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3 years ago

I need help with these two problems... PLZ HELP ME!!!!!!!

Mathematics

1 answer:

hichkok12 [17]3 years ago

3 0

Answer:

3-B,E

4-. ( 1/8)×π×100

arc length of an enclosed circle is 2πr where r is the radius

the atc length of is proportional to the angle subtended at the centre . i.e for this arc it is 1/4th (90°/360°) of the arc length of an enclosed circle.

it is the same relationship with area of a sector

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The length of the rectangular basketball court is 44 feet more than the width. If the perimeter of the basketball court is 288 f

Temka [501]

Let's first say that L=W+44

and then remember that perimeter is P=2L+2W

replace the L with W+44

we then get P=2(W+44)+2W, now I'll solve it

P=2W+88+2W
P=4W+88

substitute 288 for P

288=4W+88
200=4W
50=W

so now we now how wide the court is. add 44 to find the length which gives you L=94

as always plug the numbers back into your perimeter equation to ensure L and W are correct

8 0

3 years ago

Please awnser thase questions for my prarice test

Vinvika [58]

Answer:

Add 5 for the first one

The 2nd one is 25 and k or 25 thousand.

Step-by-step explanation:

add 5 for the first one because 5 + 5 = 10

9 + 5 = 14 . and 13 + 5 = 18.

------------------------------------------------------------------

For the 2nd one: 25k means 25 thousand or the Product of 25 and k.

5 0

2 years ago

The midpoint of AB is M (0, -3). If the coordinates of AA are (2, 1) what are the coordinates of BB?

frez [133]

Answer:

B(-2, -7)

Step-by-step explanation:

Midpoint is (0, -3)

A(x₁, y₁) B(x₂, y₂)

A(2, 1)

$Midpoint = (\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2} )\\\\Midpoint = (\frac{2+ x_{2} }{2}, \frac{1 + y_{2} }{2} )$

$(0, -3) = (\frac{2+ x_{2} }{2}, \frac{1 + y_{2} }{2} )$

$\frac{2+ x_{2} }{2} = 0\\\\2 + x_{2} = 0\ \ \ \ times\ both\ sides\ by\ 2\\x_{2} = -2\ \ \ \ subtract\ both\ sides\ by\ 2$

$\frac{1 + y_{2} }{2} = -3\\\\1 + y_{2} = -6\ \ \ \ times\ both\ sides\ by\ 2\\y_{2} = -7\ \ \ \ subtract\ both\ sides\ by\ 1$

∴ B(-2, -7)

6 0

2 years ago

How do I solve for Log(x+3)+log2=3 ?

Rzqust [24]

Answer: x = 995

Step-by-step explanation:

In this equation our log doesn't have a base so it is given the default base of 10. This makes our equation really.

$log_{10}(x+3) + log_{10}(2) = 3$

Now we put every equation as the exponent of our base 10 looks like this:

$10^{log_{10} (x+3)} + 10^{log_{10} (2)} = 10^{3}$

Now we cancel out the log base 10 and the 10. We are left with:

(x+3) + (2) = 10^3

Lastly solve for x

x + 5 = 1000

x = 995

7 0

3 years ago

Read 2 more answers

Insert geometric means in each geometric sequence.

Digiron [165]

Answer:

$\underline{192}, 24, \underline{3}, \underline{\dfrac{3}{8}}, \dfrac{3}{64}$

$\underline{\dfrac{1}8}, \dfrac{1}{4}, \dfrac{1}{2}, \underline{1}$

$81, \underline{27, 9, 3, 1},\dfrac{1}{3}$

Step-by-step explanation:

Given the Geometric sequences:

1. ___, 24, ___, ___, 3/64

2. ___, 1/4, 1/2, ___

3. 81, ___, ___, ___, ___, 1/3

To find:

The values in the blanks of the given geometric sequences.

Solution:

First of all, let us learn about the $n^{th}$ term of a geometric sequence.

$a_n=ar^{n-1}$

Where $a$ is the first term and

$r$ is the common ratio by which each term varies from the previous term.

Considering the first sequence, we are given the second and fifth terms of the sequences.

Applying the above formula:

$ar = 24\\ar^4 = \dfrac{3}{64}$

Solving the above equation:

$r = \dfrac{1}{8}$

Therefore, the sequence is:

$\underline{192}, 24, \underline{3}, \underline{\dfrac{3}{8}}, \dfrac{3}{64}$

Considering the second given sequence:

$ar = \dfrac{1}{4}\\ar^2 = \dfrac{1}{2}\\\text{Solving the above equations}, r = 2$

Therefore, the sequence is:

$\underline{\dfrac{1}8}, \dfrac{1}{4}, \dfrac{1}{2}, \underline{1}$

Considering the third sequence:

$a = 81\\ar^5=\dfrac{1}{3}\\\Rightarrow r = 3$

Therefore, the sequence is:

$81, \underline{27, 9, 3, 1},\dfrac{1}{3}$

5 0

3 years ago