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emmainna [20.7K]

3 years ago

13

What is the length of BC ?

1 answer:

MrRa [10]3 years ago

6 0

Looking at this triangle, we can see that is a form of a 45, 45, 90 triangle
This would mean that 2x-24=x-2
Solve: 2x-24=x-2
subtract x from both sides and add 24 to both sides...
x=22

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If the two lines below are perpendicular and the slope ofthe red line is -, what is the slope of the green line?51015-5-5-105

inysia [295]

If two lines are perpendicular, then the product of their slopes is -1:

$m_r\cdot m_g=-1...(1)$

From the problem, we identify:

$m_r=-\frac{4}{3}\text{ \lparen the slope of the red line\rparen}$

Then, using the equation (1), we calculate the slope of the green line:

$\begin{gathered} -\frac{4}{3}\cdot m_g=-1 \\ \\ \therefore m_g=\frac{3}{4} \end{gathered}$

Answer: Option D

4 0

1 year ago

The running tracks of two different playgrounds are shown below. Figure A is a semicircle with a diameter of 100 m. Figure B is

Trava [24]

Answer:

10157.08 m

Step-by-step explanation:

Find the area of the square, and get 10000, from multiplying 100 by 100. Then, find the circumference/area of the semi-circle as a whole circle. 314.16m. Divide it by 2 and get 157.08 m. Add everything up and get 10157.08 m. Remember, that we aren't calculating the total of both semi-circles because we subtract Figure A from Figure B.

Note:

Pls, notify me if my answer is correct, for the other users that will see this message.

<em>-kiniwih426</em>

6 0

3 years ago

Read 2 more answers

Ryan has scored 64, 75, 69, 58, and 63 on his previous five tests. What score does he need on his next test so that his average

ser-zykov [4K]

Answer:

66

Step-by-step explanation:

8 0

2 years ago

A.) pinky bought 1 1/2 kg of apples and 5 1/4 kg of mangoes and 1 1/2 Kg of oranges. Find the total weight of fruits B.) If her

Natasha2012 [34]

Answer:

a) The total weight of fruits is $8\,\frac{1}{4}$ kilograms, b) The weight of the fruits left is $4\,\frac{1}{2}$ kilograms.

Step-by-step explanation:

a) The total weight of fruits ( $m_{T}$ ) is calculated by the following formula:

$m_{T} = m_{a} + m_{m}+m_{o}$

Where:

$m_{a}$ - Total weight of apples, measured in kilograms.

$m_{m}$ - Total weight of mangoes, measured in kilograms.

$m_{o}$ - Total weight of oranges, measured in kilograms.

If $m_{a} = 1\,\frac{1}{2} \,kg$ , $m_{m} = 5\,\frac{1}{4}\,kg$ and $m_{o} = 1\,\frac{1}{2}\,kg$ , then:

$m_{T} = 1\,\frac{1}{2}\,kg + 5\,\frac{1}{4}\,kg + 1\,\frac{1}{2}\,kg$

$m_{T} = \frac{6}{4}\,kg + \frac{21}{4}\,kg + \frac{6}{4}\,kg$

$m_{T} = \frac{33}{4}\,kg$

$m_{T} = 8\,\frac{1}{4}\,kg$

The total weight of fruits is $8\,\frac{1}{4}$ kilograms.

b) The weight eaten by her family is determined by the following expression:

$m_{E} = m_{a,e} + m_{m,e} + m_{o,e}$

Where:

$m_{a,e}$ - Eaten weight of apples, measured in kilograms.

$m_{m,e}$ - Eaten weight of mangoes, measured in kilograms.

$m_{o,e}$ - Eaten weight of oranges, measured in kilograms.

Given that $m_{a,e} = \frac{3}{4}\,kg$ , $m_{m,e} = 2\,\frac{1}{2}\,kg$ and $m_{o,e} = \frac{1}{2}\,kg$ , the weight eaten by her family is:

$m_{E} = \frac{3}{4}\,kg + 2\,\frac{1}{2}\,kg + \frac{1}{2}\,kg$

$m_{E} = \frac{3}{4}\,kg + \frac{10}{4}\,kg + \frac{2}{4}\,kg$

$m_{E} = \frac{15}{4}\,kg$

$m_{E} = 3\,\frac{3}{4}\,kg$

The weight of the fruits left is found by subtraction:

$m_{R} = m_{T}-m_{E}$

$m_{R} = 8\,\frac{1}{4} \,kg -3\,\frac{3}{4}\,kg$

$m_{R} = \frac{33}{4}\,kg-\frac{15}{4}\,kg$

$m_{R} = \frac{18}{4}\,kg$

$m_{R} = 4 \,\frac{1}{2}\,kg$

The weight of the fruits left is $4\,\frac{1}{2}$ kilograms.

3 0

3 years ago

Use the binomial series series to expand the function as a power series. 3/((6+x)^3)

Degger [83]

= 3/(6^3*x^0 + 3*6^2*x^1 + 3*6^1*x^2 + 6^0*x^3)
= 3/(216 + 108x + 18x^2 + x^3)

6 0

4 years ago