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3 years ago

What is the area of the figure in the picture?

Mathematics

2 answers:

Ilia_Sergeevich [38]3 years ago

6 0

Answer:

$26.14yd^2$

Step-by-step explanation:

Triangle

$A=\frac{h_bb}{2} \\=\frac{4*6}{2} \\=12yd^2\\\\A=\frac{1}{2} \pi r^2\\=\frac{1}{2} \pi 3^2\\=14.14yd^2\\\\12+14.14=26.14yd^2$

coldgirl [10]3 years ago

6 0

Answer:

Step-by-step explanation:

The area of a circle knowing the diameter is 28.2743, now divide between 2: 14,13715

Now the area of the triangle give us: 9.238

9.238 + 14,13715 = 23.37515

The procedure is in the images

App used: CalculateMyFigure

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Consider the line shown on the graph. Enter the

Semenov [28]

<h2>Solution -</h2>

From the graph, we concluded that line passes through the points (0, 3) and (1, 4) respectively.

So, slope of the line (m) passing through the points (a, b) and (c, d) is

m = d - b/ c - a

<h3>So, on substituting the values, we get </h3>

m = 4 - 3 / 1 - 0 = 1/1 = 1

Also, line makes an intercept of 3 units in positive direction of y - axis.

So, c = 3

<h3>So, Equation of line is given by</h3>

y = mx + c

y = 1 × x + 3

<h3>Alternative Method :-</h3>

We know, Equation of line passing through the points (a,b) and (c, d) is given by

y - b = d - b / c - a (x - a)

a = 0

b = 3

c = 1

d = 4

<h3>So, on substituting these values in above result, we get :-</h3>

y - 3 = 4 - 3 /1 - 0 ( x - 0 )

y - 3 = 1/1 ( x - 0 )

y - 3 = x

<h3>Different forms of equations of a straight line :-</h3>

1. Equations of horizontal and vertical lines

Equation of line parallel to x - axis passes through the point (a, b) is y = b.

Equation of line parallel to y - axis passes through the point (a, b) is x = a.

<h3>2. Point-slope form equation of line</h3>

Equation of line passing through the point (a, b) having slope m isy - b = m(x - a)

<h3>3. Slope-intercept form equation of line</h3>

Equation of line which makes an intercept of c units on y axis and having slope m is y = mx + c.

<h3>4. Intercept Form of Line</h3>

Equation of line which makes an intercept of a and b units on x - axis and y axis respectively is x/a + y/b = 1.

<h3>5. Normal form of Line</h3>

Equation of line which is at a distance of p units from the origin and perpendicular makes an angle B with the positive X-axis is x cosB + y sinB = p.

<h2 />

8 0

2 years ago

A certain bookstore chain has two stores, one in San Francisco and one in Los Angeles. It stocks three kinds of books: hardcover

riadik2000 [5.3K]

Answer:

<h2>See the explanation.</h2>

Step-by-step explanation:

(a)

$\left[\begin{array}{cccc}T&H&S&P\\S&600&1300&2000\\L&400&300&400\end{array}\right]$ = A.

In the above matrix A, the columns refers the three type of books and the rows refers the from which stores the books are been sold.

The numbers represents the corresponding sales in the month of January.

The sale is same for the 6 months.

Hence, 6A = $\left[\begin{array}{cccc}T&H&S&P\\S&3600&7800&12000\\L&2400&1800&2400\end{array}\right]$ . This matrix 6A represents the total sales over the 6 months.

(b)

If we denote the books in stock at the starting of January by B, then

B = $\left[\begin{array}{cccc}T&H&S&P\\S&1000&3000&6000\\L&1000&6000&3000\end{array}\right]$ .

Each month, the chain restocked the stores from its warehouse by shipping 500 hardcover, 1,400 softcover, and 1,400 plastic books to San Francisco and 500 hardcover, 500 softcover, and 500 plastic books to Los Angeles.

If we represent the amount restocked books at the end of each month by another matrix C, then

C = $\left[\begin{array}{cccc}T&H&S&P\\S&500&1400&1400\\L&500&500&500\end{array}\right]$ .

This restocking will be done for 5 times before the end of June.

If there would be no sale, then the stock would be

B + 5C = $\left[\begin{array}{cccc}T&H&S&P\\S&1000+2500&3000+7000&6000+7000\\L&1000+2500&6000+2500&3000+2500\end{array}\right] \\= \left[\begin{array}{cccc}T&H&S&P\\S&3500&10000&13000\\L&3500&8500&5500\end{array}\right]$ .

Since, the total sale is given by 6A, at the end of June, the inventory in each store can be shown as following,

B+5C-6A $\left[\begin{array}{cccc}T&H&S&P\\S&3500&10000&13000\\L&3500&8500&5500\end{array}\right] - \left[\begin{array}{cccc}T&H&S&P\\S&3600&7800&12000\\L&2400&1800&2400\end{array}\right] \\= \left[\begin{array}{cccc}T&H&S&P\\S&-100&2200&1000\\L&1100&6700&3100\end{array}\right]$

6 0

3 years ago

Factor the following:<br><br> x2−6x+5

lukranit [14]

Answer:

the 6x element as there is no solution in integers.

Step-by-step explanation:

8 0

3 years ago

What the answer to this problem

Lana71 [14]

The Answer Would Be The Second Bubble. The Quotient is 6 with a remainder of 3.

Hope I Helped :)

8 0

3 years ago

If cos θ > 0, which quadrant(s) could the terminal side of θ lie?

Sergeeva-Olga [200]

Answer:

I and IV

Step-by-step explanation:

cosine is positive/greater than 0 in quadrants I and IV. Its just a rule of the unit circle.

Can remember by All Star Trig Class

meaning all the trig funtions are positive in quadrant I

sin is positive in II

tan is positive in III

and cos is positive in IV

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3 years ago