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slega [8]
2 years ago
5

Help! me find perimeter for kite help check

Mathematics
2 answers:
ozzi2 years ago
5 0

√(58)*2 + √(25)*2

answer: 2√58+10

Aleks [24]2 years ago
5 0

Answer:

10 + 2\sqrt{58}

Step-by-step explanation:

perimeter of DEFG = DE + EF + FG + DG

DE, EF, FG, and DG are all hypotenuse.

DE =\sqrt[]{4^{2} + 3^{2} }  = 5

EF = \sqrt{4^{2}+3^{2}  } = 5

FG = \sqrt{3^{2} +7^{2} } = \sqrt{58}

DG = \sqrt{3^{2}+7^{2}  } = \sqrt{58}

DE + EF + FG + DG

= 5 + 5 + \sqrt{58} + \sqrt{58}

= 10 + 2\sqrt{58}

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Help meeee please eeee
Sphinxa [80]

Answer:

112 ft³

Step-by-step explanation:

→ Length (l) = 7 ft

→ Breadth (b) = 4 ft

→ Height (h) = 4 ft

Formula we use,

→ l × b × h

The volume of cuboid will be,

→ l × b × h

→ 7 × 4 × 4

→ [ 112 ft³ ]

Hence, the volume is 112 ft³.

8 0
2 years ago
Circle X is shown. Points W, Y, and Z are on the circle. Lines are drawn through each point on the circle.
Dimas [21]

Answer:

Its point X

Step-by-step explanation:

Just did it on edge 2020

6 0
3 years ago
Read 2 more answers
Someone help me<br>10-14x-7-7x=6​
Burka [1]

Answer: your answer is going to be 1/7

Step-by-step explanation:

4 0
3 years ago
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Write the equation of a line that passes through the point (-6, 8) and has a slope of 3.
MakcuM [25]

Answer:

Step-by-step explanation:

y - 8 = 3(x + 6)

y - 8 = 3x + 18

y = 3x + 10

answer is a

5 0
3 years ago
38. Evaluate f (3x +4y)dx + (2x --3y)dy where C, a circle of radius two with center at the origin of the xy
lina2011 [118]

It looks like the integral is

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy

where <em>C</em> is the circle of radius 2 centered at the origin.

You can compute the line integral directly by parameterizing <em>C</em>. Let <em>x</em> = 2 cos(<em>t</em> ) and <em>y</em> = 2 sin(<em>t</em> ), with 0 ≤ <em>t</em> ≤ 2<em>π</em>. Then

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \int_0^{2\pi} \left((3x(t)+4y(t))\dfrac{\mathrm dx}{\mathrm dt} + (2x(t)-3y(t))\frac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt \\\\ = \int_0^{2\pi} \big((6\cos(t)+8\sin(t))(-2\sin(t)) + (4\cos(t)-6\sin(t))(2\cos(t))\big)\,\mathrm dt \\\\ = \int_0^{2\pi} (12\cos^2(t)-12\sin^2(t)-24\cos(t)\sin(t)-4)\,\mathrm dt \\\\ = 4 \int_0^{2\pi} (3\cos(2t)-3\sin(2t)-1)\,\mathrm dt = \boxed{-8\pi}

Another way to do this is by applying Green's theorem. The integrand doesn't have any singularities on <em>C</em> nor in the region bounded by <em>C</em>, so

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \iint_D\frac{\partial(2x-3y)}{\partial x}-\frac{\partial(3x+4y)}{\partial y}\,\mathrm dx\,\mathrm dy = -2\iint_D\mathrm dx\,\mathrm dy

where <em>D</em> is the interior of <em>C</em>, i.e. the disk with radius 2 centered at the origin. But this integral is simply -2 times the area of the disk, so we get the same result: -2\times \pi\times2^2 = -8\pi.

3 0
3 years ago
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