All categories

3 years ago

True or false 6 to the 3 power = 216

Mathematics

2 answers:

timofeeve [1]3 years ago

5 0

Answer:

True

Step-by-step explanation:

6^3, or 6x6x6, is equal to 216.

---

hope it helps

-Dominant- [34]3 years ago

4 0

Answer:

true

Step-by-step explanation:

You might be interested in

Use the diagram. AB is a diameter, and AB is perpendicular to CD. The figure is not drawn to scale.

Stells [14]

Answer:

i) mAB = mBD (Not True)

Step-by-step explanation:

ii) mAC = mAD , CD is a chord and AB is a diameter. CD⊥AB and hence AC = AD,

Equal chords have equal segments . Hence it is true

iii) ∠BDA = ∠BCA true

iv) ∠PBC = ∠DBA True

Hence the three options are correct except the one.

So our answer is option A mAB = mBD (Not True)

4 0

3 years ago

What is a percent? Doesn't it mean a fraction out of 100?

PIT_PIT [208]

We use the symbol % for "percent."

The word "percent" really just means "per hundred.

3 0

4 years ago

Read 2 more answers

1Write an equation in point-slope form for the line that passes through (3.5) with a slope of 1.

Sonbull [250]

Answer:

y-5=1(x-3)

Step-by-step explanation:

8 0

3 years ago

Write this phrase as an expression. “25 more than 50”

Brut [27]

25+50 I assume...I mean I could be wrong but what else would it be

6 0

3 years ago

Read 2 more answers

Please find the perimeter and area of these shapes

ladessa [460]

Answer:

ABC shaded area = 36 $\pi$ - 72 cm²

ABC shaded area perimeter = $6\pi +12\sqrt{2}$ cm

ABCD area = $\dfrac52 \pi$ cm²

ABCD perimeter = $3\pi +2$ cm

Step-by-step explanation:

Shape ABC

Assuming you want the area and perimeter of the shaded part of the shape only...

Area

Area of a sector = $\dfrac12r^2\theta$ (where r is the radius and $\theta$

⇒ area of a sector = $\dfrac12 \times 12^2\times \dfrac{\pi}{2} =36\pi \ \textsf{cm}^2$

Area of triangle = 1/2 x base x height

⇒ area of triangle = 1/2 x 12 x 12 = 72 cm²

Therefore, area of shaded area = area of sector - area of triangle

⇒ area = 36 $\pi$ - 72 cm²

Perimeter

Arc length = $r\theta$ (where r is the radius and $\theta$

⇒ arc length = $12\times\dfrac12\pi =6\pi \ \textsf{cm}$

Hypotenuse of triangle = $\sqrt{a^2+b^2}$ (where a and b are the legs of the right triangle)

⇒ hypotenuse = $\sqrt{12^2+12^2} =12\sqrt{2}$ cm

Therefore, perimeter = arc length + hypotenuse

⇒ perimeter = $6\pi +12\sqrt{2}$ cm

Shape ABCD

Area

Area of a semicircle = $\dfrac12 \pi r^2$ (where r is the radius)

⇒ area of large semicircle ABC = $\dfrac12 \times \pi \times 2^2=2\pi \ \textsf{cm}^2$

⇒ area of small semicircle AD = $\dfrac12 \times \pi \times 1^2=\dfrac12\pi \ \textsf{cm}^2$

⇒ area of shape ABCD = $\dfrac12 \pi + 2 \pi=\dfrac52 \pi \ \textsf{cm}^2$

Perimeter

1/2 circumference = $\pi r$

⇒ perimeter = $2\pi +2+\pi=3 \pi+2 \ \textsf{cm}$

7 0

2 years ago