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3 years ago

What is the area of the trapezoid?

Mathematics

1 answer:

MA_775_DIABLO [31]3 years ago

4 0

Answer:

62 square mm.

Step-by-step explanation:

First find the area of the rectangle

4*10=40

Find the area of the Triangle

11*4=44

44÷2=22

Add

40+22=62

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Find the domain and range of f(x)= 2|x+2|-8

nordsb [41]

Answer:

Domain (-∞,∞)

Range: [-8,∞)

3 0

3 years ago

Please answer question 4

olasank [31]

#4)
cross multiply
2(8x+10) = 4 * 5x
16x + 20 = 20x
4x = 20
x = 5

3 0

3 years ago

A and B are independent events. P(A) = 0.50 and P(B) = 0.30. What is<br> PA and B)?

katen-ka-za [31]

Answer:

0.15

Step-by-step explanation:

5 0

3 years ago

Sage and Tom started the month with the same number of talk minutes on their cell phones. Sage talked for 7 minutes with her dad

Paraphin [41]

Answer:

a. Sage and Tom have the same number of talk minutes left on their cell phone plans.

b. y = x - 7

Explanation:

Sage talked for 7 minutes with her dad.

Tom talked for 4 minutes with a friend and 3 more minutes with his mom which means he spoke for :

= 4 + 3

= 7 minutes

They both spoke for 7 minutes and therefore have the same amount of time left.

Algebraic expression:

Assuming the time Tom started with is denoted by x and the time he is left with is denoted by y.

The expression is:

y = x - 7

6 0

3 years ago

Rewrite the system of linear equations as a matrix equation AX = B.

iren2701 [21]

Answer:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

Step-by-step explanation:

Let's find the answer.

Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

Equations:

Eq. 1 : x1 + 2x2 + 5x3 = 5

Eq. 2 : x1 + x2 + x3 = 6

E1. 3 : 4x1 + 6x2 + 5x3 = 7

Coefficients for x1 ; x2 ; x3

From eq. 1 : 1 ; 2 ; 5

From eq. 2 : 1 ; 1 ; 1

From eq. 3 : 4 ; 6 ; 5

So matrix A is:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]$

And the vector of vriables (X) is:

$\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]$

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

$\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

In conclusion, AX=B is:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

7 0

4 years ago