i^2}{4}$

$(IL)^2 = \dfrac{3i^2}{4}$

$(IL) =\sqrt{ \dfrac{3i^2}{4}}$

$IL =\sqrt{3} \dfrac{i}{2}$

Area of triangle IJK can be expressed as:

$Area = \dfrac{1}{2}\times Base \times Height$

$Area = \dfrac{1}{2}\times i \times \sqrt{3}\dfrac{i}{2}$

$Area = \sqrt{3}\dfrac{i^2}{4}$

where, i = 4

Then:

$Area = \sqrt{3}\dfrac{4^2}{4}$

$Area = \sqrt{3} \dfrac{16}{4}$

$Area =4 \sqrt{3}$

when the area is exactly half of the original triangle's area, the new height is :

$A = \dfrac{1}{2} \times Area$

$\sqrt{3} \dfrac{i^2}{4} = \dfrac{1}{2} \times 4\sqrt{3}$

$\dfrac{i^2}{4} = \dfrac{1}{2} \times 4$

$\dfrac{i^2}{4} =2$

$i^2 = 8$

$i= \sqrt{8}$

$i= \sqrt{4 \times 2}$

$i= 2 \sqrt{2}$

Finally, the new height of the new triangle is:

$IL =\sqrt{3} \dfrac{i}{2}$

$IL =\sqrt{3} \dfrac{2 \sqrt{2}}{2}$

$IL =\sqrt{3 \times 2}$

$IL =\sqrt{6}$

IL = 2.449 m

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4 years ago

An envelope from the post office is 3 inches wide with a total area of 30 square inches.what is the height of the envelope

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The height of the envelope is 10. This is because you divide the area by 3. So, the missing side is 10 inches.

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