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Vikki [24]
3 years ago
7

The sum of three consecutive intergers is 36.if the first integer is y,find the value of y.

Mathematics
1 answer:
wariber [46]3 years ago
6 0
Hi,
Since they are consecutive, the equation will be:
y + (y + 1) + ( y + 2)=36
y + y + y +1 +2 = 36
3y + 3 = 36
3y = 36-3
3y = 33
y = 33/3
y=11
The three consecutive integers are 11, 12(11+1) and 13(11+2).
Hope this helps you.
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solong [7]
Foil the binomials

First: n*n=n^2
Outer: n*-5=-5n
Inner: -3*n=-3n
Last: -3*-5=15

Put it together and simplify
n^2-5n-3n+15
n^2-8n+15

Final answer: A
5 0
3 years ago
Find the midpoint of a and b when a has coordinates (-5,3) and b has coordinates (3,-1)
BlackZzzverrR [31]

Answer:

The midpoint is (2 , 1)

Step-by-step explanation:

To find the midpoint, we have to add the corresponding coordinates

[(-5 , 3) + (3 , -1)] / 2

we separate into the corresponding

(-5 + 3) / 2 =

-2 / 2 = -1

(3 - 1) / 2 =

2 / 2 = 1

The midpoint is (2 , 1)

3 0
3 years ago
Find the two square roots of each complex number by creating and solving polynomial equations.
son4ous [18]

Answer:

1) w₁=4 - i w₂= -4 + i

2) w₁= 3 - i w₂= -3  + i

3) w₁= 1 + 2i w₂= - 1 - 2i

4) w₁= 2- 3i w₂= -2 + 3i

5) w₁= 5 - 2i w₂= -5 + 2i

6) w₁= 5 - 3i w₂= -5 + 3i

Step-by-step explanation:

The root of a complex number is given by:

\sqrt[n]{z}=\sqrt[n]{r}(Cos(\frac{\theta+2k\pi}{n}) + i Sin(\frac{\theta+2k\pi}{n}))

where:

r: is the module of the complex number

θ: is the angle of the complex number to the positive axis x

n: index of the root

1) z = 15 − 8i  ⇒ r=17 θ= -0.4899 rad

w₁=\sqrt{17}(Cos(\frac{-0.4899}{2}) + i Sin(\frac{-0.4899}{2}))=4-i

w₂=\sqrt{17}(Cos(\frac{-0.4899+2\pi}{2}) + i Sin(\frac{-0.4899+2\pi}{2}))=-1+i

2) z = 8 − 6i  ⇒ r=10 θ= -0.6435 rad

w₁=\sqrt{10}(Cos(\frac{ -0.6435}{2}) + i Sin(\frac{ -0.6435}{2}))= 3 - i

w₂=\sqrt{10}(Cos(\frac{ -0.6435+2\pi}{2}) + i Sin(\frac{ -0.6435+2\pi}{2}))= -3  + i

3) z = −3 + 4i  ⇒ r=5 θ= -0.9316 rad

w₁=\sqrt{5}(Cos(\frac{-0.9316}{2}) + i Sin(\frac{-0.9316}{2}))= 1 + 2i

w₂=\sqrt{5}(Cos(\frac{-0.9316+2\pi}{2}) + i Sin(\frac{-0.9316+2\pi}{2}))= -1 - 2i

4) z = −5 − 12i  ⇒ r=13 θ= 0.4426 rad

w₁=\sqrt{13}(Cos(\frac{0.4426}{2}) + i Sin(\frac{0.4426}{2}))= 2- 3i

w₂=\sqrt{13}(Cos(\frac{0.4426+2\pi}{2}) + i Sin(\frac{0.4426+2\pi}{2}))= -2 + 3i

5) z = 21 − 20i  ⇒ r=29 θ= -0.8098 rad

w₁=\sqrt{29}(Cos(\frac{-0.8098}{2}) + i Sin(\frac{-0.8098}{2}))= 5 - 2i

w₂=\sqrt{29}(Cos(\frac{-0.8098+2\pi}{2}) + i Sin(\frac{-0.8098+2\pi}{2}))= -5 + 2i

6) z = 16 − 30i ⇒ r=34 θ= -1.0808 rad

w₁=\sqrt{34}(Cos(\frac{-1.0808}{2}) + i Sin(\frac{-1.0808}{2}))= 5 - 3i

w₂=\sqrt{34}(Cos(\frac{-1.0808+2\pi}{2}) + i Sin(\frac{-1.0808+2\pi}{2}))= -5 + 3i

6 0
3 years ago
Is lmn congruent to opq if so name the postulate
insens350 [35]

Answer:

Option (A)

Step-by-step explanation:

Given:

LM ≅ OP

MN ≅ PQ

∠M ≅ ∠P

To Prove:

ΔLMN ≅ ΔOQP

                   Statements                      Reasons

1). LM ≅ OP                                   1). Given

2). MN ≅ PQ                                 2). Given

3). ∠P ≅  ∠M                                3). Given

4). ΔLNM ≅ ΔOQP                      4). By the SAS postulate of congruence.

                                                         [Side - Angle - Side]                          

Therefore, Option (A) will be the answer.

5 0
3 years ago
What is the product?<br><br> (5r − 4)(r2 − 6r + 4)
maxonik [38]

Answer:

1r

Step-by-step explanation:

(5r-4)(2r-6r+4)

1(5r-4)+1(2r-6r+4)

5r-4+2r-6r+4

5r+2r-6=1r

1r-4+4

-4+4=0

1r

7 0
2 years ago
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