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Pachacha [2.7K]
3 years ago
14

(9.03)×10 3rd power -(2.03×10

2.03%20%5Ctimes%20%20%7B10%7D%5E%7B2%7D%20%29%20%20%3D%20" id="TexFormula1" title="(9.03 \times {10}^{3} ) - (2.03 \times {10}^{2} ) = " alt="(9.03 \times {10}^{3} ) - (2.03 \times {10}^{2} ) = " align="absmiddle" class="latex-formula">
Mathematics
1 answer:
alina1380 [7]3 years ago
5 0
The answer is 8827, I hope that this helps.

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If the length of a pool is 8 more than twice the width and the total area is 960 square feet, find the length and width of the r
melomori [17]

Answer:

l=48, w=20

Step-by-step explanation:

Let the width of the pool be w feet, then the length will be:

l = 2w + 8

The area of the pool is 960 square feet.

This means that:

w(2w + 8) = 960

Expand and rewrite in standard form to get:

2 {w}^{2}  + 8w - 960 = 0

{w}^{2}  + 4w  - 480 = 0

{w}^{2}  + 24w   - 20w- 480 = 0

{w}(w+ 24) - 20(w + 24) = 0

Factor to get:

(w+ 24)(w  - 20) = 0

w =  - 24 \: or \: w   =  20

We discard the negative value because magnitude is positive.

This means

l = 2 \times 20 + 8 = 48

3 0
4 years ago
Let $m$ be the number of integers $n$, $1 \le n \le 2005$, such that the polynomial $x^{2n} + 1 + (x + 1)^{2n}$ is divisible by
Kisachek [45]

x^2+x+1=\dfrac{x^3-1}{x-1}

so we know x^2+x+1 has roots equal to the cube roots of 1, not including x=1 itself, which are

\omega=e^{2i\pi/3}\text{ and }\omega^2=e^{4i\pi/3}

Any polynomial of the form p_n(x)=x^{2n}+1+(x+1)^{2n} is divisible by x^2+x+1 if both p(\omega)=0 and p(\omega^2)=0 (this is the polynomial remainder theorem).

This means

p(\omega)=\omega^{2n}+1+(1+\omega)^{2n}=0

But since \omega is a root to x^2+x+1, it follows that

\omega^2+\omega+1=0\implies1+\omega=-\omega^2

\implies\omega^{2n}+1+(-\omega^2)^{2n}=0

\implies\omega^{4n}+\omega^{2n}+1=0

and since \omega^3=1, we have \omega^{4n}=\omega^{3n}\omega^n=\omega^n so that

\implies\omega^{2n}+\omega^n+1=0

From here, notice that if n=3k for some integer k, then

\omega^{2(3k)}+\omega^{3k}+1=1+1+1=3\neq0

\omega^{4(3k)}+\omega^{2(3k)}+1=1+1+1=3\neq0

which is to say, p(x) is divisible by x^2+x+1 for all n in the given range that are *not* multiples of 3, i.e. the integers 3k-2 and 3k-1 for k\ge1.

Since 2005 = 668*3 + 1, it follows that there are m=668 + 669 = 1337 integers n such that x^2+x+1\mid p(x).

Finally, m\equiv\boxed{337}\pmod{1000}.

8 0
4 years ago
Find the area of the triangle.
vodka [1.7K]
if you multiply 18 and 11 you get 198 the you divide it by two and get the final answer of 99
5 0
3 years ago
An engineer is designing a play set for an elementary school. The drawing above shows the detailed view for the pre-drilled hole
Kazeer [188]
This is a case of counter bore holes. The correct answer is C.

Where:
2 ---- Indicates two holes.
∅1.25 THRU ---- Shows that the holes have a diameter of 1.25 through the material.
Then, there is a counterbore of diameter 2.25 and a depth of 1.00.
 
3 0
4 years ago
Solve −5x = 15. (1 point)<br><br><br> −3<br> 3<br> 10<br> 20
Luba_88 [7]
Remember you can do anything to an eqautoin as long as you do oit to both sides

-5x=15
try to get 1x by itself

remember
(ax)/a=x when a=a

so
-5x=15
get x
divide both sides by -5
remember to flip sign
(-5x)/(-5)=15/(-5)
x=-3

answer is first one
x=-3 

8 0
3 years ago
Read 2 more answers
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