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STatiana [176]
3 years ago
15

I need help figuring this out asap please

Mathematics
1 answer:
Vera_Pavlovna [14]3 years ago
4 0

Answer:

38.13 ft squared

Step-by-step explanation:

Since it is half of a circle we first find the area of a normal circle then divide by two.

So to find the area of  a circle it is pi times radius squared which in this case it would be 3.14 times 9 = 28.26

Then since it is a half circle we divide by two so

28.26/2 = 14.13

Then to find the area of the triangle we do 4 times 6 which is 24.

now we add the two which is 14.13 + 24 which is

38.13

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what is the equation of a line, in point-slope form, that passes through (5, -3) and has a slope of 2/3? please help me
Brums [2.3K]

y + 3 = \frac{2}{3} (x - 5)

the equation of a line in point- slope form is

y - b = m ( x - a )

where m is the slope and (a, b) a point on the line

here m =\frac{2}{3} and (a,b) = (5 , - 3 )

y + 3 = \frac{2}{3} (x - 5 ) ← in point-slope form


3 0
3 years ago
Given the Standard Form equation y - 6x = 4
malfutka [58]

Answer: <u>6 is the slope and 4 is the y-intercept.</u>

Step-by-step explanation:

Based on the question, I feel this is the best way to answer your question.  I'm assuming you are in a basic graphing class.

The best, most useful thing for you right now would be to learn slope-intercept form, y = mx + b, where m and b are constants.

Simply add 6x to both sides to get into this form

y = 6x + 4.

In slope-intercept form, m is the slope, and b is the y-intercept.  Thus, 6 is the slope and 4 is the y-intercept.

Hope it helps and lmk if you need more <3 :)

7 0
2 years ago
If B = { p : p is a factor of 12 } list the elements of this set and find n ( B )​
anastassius [24]

Answer:

B={1,2,3,4,6and12}

n (B) =6

Step-by-step explanation:

<h3>Greetings !</h3>

Factor, in mathematics, a number or algebraic expression that divides another number or expression evenly—i.e., with no remainder. For example, 3 and 6 are factors of 12 because 12 ÷ 3 = 4 exactly and 12 ÷ 6 = 2 exactly. The other factors of 12 are 1, 2, 4, and 12.

Hope it helps!

3 0
2 years ago
Jkl is an equilateral triangle jk= 13x+5,KL= 17X-19 AND JL= 8X+35​
stira [4]

Answer:

x = 6

Step-by-step explanation:

Since the triangle is equilateral then sides are congruent.

Equate any 2 sides and solve for x

13x + 5 = 8x + 35 ( subtract 8x from both sides )

5x + 5 = 35 ( subtract 5 from both sides )

5x = 30 ( divide both sides by 5 )

x = 6

JK = (13 × 6) + 5 = 78 + 5 = 83

KL = (17 × 6) - 19 = 102 - 19 = 83

JL = (8 × 6) + 35 = 48 + 35 = 83

ΔJKL is equilateral with side = 83

4 0
3 years ago
1)A System of equations is shown below . What is the solution to the system of equations? 5x+2y=-15 2x-2y=-6
meriva

Answer:

x= -3 and y= 0

Step-by-step explanation:

5x+2y=-15

<u>2x-2y=-6     </u>

<u>7x        =-21</u>

x= -3

Putting value of x in equation 1  

5(-3) +2y=-15

-15+2y= -15

2y= 0

y= 0

This can be solved with the help of matrices

In matrix form the above equations can be written in the form

\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]  \left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]

Let

\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right] = A  \left[\begin{array}{ccc}x\\y\\\end{array}\right]  = X  and  \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]= B

Then AX= B

or X= A⁻¹ B

where  A⁻¹= adj A/ ║A║   where mod A≠ 0

adj A=  \left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]

║A║= ( 5*-2- 2*2)= -10-4= -14≠0

X= A⁻¹ B

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]    =- 1/14  \left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]   \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]    =- 1/14     \left[\begin{array}{ccc}-2*-15&+ -2*-6\\-2*-15&+ 5*-6\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]  =- 1/14 \left[\begin{array}{ccc} 30&+12\\30&+-30\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]  =- 1/14 \left[\begin{array}{ccc}42\\0\\\end{array}\right]

\left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-42/14\\0/-14\\\end{array}\right]

\left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-3\\0\\\end{array}\right]

From here x= -3 and y= 0

Solution Set = [(-3,0)]

3 0
2 years ago
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