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3 years ago

12

The circumference of circle A is three times the circumference of circle B. Which statement about the areas of the circles is tr

ue?

1 answer:

Valentin [98]3 years ago

7 0

Answer:

A=3B

Step-by-step explanation:

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What do the numbers listed for an element on the periodic table represent?

maria [59]

Im sure you have your answer by now but would you mind posting the answer

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3 years ago

Using (x + a) (x + b) = x² + (a + b) x + ab, evaluate 75 x 77

Gre4nikov [31]

(x + a) (x + b) = x² + (a + b) x + ab

75 * 77=

=(70+5)(70+7)

=70²+(5+7)*70+5*7

=4900+12*70+35

=4900+840+35

=5775

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2 years ago

5trig review please help me

snow_lady [41]

Answer:

Is A

Step-by-step explanation:

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2 years ago

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I will be marking brainliest please help me with these questions.

krek1111 [17]

Answer/Step-by-step explanation:

1. To find the area of the shaded region, you'd find the area of the white rectangular shape, next, find the area of the whole triangular shape, then find the difference of their areas to get the area of the shaded region. Thus, the formula to use would be:

Area of shaded region = area of triangle - area of rectangle

Area of shaded region = ½*base*height - length*width

1. a. Volume of triangular prism = area of triangular base * height of prism

Volume of triangular prism = ½bh * H

Where,

b = 6 m

h = 4 m

H = 8 m

Substitute

Volume of prism = ½*6*4*8

Volume of prism = 96 m³

b. Volume of sphere = ⁴/3πr³

Where,

r = 9 cm

Substitute

Volume = ⁴/3*π*9³

Volume = ⁴/3*π*729

Volume ≈ 3,053.6 cm³ (nearest tenth)

2. Use Pythagorean theorem to find the height of the cone

radius of the cone (r) = ½(16) = 8 cm

Slant height (l) = 11 cm

height (h) = ?

Using Pythagorean theorem, we have:

h = √(l² - r²)

Substitute

h = √(11² - 8²)

h = √(57)

h ≈ 7.5 cm (nearest tenth)

b. Volume of the cone = ⅓πr²h

where,

r = 8 cm

h = 7.5 cm

Volume = ⅓*π*8²*7.5

Volume = 502.7 cm³ (nearest tenth)

7 0

2 years ago

Which answer best describes the complex zeros of the polynomial function?

marysya [2.9K]

<u>Answer-</u>

<em>D. The function has one real zero and two nonreal zeros. The graph of the function intersects the x-axis at exactly one location.</em>

<u>Solution-</u>

The given polynomial is,

$f(x)=x^3-x^2+6x-6$

The zeros of the polynomials are,

$\Rightarrow f(x)=0$

$\Rightarrow x^3-x^2+6x-6=0$

$\Rightarrow x^2(x-1)+6(x-1)=0$

$\Rightarrow (x^2+6)(x-1)=0$

$\Rightarrow (x^2+6)=0,(x-1)=0$

$\Rightarrow x=\sqrt{-6},\ x=1$

$\Rightarrow x=1,\ \pm \sqrt6i$

Therefore, this function has only one real zero i.e 1 and two nonreal zeros i.e ±√6i . The graph of the function intersects the x-axis at exactly one location i.e at x = 1

3 0

3 years ago