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cricket20 [7]
3 years ago
8

Y is less than or equal to

Mathematics
1 answer:
Snezhnost [94]3 years ago
8 0

Answer:

less than I think

not sure tho

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Find the square roots of 119 + 120i algebraically.
blondinia [14]

Answer:

Step-by-step explanation:

Given

z=119+120 i

Let \sqrt{119+120 i}=p+iq

Squaring both sides

119+120 i=p^2-q^2+2ipq

Comparing real and imaginary part

Re(LHS)=Re(RHS)

119=p^2-q^2-----------1

comparing Im(LHS)=Im(RHS)

120=2pq

q=\frac{60}{p}

Substitute q in 1

119=p^2-(\frac{60}{p})^2

p^4-119p^2-(68)^2=0

Let x=p^2

x^2-119x-4624=0

x=\frac{119\pm \sqrt{119^2+4\times 4624}}{2}

x=\frac{119\pm 180.71}{2}

we take only Positive value because p^2=x

x=149.85

p^2=149.85

thus p=\pm 12.24

q=\mp 4.90

thus \sqrt{119+120 i}=\pm (12.24+i 4.90)

8 0
3 years ago
1<br>1<br>1<br>+<br>+<br>1+ a + 6-1<br>1+b+c-1<br>1+ C + a<br>= 1.​
hram777 [196]

Answer:

plz write the question correctly

4 0
3 years ago
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A grade of B is worth ___ grade points<br><br>A. 2.0<br>B. 4.0<br>C. 3.0<br>D. 80
scZoUnD [109]

The correct answer is C. 3.0

Hope this helps

-AaronWiseIsBae

7 0
3 years ago
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Find the coordinates of the intersection of the diagonals of parallelogram HJKL with the given vertices: H(-1, 4), J(3, 3), K(3,
lakkis [162]

Answer:

(1, 1)

Step-by-step explanation:

Given vertices of the <u>parallelogram</u>:

  • H = (-1, 4)
  • J = (3, 3)
  • K = (3, -2)
  • L = (-1, -1)

Therefore the <u>parallel sides</u> are:

\sf \overline{HJ} \parallel \overline{LK}\:\: \textsf{ and }\:\: \overline{LK} \parallel \overline{HL}

Therefore, the <u>diagonals</u> of the parallelogram are:

\sf \overline{LJ} \:\: \textsf{ and }\:\:\overline{HK}

To find the <u>coordinates of the intersection of the diagonals</u>, either:

  1. draw a diagram (see attached) and determine the point of intersection of the diagonals from the diagram, or
  2. determine the midpoint of either diagonal (as the diagonals of a parallelogram bisect each other, i.e. divide into 2 equal parts).

<u>Midpoint between two points</u>

\textsf{Midpoint}=\left(\dfrac{x_2+x_1}{2},\dfrac{y_2+y_1}{2}\right)\quad \textsf{where}\:(x_1,y_1)\:\textsf{and}\:(x_2,y_2)\:\textsf{are the endpoints}}\right)

To find the <u>midpoint of diagonal LJ</u>, define the endpoints:

  • (x_1,y_1)=L=(-1, -1)
  • (x_2,y_2)=J=(3,3)

Substitute the defined endpoints into the formula and solve:

\begin{aligned} \implies \textsf{Midpoint of LJ} & =\left(\dfrac{3-1}{2},\dfrac{3-1}{2}\right)\\ & =\left(\dfrac{2}{2},\dfrac{2}{2}\right)\\ & =\left(1,1\right) \end{aligned}

Therefore, the coordinates off the intersection of the diagonals of parallelogram HJKL are (1, 1).

Learn more about midpoints here:

brainly.com/question/27962681

8 0
2 years ago
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A discount of 16% saved jackie 38.00 on office supplies what was the price of the supplies before the discount
Oksanka [162]

16 %  38 dollars.  By proportion:-

100 %   =   100*38  /  16  =   237.5   answer

8 0
4 years ago
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