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DaniilM [7]
1 year ago
7

Calculate the area of this trapezium:

Mathematics
2 answers:
QveST [7]1 year ago
8 0

Answer:

72

Step-by-step explanation:

hopefully you dont need work since you need it asap :(

poizon [28]1 year ago
6 0

Answer:

Step-by-step explanation:

1/2 (a+b) x h

a+b = 9+15= 24 / 2 = 12

12 x 6 = 72cm2

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What is the value of (4-2): – 3 x 4?<br> -20<br> -4<br> 4<br> 20
jarptica [38.1K]

Answer:

4

Step-by-step explanation:

5 0
3 years ago
F(x) = 9 + 4x f(0) = f(-1) = Find the value of x for which f(x) =6 x=
puteri [66]

Answer: x=-3/4

Step-by-step explanation:

Since we know f(x)=6, we can set it equal to the equation.

6=9+4x           [subtract 9 on both sides]

-3=4x              [divide both sides by 4]

x=-3/4

8 0
3 years ago
Can someone explain to me how to turn a fraction into a decimal..? And a decimal into a fraction? Be specific please! I have a t
cricket20 [7]
To turn a decimal into a fraction move the decimal point two spots over (Ex: 0.34=34%). To turn a fraction into a decimal get the denominator to 100 by multiplying both the top and bottom by whatever number needed (must both be multiplied by the same number) or divide to top number into the bottom number of you can’t multiply.
8 0
3 years ago
12) Josh has $252 in his bank account. This is $12 more than 6 times the amount of money
OLga [1]

Answer:

40

Step-by-step explanation:

(252-12)/6=40

7 0
3 years ago
let t : r2 →r2 be the linear transformation that reflects vectors over the y−axis. a) geometrically (that is without computing a
tangare [24]

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b)  two eigen values of 'k' = 1, -1

for k = 1, eigen vector is \left[\begin{array}{c}0\\1\end{array}\right]

for k = -1 eigen vector is \left[\begin{array}{c}1\\0\end{array}\right]

See the figure for the graph:

(a) for any (x, y) ∈ R² the reflection of (x, y) over the y - axis is ( -x, y )

∴ x → -x hence '-1' is the eigen value.

∴ y → y hence '1' is the eigen value.

also, ( 1, 0 ) → -1 ( 1, 0 ) so ( 1, 0 ) is the eigen vector for '-1'.

( 0, 1 ) → 1 ( 0, 1 ) so ( 0, 1 ) is the eigen vector for '1'.

(b) ∵ T(x, y) = (-x, y)

T(x) = -x = (-1)(x) + 0(y)

T(y) =  y = 0(x) + 1(y)

Matrix Representation of T = \left[\begin{array}{cc}-1&0\\0&1\end{array}\right]

now, eigen value of 'T'

T - kI =  \left[\begin{array}{cc}-1-k&0\\0&1-k\end{array}\right]

after solving the determinant,

we get two eigen values of 'k' = 1, -1

for k = 1, eigen vector is \left[\begin{array}{c}0\\1\end{array}\right]

for k = -1 eigen vector is \left[\begin{array}{c}1\\0\end{array}\right]

Hence,

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b)  two eigen values of 'k' = 1, -1

for k = 1, eigen vector is \left[\begin{array}{c}0\\1\end{array}\right]

for k = -1 eigen vector is \left[\begin{array}{c}1\\0\end{array}\right]

Learn more about " Matrix and Eigen Values, Vector " from here: brainly.com/question/13050052

#SPJ4

6 0
1 year ago
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