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3 years ago

Problem 3

Mathematics

1 answer:

Murljashka [212]3 years ago

8 0

Answer:

$-\frac{2}{5}$

Step-by-step explanation:

Given expression:

2x + 5y = -10

The equation of a straight line is;

y = mx + c

y and x are the coordinates

m is the slope

c is the intercept

Now;

let us write the given expression in slope intercept format;

2x + 5y = -10

5y = -2x - 10

y = $-\frac{2x}{5}$ - 2

So, the slope of the line is $-\frac{2}{5}$

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What is the value of x if x over 5=3

Zolol [24]

Answer:

x = 15

Step-by-step explanation:

the easiest way to do this is 5 times 3 and with that we get 15 :)

Have an amazing day!!

Please rate and mark brainliest!!

3 0

2 years ago

Read 2 more answers

Find k so that the distance from (–1, 1) to (2, k) is 5 units. k= k= *there are two solutions for 2*

dalvyx [7]

Answer:

k = -3

k =5

Step-by-step explanation:

$d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\d = 5\\(-1,1) =(x_1,y_1)\\(2,k)=(x_2,y_2)\\$

$5=\sqrt{\left(2-\left(-1\right)\right)^2+\left(k-1\right)^2}\\\\\mathrm{Square\:both\:sides}:\quad 25=k^2-2k+10\\25=k^2-2k+10\\\\\mathrm{Solve\:}\:25=k^2-2k+10:\\k^2-2k+10=25\\\\\mathrm{Subtract\:}25\mathrm{\:from\:both\:sides}\\k^2-2k+10-25=25-25\\k^2-2k-15=0\\\\\mathrm{Solve\:by\:factoring}\\\\\mathrm{Factor\:}k^2-2k-15:\quad \left(k+3\right)\left(k-5\right)\\\mathrm{Solve\:}\:k+3=0:\quad k=-3\\$

$\mathrm{Solve\:}\:k-5=0:\quad k=5\\\\k =5 , k=-3$

7 0

3 years ago

Read 2 more answers

The perimeter of equilateral triangle ABC is 81/3 centimeters, find the length of the radius and apothem.

MAXImum [283]

There is a typo error, the perimeter of equilateral triangle ABC is 81/√3 centimeters.

Answer:

Radius = OB= 27 cm

Apothem = 13.5 cm

A diagram is attached for reference.

Step-by-step explanation:

Given,

The perimeter of equilateral triangle ABC is 81/√3 centimeters.

Substituting this in the formula of perimeter of equilateral triangle = $3\times\ side$

$3\times\ side$ $=[tex]81\sqrt{3}$

$Side = \frac{81\sqrt{3} }{3} =27\sqrt{3} \ cm$

Thus from the diagram , Side $AB=BC=AC= 27\sqrt{3} \ cm$

We know each angle of an equilateral triangle is 60°.

From the diagram, OB is an angle bisector.

Thus $\angle OBC = 30$ °

Apothem is the line segment from the mid point of any side to the center the equilateral triangle.

Therefore considering ΔOBE, and applying tan function.

$tan\theta =\frac{perpendicular}{base} \\tan\theta=\frac{OE}{BE} \\tan\theta=\frac{OE}{\frac{27\sqrt{3}}{2} } \\tan30\times {\frac{27\sqrt{3} }{2} }= OE\\\frac{1}{\sqrt{3} } \times\frac{27\sqrt{3} }{2} =OE\\$

Thus ,apothem OE= 13.5 cm

Now for radius,

We consider ΔOBE

$cos\theta=\frac{base}{hypotenuse} \\cos30= \frac{BE}{OB} \\Cos30 = \frac{\frac{27\sqrt{3} }{2}}{OB} \\OB= \frac{\frac{27\sqrt{3} }{2}}{cos30} \\OB= \frac{\frac{27\sqrt{3} }{2}}{\frac{\sqrt{3} }{2} } \\OB =27 \ cm$

Thus for

Perimeter of equilateral triangle ABC is 81/√3 centimeters,

The radius of equilateral triangle ABC is 27 cm

The apothem of equilateral triangle ABC is 13.5 cm

4 0

3 years ago

A black tailed Jackrabbit hops about 7 feet in a single hop how far can it hop in 5 seconds

Sever21 [200]

If it hops 7 feet a second then it would hop 35 times in 5 seconds. 5 x 7 = 35.

Happy studying ^-^

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3 years ago

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Mars2501 [29]

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0 = cos⁴(x)[1] + cos⁴(x)[2tan²(x)] + cos⁴(x)[tan⁴(x)]
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⁴√0 = ⁴√cos⁴(x)    or   √0 = (√1 + tan²(x))²
0 = cos(x)   or   0 = 1 + tan²(x)
cos⁻¹(0) = cos⁻¹(cos(x)) or -1 = tan²(x)
90 = x       or        √-1 = √tan²(x)
i = tan(x)
(No Solution)

2. sin(x)[tan(x)cos(x) - cot(x)cos(x)] = 1 - 2cos²(x)
  sin(x)[sin(x) - cos(x)cot(x)] = 1 - cos²(x) - cos²(x)
sin(x)[sin(x)] - sin(x)[cos(x)cot(x)] = sin²(x) - cos²(x)
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+ cos²(x) + cos²(x)
sin²(x) = sin²(x)
- sin²(x) - sin²(x)
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sec²(x) sec²(x)
cos²(x) + sin²(x) = 1
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√cos²(x) = √(1 - sin²(x))
cos(x) = √(1 - sin²(x))
cos⁻¹(cos(x)) = cos⁻¹(√1 - sin²(x))
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tan(x) = √(-1 + sec²(x))
  tan⁻¹(tan(x)) = tan⁻¹(√(-1 + sec²(x))
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5 0

3 years ago