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kipiarov [429]
2 years ago
14

If AB= 4 units and DC=4 units , what is the length of BC​

Mathematics
1 answer:
Degger [83]2 years ago
4 0

Answer:

4 units i think

Step-by-step explanation:

i dont know how beacuse 4 is answer

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A principal of $450 is deposited in an account that pays 2.5% interest compounded yearly. Find the account balance after 3 years
KonstantinChe [14]
A=p(1+r)^t
A=450×(1+0.025)^(3)
A=484.60
4 0
3 years ago
Read 2 more answers
Solve the following word problem: Kay made 5 less phone calls today than Stephanie. If together they made 21 calls, how many cal
professor190 [17]
Answer:
Kay made 8 phone calls
Stephanie made 13 phone calls

Explanation:
Assume the number of phone calls that Stephanie made is x.
We are given that:
Kay made 5 phone calls less than Stephanie.
This means that:
Kay made x-5 phone calls

We know that the total number of phones is 21.
This means that:
21 = x + x - 5

Now we solve for x as follows:
21 = x + x - 5
21 = 2x - 5
2x = 26
x = 26/2
x = 13

This means that:
Stephanie made x = 13 phone calls
Kay made x - 5 = 13 - 5 = 8 phone calls

Hope this helps :)
5 0
3 years ago
According to a college survey 32% of all students work fulltime. Find the mean for the number of students who work full time in
VLD [36.1K]

Answer:

mean is equal to 5.12

Step-by-step explanation:

<em>We are given that 32% of college students work fulltime. We have to find the mean for the number of student s who are working full time in a sample of 16</em>

success rate, p = 32% = 0.32

Sample size is denoted as n = 16

The forumla of mean is given as

mean = sample size × success rate

mean = n × p

         = 16 × 0.32

         = 5.12

3 0
3 years ago
I guess u have to find the area
Nesterboy [21]
18
Because 6•3= 18
Hope this helped
7 0
3 years ago
Read 2 more answers
Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}
miskamm [114]

Answer:

-3+i\sqrt{3} , 1+\sqrt{3}

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

\alpha = x+iy\\\beta = x-iy

since they are conjugates

\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}

\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}

Imaginary part of the above =0

i.e. \sqrt{3} (x^2-3)+2x\sqrt{3} =0\\x^2+2x-3=0\\(x+3)(x-1) =0\\x=-3,1

So the value of alpha = -3+i\sqrt{3} , 1+\sqrt{3}

3 0
3 years ago
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