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3 years ago

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 1 1 + + 1+ a + 6-1 1+b+c-1 1+ C + a = 1.

hram777 [196]

Answer:

plz write the question correctly

4 0

3 years ago

Read 2 more answers

A grade of B is worth ___ grade points A. 2.0 B. 4.0 C. 3.0 D. 80

scZoUnD [109]

The correct answer is C. 3.0

Hope this helps

-AaronWiseIsBae

7 0

3 years ago

Read 2 more answers

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 parallelogram:

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

Therefore the parallel sides are:

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

Therefore, the diagonals of the parallelogram are:

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

To find the coordinates of the intersection of the diagonals, either:

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

Midpoint between two points

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

To find the midpoint of diagonal LJ, 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