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Sergeeva-Olga [200]
3 years ago
14

Any one know this I’ll put the most brilliant answer there’s the picture

Mathematics
1 answer:
Lelechka [254]3 years ago
5 0

Answer:

ms. sempe is right.

.

first we'll find area of trapezium.

area of trapezium = 1/2×height×(sum of parellel sides of the trapezium)

area of trapezium =1/2×4×(13+24)

area of trapezium =1/2×4×37

area pf trapezium = 74in²

now area of rectangle.

area of rectangle = length ×width

area of rectangle= 24in×7in

area of rectangle =168 in²

area of the figure= 74in²+168in²

area of figure=242in²

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In how many ways can 4 people sit next to each other in a stadium row​
Viktor [21]

Answer:

They can sit next to each other in 24 different ways.

Step-by-step explanation:

In order to solve this problem, use the factorial function (!).

4! = 4 * 3 * 2 * 1 = 24

8 0
3 years ago
Find the total surface area of a trapezoidal prism. <br> Please help. Picture below.
sineoko [7]
The total surface area would be 370.58 square units.
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3 years ago
Prove that the roots of x2+(1-k)x+k-3=0 are real for all real values of k​
masha68 [24]

Answer:

Roots are not real

Step-by-step explanation:

To prove : The roots of x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0 are real for all real values of k ?

Solution :

The roots are real when discriminant is greater than equal to zero.

i.e. b^2-4ac\geq 0b

2

−4ac≥0

The quadratic equation x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0

Here, a=1, b=1-k and c=k-3

Substitute the values,

We find the discriminant,

D=(1-k)^2-4(1)(k-3)D=(1−k)

2

−4(1)(k−3)

D=1+k^2-2k-4k+12D=1+k

2

−2k−4k+12

D=k^2-6k+13D=k

2

−6k+13

D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))

For roots to be real, D ≥ 0

But the roots are imaginary therefore the roots of the given equation are not real for any value of k.

6 0
3 years ago
If Jessica can bike 20kilometers in 0.8hours how far can she travel in an hour
likoan [24]

Answer:16 hours

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
in quadrilateral ABCD, AE=x+10, and BE=3x-18. for what value of x is ABCD a rectangle? a.24 b.14 c.18 d.16
Evgen [1.6K]

Answer:

for x  = 14, the given ABCD is a RECTANGLE.

Step-by-step explanation:

Given : ABCD is  a quadrilateral.

E is the intersection points of the diagonal AC and BD.

AE  =  x+10, and BE=  3x -18

Now, as we know:

IN A RECTANGLES , DIAGONALS BISECT EACH OTHER.

⇒  E is the mid point of AC and BD respectively.

⇒ AE = BE

⇒ x+10 = 3 x - 18

or, 10 +  18 = 3 x - x

or, 2  x = 28

or, x  = 28/2 = 14   ⇒  x = 14

Hence, for x  = 14, the given ABCD is a RECTANGLE.

4 0
3 years ago
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