Please help!!!! :( will give brainliest

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adoni [48]

Answers:

Plane EFGH
Angle 2 and angle 3
Alternate exterior angles
See the explanation below

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Explanation:

Problem 1

The plane ABCD is the floor of the box or room.

The ceiling plane EFGH is parallel to the floor.

Parallel planes never intersect, must like how parallel lines in 2D never intersect. As such, parallel planes are the same distance apart.

In contrast, something like plane ABFE intersects with ABCD along the segment AB. This shows ABFE is not parallel to ABCD.

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Problem 2

Same side interior angles, aka consecutive interior angles, are inside the parallel (or nearly parallel) lines. One such example is angle 2 and angle 3. The other pair being angle 6 and angle 7. They must be on the same side of the transversal.

In contrast, alternate interior angles would be something like the pair angle 2 and angle 7. The other pair being angle 3 and angle 6. This time the angles are on alternating or opposite sides of the transversal.

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Problem 3

Angles 1 and 8 are exterior of the parallel (or nearly parallel) lines. They are on alternating sides of the transversal. Therefore, we consider them to be alternate exterior angles.

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Problem 4

The use of "alternate" refers to the idea the angles are on opposite or alternating sides of the transversal line. The transversal line is the line that crosses the two other lines that are almost parallel.

I like to think of the parallel (or nearly parallel) lines as train tracks. The transversal is the roadway that crosses both train tracks. Then we could have locations 1 and 8 on opposite sides of the roadways to represent the alternate exterior angles.

As you can probably guess or know by now, the "exterior" is from the angles being outside or exterior of the parallel (or nearly parallel) lines.

4 0

2 years ago

-10=x/2+6<br> Answers <br> -8 -14 -32 18

Yanka [14]

Answer:

$x=-32$

6 0

3 years ago

Plz help I'll give brainlest

jarptica [38.1K]

<h3>1.Area of the parallelogram= 288 square units</h3><h3>2.Area of the parallelogram=45 $m^2$ </h3><h3>3.Area of the trapezoid = 34 square in.</h3><h3>4.Area of the trapezoid = 8 square ft</h3><h3>5.Area of the rhombus= 27 square cm</h3><h3>6.Area of the rhombus= 108 square in</h3><h3>7.The area of the desktop is = 1200 square in</h3><h3>8.The area of the rhombus is =84 $cm^2$ </h3><h3>9.Area of the trapezoid = 240 square ft</h3>

Step-by-step explanation:

1.

Base =16 ft and Height = 18 ft

Area of the parallelogram = base × height

=16× 18 square units

= 288 square units

2.

Base = 9 m and height = 5 m

Area of the parallelogram = base × height

=(9×5) $m^2$

=45 $m^2$

3 .

Height = 4 in and parallel sides are 12 in and 5 in

Area of the trapezoid = $\frac {1}{2} \times( {\textrm{sum of the parallel sides}}) \times height$

$=\frac{1}{2} \times (12+5) \times 4$ square in.

= 34 square in.

4.

Height = 2 ft and parallel sides are 2 ft and 6 ft

Area of the trapezoid = $\frac {1}{2} \times( {\textrm{sum of the parallel sides}}) \times height$

$=\frac{1}{2} \times (2+6) \times 2$ square ft

= 8 square ft

5.

Diagonals are 6 cm and 9 cm.

Area of the rhombus $= \frac{1}{2}\times {\textrm{product of diagonals}$

$=\frac{1}{2 } \times 6\times 9$ square cm

= 27 square cm

6. Diagonals are 12 in and 18 in

Area of the rhombus $= \frac{1}{2}\times {\textrm{product of diagonals}$

$=\frac{1}{2 } \times 12\times 18$ square in

= 108 square in

7. Given a desktop in the shape of a parallelogram has a base 30 in. and a height of 40 in

The area of the desktop is = (30 × 40 ) square in

= 1200 square in

8. Given , a rhombus has one diagonal that is 14 cm and other diagonal 12 cm.

The area of the rhombus = $\frac {1}{2} \times 14 \times 12$ $cm^2$

=84 $cm^2$

9.Given , the base of trapezoid are 24 ft and 16 ft and height is 12 ft

Area of the trapezoid = $\frac {1}{2} \times( {\textrm{sum of the parallel sides}}) \times height$

= $\frac{1}{2} \times (24+16) \times 12$ square ft

= 240 square ft

8 0

3 years ago

Evaluate each expression<br> 6! =<br> 3! • 2=<br> 31

Airida [17]

<h2><u>N-FACTORIAL!</u></h2>

<h3> $\mathbb{ \color{brown} \: PROBLEM } \: \: \bold{{ \color{brown}{1}}} :$ </h3>

$\mathcal{SOLUTION: }$

$\: \bold{ \red{6!}= 6×5×4×3×2×1= \boxed{ \bold{ \blue{720}}}}$

Therefore, <u>the value of 6! is 720</u>.

━┈─────────────────────┈━

<h3> $\mathbb{ \color{brown} \: PROBLEM } \: \: \bold{{ \color{brown}{2}}} :$ </h3>

$\mathcal{SOLUTION: }$

$\bold{3! \cdot 2! }\implies \bold{(3×2×1) \cdot( 2×1 )}$

$\: \: \: \: \: \: \: \: \: \: \: \: \implies \: \bold{ 6 \times 2} = \boxed{ \bold{ \blue{12}}}$

Therefore, <u>the value of the given expression is 12.</u>

━┈─────────────────────┈━

<h3><u>EXPLANATION</u><u>:</u></h3>

The mathematical symbol n! is read as "n factorial". The exclamation point "!" is read as factorial. And please remember or take note that 0! is equal to 1, and 1! is also equal to 1.

_______________∞_______________

6 0

2 years ago

The Wilson family had 5 children. Assuming that the probability of a child being a girl is 0.5, find the probability that the Wi

Dafna11 [192]

Answer:

0.813

0.500

Step-by-step explanation:

Use binomial probability.

P = nCr p^r q^(n−r)

where n is the number of trials,

r is the number of successes,

p is the probability of success,

and q is the probability of failure (1−p).

In this problem, n = 5, p = 0.5, and q = 0.5.

"At least 2 girls" means r = 2, 3, 4, or 5.

Or, we can use the complement.

P(at least 2 girls) = 1 − P(at most 1 girl)

P(at least 2 girls) = 1 − P(r=0 or r=1)

P(at least 2 girls) = 1 − ₅C₁ (0.5)¹ (0.5)⁵⁻¹ − ₅C₀ (0.5)⁰ (0.5)⁵⁻⁰

P(at least 2 girls) = 1 − 5 (0.5) (0.5)⁴ − 1 (1) (0.5)⁵

P(at least 2 girls) = 1 − 6 (0.5)⁵

P(at least 2 girls) ≈ 0.813

"At most 2 girls" means r = 0, 1, or 2.

P(at most 2 girls) = P(r=0, r=1, or r=2)

P(at most 2 girls) = ₅C₀ (0.5)⁰ (0.5)⁵⁻⁰ + ₅C₁ (0.5)¹ (0.5)⁵⁻¹ + ₅C₂ (0.5)² (0.5)⁵⁻²

P(at most 2 girls) = 1 (1) (0.5)⁵ + 5 (0.5) (0.5)⁴ + 10 (0.5)² (0.5)³

P(at most 2 girls) = 16 (0.5)⁵

P(at most 2 girls) = 0.500

4 0

3 years ago