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stepan [7]
2 years ago
7

The length of a rectangle is 3 cm greater than the

Mathematics
1 answer:
Nitella [24]2 years ago
7 0
I wowowowowowowowowowow
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Kite A B C D is shown. Lines are drawn from point A to point C and from point B to point D and intersect. In the kite, AC = 10 a
Oduvanchick [21]

Answer:

30 u^{2}

Step-by-step explanation:

The computation of the area of kite ABCD is shown below:

Given data

AC = 10 ;

BD = 6

As we can see from the attached figure that the Kite is a quadrilateral as it involves two adjacent sides i.e to be equal

Now the area of quadrilateral when the diagonals are given

So, it is

\text { area of kite }=\frac{1}{2} \times d_{1} d_{2}

where,

d_{1}=10\ and\ d_{2}=6

So, the area of the quadrilateral is

=\frac{1}{2}(10)(6)\\\\=30 u^{2}

4 0
3 years ago
A school purchases boxes of candy bars each box contains 50 candy bars each box cost $20. how much does the school have to charg
Butoxors [25]
20/50=.4  .6x50=30 so ur answer is $0.60
6 0
3 years ago
Read 2 more answers
Find the value of x and explain how you got the answer
pashok25 [27]
5x + 3 = 38 
5x = 38-3 
5x = 35 
x= 7 
8 0
2 years ago
Read 2 more answers
Theresa bought three containers of tennis balls at $2.98 each. She had a coupon for $1 off. Her mom paid for half of the remaini
Rzqust [24]

Answer:

$3.97

Step-by-step explanation:

Three containers @$2.98 = 3 × 2.98 = $8.94

                                               Less $1 =   <u>-1.00 </u>

                                                                  7.94

Less 50% paid my mom = ½ × 7.94 =   <u>-3.97 </u>

                               Paid by Theresa =  $3.97

[(3 × 2.98) - 1] ÷ 2 = (8.94 - 1) ÷ 2 = 7.94 ÷ 2 = 3.97

6 0
3 years ago
Can anyone pls help me to solve question 2 f and g and pls provide me a explanation I’m with that questions for three days
zvonat [6]

Answer:

  f)  a[n] = -(-2)^n +2^n

  g)  a[n] = (1/2)((-2)^-n +2^-n)

Step-by-step explanation:

Both of these problems are solved in the same way. The characteristic equation comes from ...

  a[n] -k²·a[n-2] = 0

Using a[n] = r^n, we have ...

  r^n -k²r^(n-2) = 0

  r^(n-2)(r² -k²) = 0

  r² -k² = 0

  r = ±k

  a[n] = p·(-k)^n +q·k^n . . . . . . for some constants p and q

We find p and q from the initial conditions.

__

f) k² = 4, so k = 2.

  a[0] = 0 = p + q

  a[1] = 4 = -2p +2q

Dividing the second equation by 2 and adding the first, we have ...

  2 = 2q

  q = 1

  p = -1

The solution is a[n] = -(-2)^n +2^n.

__

g) k² = 1/4, so k = 1/2.

  a[0] = 1 = p + q

  a[1] = 0 = -p/2 +q/2

Multiplying the first equation by 1/2 and adding the second, we get ...

  1/2 = q

  p = 1 -q = 1/2

Using k = 2^-1, we can write the solution as follows.

The solution is a[n] = (1/2)((-2)^-n +2^-n).

 

5 0
2 years ago
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