For some JKL the side lengths are such that LJ< JK< KL what must be true about angles J, K, L

A square piece of paper has an area of 36 square inches what is the perimeter of the paper

Gekata [30.6K]

Lets represent the side of the square as s , since we know the area is 36. write s

${s}^{2} = 36$

$s = \sqrt{36}$

the square root of 36 is 6. So, we know the side is 6. The formula for perimeter is P= s+s+s+s , but since it is a square and all the sides are equal, for a perimeter of a square you can do p= 4s , now substitute 6 for s to get 4(6)=24 so the perimeter of a square would be 24 inches

6 0

3 years ago

I’m not sure how to slove this problem it’s a equilateral triangle and I’m trying to find out what the side is equal to.

Lerok [7]

Answer:

AB = 5/4 y - 1 = 5/4 (12/13) -1 = 2/13

AC= 7/3 y - 2 = 7/3 (12/13) -2 = 2/13

Step-by-step explanation:

Because it is a equilateral triangle, it means all sides have the same length.

AB = AC

5/4 y - 1 = 7/3 y - 2

5/4 y - 7/3 y = -2 +1

-13/12 y = -1

y = 12/13

input y to the equation.

AB = 5/4 y - 1 = 5/4 (12/13) -1 = 2/13

AC= 7/3 y - 2 = 7/3 (12/13) -2 = 2/13

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6 0

3 years ago

Evaluate the indefinite integral. <br> integar x4/1 + x^10 dx

ivann1987 [24]

Answer:

$\int\ {\frac{x^4}{1 + x^{10}}} \, dx = \frac{1}{5}( \arctan(x^5)) + c$

Step-by-step explanation:

Given

$\int\ {\frac{x^4}{1 + x^{10}}} \, dx$

Required

Integrate

We have:

$\int\ {\frac{x^4}{1 + x^{10}}} \, dx$

Let

$u = x^5$

Differentiate

$\frac{du}{dx} = 5x^4$

Make dx the subject

$dx = \frac{du}{5x^4}$

So, we have:

$\int\ {\frac{x^4}{1 + x^{10}}} \, dx$

$\int\ {\frac{x^4}{1 + x^{10}}} \, \frac{du}{5x^4}$

$\frac{1}{5} \int\ {\frac{1}{1 + x^{10}}} \, du$

Express x^(10) as x^(5*2)

$\frac{1}{5} \int\ {\frac{1}{1 + x^{5*2}}} \, du$

Rewrite as:

$\frac{1}{5} \int\ {\frac{1}{1 + x^{5)^2}}} \, du$

Recall that: $u = x^5$

$\frac{1}{5} \int\ {\frac{1}{1 + u^2}}} \, du$

Integrate

$\frac{1}{5} * \arctan(u) + c$

Substitute: $u = x^5$

$\frac{1}{5} * \arctan(x^5) + c$

Hence:

$\int\ {\frac{x^4}{1 + x^{10}}} \, dx = \frac{1}{5}( \arctan(x^5)) + c$

7 0

3 years ago

*

Phoenix [80]

Answer:

Step-by-step explanation:

Are the sides consistent with the Pythagorean Theorem?

6² + 8² ≟ 9²

36 + 64 ≠ 81, so it is not a right triangle.

6 0

3 years ago

Find the values of y for which the distance between the point p2,-3 and q 10, y is 10 units

lina2011 [118]

Answer:

y = - 9, y = 3

Step-by-step explanation:

Calculate distance d using the distance formula

d = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }$

with (x₁, y₁ ) = (2, - 3) and (x₂, y₂ ) = (10, y)

d = $\sqrt{(10-2)^2+(y+3)^2}$

= $\sqrt{8^2+(y+3)^2}$

Given distance between points is 10, then

$\sqrt{64 +(y+3)^2}$ = 10 ( square both sides )

64 + (y + 3)² = 100 ( subtract 64 from both sides )

(y + 3)² = 36 ( take the square root of both sides )

y + 3 = ± $\sqrt{36}$ = ± 6 ( subtract 3 from both sides )

y = - 3 ± 6 , thus

y = - 3 - 6 = - 9

y = - 3 + 6 = 3

7 0

3 years ago