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dedylja [7]
3 years ago
11

Need help asap plz

Mathematics
1 answer:
Maksim231197 [3]3 years ago
8 0

Answer:

I'm sorry but i only know a) so yeah

anyway a) is 20 in.

Step-by-step explanation:

Hope it helps!

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Simplify 12 square root of 1764
Marina CMI [18]
You need 12 square roots of 1764 ?

<span>Why not find the square root of 1764 first (it's 42) and then take 12 of those ?

Hope this helps!</span>
7 0
4 years ago
In the system shown below, what are the coordinates of the solution that lies in quadrant I?
Arisa [49]

Answer:

(4,3)

Step-by-step explanation:

Given

x^2+y^2=25

x-y^2=-5

In order to solve the equations, from equation 2 we get

-y^2= -5-x

y^2=5+x

Putting the value of y^2 in equation 1

x^2+5+x=25

x^2+5-25+x=0

x^2+x-20=0

x^2+5x-4x-20= 0

x(x+5)-4(x+5)=0

(x+5)(x-4)=0

So

x+5=0  x-4=0

x=-5   x=4

Now for x=-5

x^2+y^2=25

(-5)^2+y^2=25

25+y^2=25

y^2=25-25

y^2=0

so Y=0  

And for x = 4

x^2+y^2=25

(4)^2+y^2=25

16+y^2=25

y^2=25-16

y^2=9

y= ±3

So the solution to the system of equations is

(-5,0) , (4,3), (4,-3)

The only solution that belongs to first quadrant is (4,3)

4 0
4 years ago
Read 2 more answers
Write the equation 4X -3Y = 12 in the form Y =MX +b
USPshnik [31]

Answer:

Step-by-step explanation:

we have 4x-3y=12

-3y=-4x+12

-y=-4x+12/3

we have 12=3*4  so we remove 3 that is up and 3 that is down

-y=-4x+4

then we get y=+4x+4

4 0
4 years ago
Given: angle 1 is supplementary to angle 2 and angle 1 is congruent to angle 3.
gregori [183]
 ∠1 and ∠2 are supplementary    // given∠3 and ∠4 are supplementary    // given∠1 ≅ ∠3    // given m∠1 + m∠2 = 180°    // definition of supplementary anglesm∠3 + m∠4 = 180°    // definition of supplementary angles m∠1 + m∠2 = m∠3 + m∠4    // transitive property of equality m∠1 = m∠3    // definition of congruent angles m∠1 + m∠2 = m∠1 + m∠4    // substitution property of equality (replaced m∠3 with m∠1) m∠2 = m∠4    // subtraction property of equality (subtracted m∠1 from both sides) ∠2 ≅ ∠4    // definition of congruent angles


8 0
4 years ago
What is the smallest integer $n$, greater than $1$, such that $n^{-1}\pmod{130}$ and $n^{-1}\pmod{231}$ are both defined?
olasank [31]

First of all, the modular inverse of n modulo k can only exist if GCD(n, k) = 1.

We have

130 = 2 • 5 • 13

231 = 3 • 7 • 11

so n must be free of 2, 3, 5, 7, 11, and 13, which are the first six primes. It follows that n = 17 must the least integer that satisfies the conditions.

To verify the claim, we try to solve the system of congruences

\begin{cases} 17x \equiv 1 \pmod{130} \\ 17y \equiv 1 \pmod{231} \end{cases}

Use the Euclidean algorithm to express 1 as a linear combination of 130 and 17:

130 = 7 • 17 + 11

17 = 1 • 11 + 6

11 = 1 • 6 + 5

6 = 1 • 5 + 1

⇒   1 = 23 • 17 - 3 • 130

Then

23 • 17 - 3 • 130 ≡ 23 • 17 ≡ 1 (mod 130)

so that x = 23.

Repeat for 231 and 17:

231 = 13 • 17 + 10

17 = 1 • 10 + 7

10 = 1 • 7 + 3

7 = 2 • 3 + 1

⇒   1 = 68 • 17 - 5 • 231

Then

68 • 17 - 5 • 231 ≡ = 68 • 17 ≡ 1 (mod 231)

so that y = 68.

3 0
3 years ago
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