*pic attached* can someone please help me with this

Question 1

allochka39001 [22]

Answer:

Step-by-step explanation:

Perimeter of quarter circle = r + r + $\frac{1}{4}*2*\pi *r$

$= 10 + 10 + \frac{1}{4}*2*3.14*10\\\\= 10 + 10 + 3.14*5\\\\\\= 10 + 10 + 15.70\\\\= 35.70 cm$

2) d = 2*10 = 20 cm

Perimeter of semicircle = diameter + (1/2) * circumference of circle

$= 20 + \frac{1}{2}* 2*\pi *r\\\\= 20 + \frac{1}{2}*2*3.14*10\\\\= 20 + 3.14*10\\\\= 20 + 31.4\\\\= 51.40 cm$

3) Perimeter of three-quarter = r + r + (3/4)* circumference of circle

$= 10 + 10 + \frac{3}{4}*2*\pi *r\\\\= 10 + 10 + \frac{3}{4} * 2 * 3.14 * 10\\\\= 10 + 10 + 3 * 3.14 * 5\\\\= 10 + 10 + 47.1\\\\= 67.1 cm$

5 0

2 years ago

Boys to girls ratio is 2 to 3. There are 18 girls. What is total number of students

alexdok [17]

$\frac{2}{3}=\frac{boys}{18}$

3*boys=2*18

3*boys=36

boys=12

12+18=30

total number of students: 30

8 0

2 years ago

Read 2 more answers

If the largest angle of a triangle is 120° and it is included between sides of 1.5 and 0.5, then (to the nearest tenth) the larg

12345 [234]

Answer:

1.8 units.

Step-by-step explanation:

The questions which involve calculating the angles and the sides of a triangle either require the sine rule or the cosine rule. In this question, the two sides that are given are adjacent to each other and the given angle is the included angle. This means that the angle is formed by the intersection of the two lines. Therefore, cosine rule will be used to calculate the length of the largest side of the triangle. The cosine rule is:

b^2 = a^2 + c^2 - 2*a*c*cos(B).

The question specifies that a=0.5, B=120°, and c=1.5. Plugging in the values:

b^2 = 0.5^2 + 1.5^2 - 2(0.5)(1.5)*cos(120°).

Simplifying gives:

b^2 = 3.25.

Taking square root on the both sides gives b = 1.8 (rounded to the nearest tenth).

This means that the length of the third side is 1.8 units!!!

3 0

2 years ago

What is the simplified form of StartRoot StartFraction 72 x Superscript 16 Baseline Over 50 x Superscript 36 Baseline EndFractio

aalyn [17]

Answer:

StartFraction 6 Over 5 x Superscript 10 Baseline EndFraction

Step-by-step explanation:

Apparently you want to simplify ...

$\sqrt{\dfrac{72x^{16}}{50x^{36}}}$

The applicable rules of exponents are ...

(a^b)(a^c) = a^(b+c)

1/a^b = a^-b

(a^b)^c = a^(bc)

__

So the expression simplifies as ...

$\sqrt{\dfrac{72x^{16}}{50x^{36}}}=\sqrt{\dfrac{36}{25x^{36-16}}}=\sqrt{\dfrac{36}{25x^{20}}}\\\\\sqrt{\left(\dfrac{6}{5x^{10}}\right)^2}=\boxed{\dfrac{6}{5x^{10}}}$

9 0

2 years ago

Person A can paint the neighbor's house 6 times as fast as Person B. The year A and B worked together it

uranmaximum [27]

Person A takes 2.3 days to paint the house working alone

Person B takes 13.8 days to paint the house working alone

<em><u>Solution:</u></em>

Let "x" be the number of days it takes person A to paint the house

Person A can paint the neighbor's house 6 times as fast as Person B

Therefore.

Number of days it takes person B to paint the house = 6x

Their rates of painting are added to get the rate working together

$\frac{\text{1 house}}{\text{x days}} + \frac{\text{1 house}}{\text{6x days}} = \frac{\text{1 house}}{\text{2 days}}$

$\frac{1}{x} + \frac{1}{6x} = \frac{1}{2}\\\\\frac{1 \times 6}{6x} + \frac{1}{6x} = \frac{1}{2}\\\\\frac{7}{6x} = \frac{1}{2}\\\\\ x = \frac{7}{3}= 2.3$

Thus person A takes 2.3 days to paint the house working alone

Person B = 2x = 6(2.3) = 13.8

Thus person B takes 13.8 days to paint the house working alone

3 0

2 years ago