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BartSMP [9]
3 years ago
6

In a parallelogram, two sides are each length 12, and the perimeter is 36, what is the length of the remaining sides?

Mathematics
1 answer:
tankabanditka [31]3 years ago
5 0
P=2a+2b
P=36
a=12
36=2*12+2b
36=24+2b
36-24=2b
12=2b
b=6
So, the length of the remaining sides is 6 each.
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The area of ABED is 49 square units. Given AGequals9 units and ACequals10 ​units, what fraction of the area of ACIG is represent
stiks02 [169]

Answer:

The fraction of the area of ACIG represented by the shaped region is 7/18

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

In the square ABED find the length side of the square

we know that

AB=BE=ED=AD

The area of s square is

A=b^{2}

where b is the length side of the square

we have

A=49\ units^2

substitute

49=b^{2}

b=7\ units

therefore

AB=BE=ED=AD=7\ units

step 2

Find the area of ACIG

The area of rectangle ACIG is equal to

A=(AC)(AG)

substitute the given values

A=(9)(10)=90\ units^2

step 3

Find the area of shaded rectangle DEHG

The area of rectangle DEHG is equal to

A=(DE)(DG)

we have

DE=7\ units

DG=AG-AD=9-7=2\ units

substitute

A=(7)(2)=14\ units^2

step 4

Find the area of shaded rectangle BCFE

The area of rectangle BCFE is equal to

A=(EF)(CF)

we have

EF=AC-AB=10-7=3\ units

CF=BE=7\ units

substitute

A=(3)(7)=21\ units^2

step 5

sum the shaded areas

14+21=35\ units^2

step 6

Divide the area of  of the shaded region by the area of ACIG

\frac{35}{90}

Simplify

Divide by 5 both numerator and denominator

\frac{7}{18}

therefore

The fraction of the area of ACIG represented by the shaped region is 7/18

5 0
3 years ago
One angle of a triangle measures 136º. The other two angles are congruent.
WINSTONCH [101]

Answer:

22°

Step-by-step explanation:

The sum of the 3 angles in a triangle = 180° , then

x + x + 136° = 180°

2x + 136° = 180° ( subtract 136° from both sides )

2x = 44° ( divide both sides by 2 )

x = 22°

3 0
3 years ago
The opposite of which number is closest to -5
Oliga [24]
The answer would be positive 5
8 0
2 years ago
Which theorem or postulate justifies that angle HEF ~ angle HGE ?
Levart [38]
Based on the picture above,
The theorem or postulate that justifies that Angle HEF ~ angle HGE is :
A. AA similarity postulate
(s is used for equal side, that is used for congruent)

hope this helps
5 0
3 years ago
Use spherical coordinates. Find the volume of the solid that lies within the sphere x^2 + y^2 + z^2 = 81, above the xy-plane, an
Natasha_Volkova [10]

Answer:

The volume of the solid is 243\sqrt{2} \ \pi

Step-by-step explanation:

From the information given:

BY applying sphere coordinates:

0 ≤ x² + y² + z² ≤ 81

0  ≤ ρ²   ≤   81

0  ≤ ρ   ≤  9

The intersection that takes place in the sphere and the cone is:

x^2 +y^2 ( \sqrt{x^2 +y^2 })^2  = 81

2(x^2 + y^2) =81

x^2 +y^2 = \dfrac{81}{2}

Thus; the region bounded is: 0 ≤ θ ≤ 2π

This implies that:

z = \sqrt{x^2+y^2}

ρcosФ = ρsinФ

tanФ = 1

Ф = π/4

Similarly; in the X-Y plane;

z = 0

ρcosФ = 0

cosФ = 0

Ф = π/2

So here; \dfrac{\pi}{4} \leq \phi \le \dfrac{\pi}{2}

Thus, volume: V  = \iiint_E \ d V = \int \limits^{\pi/2}_{\pi/4}  \int \limits ^{2\pi}_{0} \int \limits^9_0 \rho   ^2 \ sin \phi \ d\rho \   d \theta \  d \phi

V  = \int \limits^{\pi/2}_{\pi/4} \ sin \phi  \ d \phi  \int \limits ^{2\pi}_{0} d \theta \int \limits^9_0 \rho   ^2 d\rho

V = \bigg [-cos \phi  \bigg]^{\pi/2}_{\pi/4}  \bigg [\theta  \bigg]^{2 \pi}_{0} \bigg [\dfrac{\rho^3}{3}  \bigg ]^{9}_{0}

V = [ -0+ \dfrac{1}{\sqrt{2}}][2 \pi -0] [\dfrac{9^3}{3}- 0 ]

V = 243\sqrt{2} \ \pi

4 0
3 years ago
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