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3 years ago

PLEASE HELP

Mathematics

1 answer:

pogonyaev3 years ago

6 0

Answer:

b and c

Step-by-step explanation:

both slopes are 2/3

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. A soccer team played 32 games. If they won 25% of them, how many games did the team win?V

Bas_tet [7]

Answer:

$8\ games$

Step-by-step explanation:

Let

x----->number of games won by the football team

we know that

$25\%=25/100=0.25$

$x=0.25(32)=8\ games$

8 0

3 years ago

I don’t understand what to do on these question, some help would be nice (trigonometry)

Bumek [7]

Step-by-step explanation:

> find the angle of center circle:

cos~¹ 5/(5+12) = cos~¹ 5/17 = 72°

so the sector area of the shaded

region =

(360-72)/360 × 3.14 × 5²

= 288/360 × 78.5

= 62.8

> the height of triangle (red Line) =

4×tan 72° = 4× 3.078 = 12.31

the height of pyramid (green

Line) = √12.31²-8.5² =√74.16

= 8.61

so the volume of pyramid =

⅓×(17×8)×8.61 = 390.32

8 0

3 years ago

What does (-x)+(-x)+(-x)=

Mazyrski [523]

Answer:

-3x

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

The system with one and only one solution set is called a/an system

beks73 [17]

Answer:

Consistent and Independent System

Step-by-step explanation:

A system with at least one solution is a consistent system. A consistent system is said to be independent if it has <u>exactly one solution</u> (often referred to as the <em>unique</em> solution). The graphs intersect at exactly one point, which gives an ordered-pair (x, y) as the solution of the system.

6 0

3 years ago

Explain how to multiply the following whole numbers 21 x 14

Lesechka [4]

Answer:

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

________

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

Step-by-step explanation:

Given

$21\:\times \:14$

Line up the numbers

$\begin{matrix}\space\space&2&1\\ \times \:&1&4\end{matrix}$

Multiply the top number by the bottom number one digit at a time starting with the ones digit left(from right to left right)

Multiply the top number by the bolded digit of the bottom number

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

Multiply the bold numbers: 1×4=4

$\frac{\begin{matrix}\space\space&2&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&\space\space&4\end{matrix}}$

Multiply the bold numbers: 2×4=8

$\frac{\begin{matrix}\space\space&\textbf{2}&1\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}$

Multiply the top number by the bolded digit of the bottom number

$\frac{\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}$

Multiply the bold numbers: 1×1=1

$\frac{\begin{matrix}\space\space&\space\space&2&\textbf{1}\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&\space\space&1&\space\space\end{matrix}}$

Multiply the bold numbers: 2×1=2

$\frac{\begin{matrix}\space\space&\space\space&\textbf{2}&1\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&2&1&\space\space\end{matrix}}$

Add the rows to get the answer. For simplicity, fill in trailing zeros.

$\frac{\begin{matrix}\space\space&\space\space&2&1\\ \space\space&\times \:&1&4\end{matrix}}{\begin{matrix}\space\space&0&8&4\\ \space\space&2&1&0\end{matrix}}$

adding portion

$\begin{matrix}\space\space&0&8&4\\ +&2&1&0\end{matrix}$

Add the digits of the right-most column: 4+0=4

$\frac{\begin{matrix}\space\space&0&8&\textbf{4}\\ +&2&1&\textbf{0}\end{matrix}}{\begin{matrix}\space\space&\space\space&\space\space&\textbf{4}\end{matrix}}$

Add the digits of the right-most column: 8+1=9

$\frac{\begin{matrix}\space\space&0&\textbf{8}&4\\ +&2&\textbf{1}&0\end{matrix}}{\begin{matrix}\space\space&\space\space&\textbf{9}&4\end{matrix}}$

Add the digits of the right-most column: 0+2=2

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

Therefore,

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

________

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

6 0

3 years ago