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2 years ago

If hexagon JGHFKI Is congruent to hexagon SRTQUV, which pair of angles must be congruent?

Mathematics

2 answers:

vagabundo [1.1K]2 years ago

6 0

Answer:

j and s

Step-by-step explanation:

i just took the test

Dima020 [189]2 years ago

3 0

Hey there!

If these hexagons are congruent, then each angle is congruent to the one that is in the same position as it is. In this case, I and V are congruent because they are both last. K and U are congruent because they are both second to last.

While looking over the answer options, you should be able to see that Angle J and Angle S is the only answer with angles that match up, because they are both the first angle in the set.

I hope this helps!

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Suri made 6 gallons of lemonade to sell at her lemonade stand. in one day she sold 2/3 of the lemonade. how much lemonade did sh

Zina [86]

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7 0

2 years ago

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Consider the initial value problem: 2ty′=8y, y(−1)=1. Find the value of the constant C and the exponent r so that y=Ctr is the s

VikaD [51]

The correct question is:

Consider the initial value problem

2ty' = 8y, y(-1) = 1

(a) Find the value of the constant C and the exponent r such that y = Ct^r is the solution of this initial value problem.

b) Determine the largest interval of the form a < t < b on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution.

c) What is the actual interval of existence for the solution obtained in part (a) ?

Step-by-step explanation:

Given the differential equation

2ty' = 8y

a) We need to find the value of the constant C and r, such that y = Ct^r is a solution to the differential equation together with the initial condition y(-1) = 1.

Since Ct^r is a solution to the initial value problem, it means that y = Ct^r satisfies the said problem. That is

2tdy/dt - 8y = 0

Implies

2td(Ct^r)/dt - 8(Ct^r) = 0

2tCrt^(r - 1) - 8Ct^r = 0

2Crt^r - 8Ct^r = 0

(2r - 8)Ct^r = 0

But Ct^r ≠ 0

=> 2r - 8 = 0 or r = 8/2 = 4

Now, we have r = 4, which implies that

y = Ct^4

Applying the initial condition y(-1) = 1, we put y = 1 when t = -1

1 = C(-1)^4

C = 1

So, y = t^4

b) Let y = F(x,y)................(1)

Suppose F(x, y) is continuous on some region, R = {(x, y) : x_0 − δ < x < x_0 + δ, y_0 −ę < y < y_0 + ę} containing the point (x_0, y_0). Then there exists a number δ1 (possibly smaller than δ) so that a solution y = f(x) to (1) is defined for x_0 − δ1 < x < x_0 + δ1.

Now, suppose that both F(x, y)

and ∂F/∂y are continuous functions defined on a region R. Then there exists a number δ2

(possibly smaller than δ1) so that the solution y = f(x) to (1) is

the unique solution to (1) for x_0 − δ2 < x < x_0 + δ2.

c) Firstly, we write the differential equation 2ty' = 8y in standard form as

y' - (4/t)y = 0

0 is always continuous, but -4/t has discontinuity at t = 0

So, the solution to differential equation exists everywhere, apart from t = 0.

The interval is (-infinity, 0) n (0, infinity)

n - means intersection.

7 0

2 years ago

Points H and F lie on circle c What is the length of line segment GH?

Nina [5.8K]

Answer:

6 units

Step-by-step explanation:

Given: Points H and F lie on circle with center C. EG = 12, EC = 9 and ∠GEC = 90°.

To find: Length of GH.

Sol: EC = CH = 9 (Radius of the same circle are equal)

Now, GC = GH + CH

GC = GH + 9

Now In ΔEGC, using pythagoras theorem,

$(Hypotenuse)^{2} = (Base)^{2} +(Altitude)^{2}$ ......(ΔEGC is a right triangle)

$(GC)^{2} = (GE)^{2} +(EC)^{2}$

$(GH + 9)^{2} = (9)^{2} +(12)^{2}$

$(GH )^{2} + (9)^{2} + 18GH = 81 + 144$

$(GH )^{2} + 81 + 18GH = 81 + 144$

$(GH )^{2} + 18GH = 144$

Now, Let GH = <em>x</em>

$x^{2} +18x = 144$

On rearranging,

$x^{2} +18 x - 144 = 0$

$x^{2} - 6x +24x + 144 = 0$

$x (x-6) + 24 (x - 6) =0$

$(x - 6) (x + 24) = 0$

So x = 6 and x = - 24

∵ x cannot be - 24 as it will not satisfy the property of right triangle.

Therefore, the length of line segment GH = 6 units. so, Option (D) is the correct answer.

3 0

2 years ago

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A test tube contains 25 bacteria, 5 of which are can stay alive for atleast 30 days, 10 of which will die in their second day. 1

boyakko [2]

Answer:

c. $\frac{1}{5}$

Step-by-step explanation:

Given,

Number of bacterias who alive for at least 30 days = 5,

Bacterias who alive for 2 days = 10,

Died bacterias = 10,

Total bacterias = 5 + 10 + 10 = 25,

Ways of choosing a bacteria = $^{25}C_1$ = $\frac{25!}{1! 24!}$ = 25,

While, ways of choosing of a bacteria who will live after 1 week = $^5C_1$ = $\frac{5!}{1!4!}$ = 5,

Hence, the probability it will still be alive after one week = $\frac{5}{25}$ = $\frac{1}{5}$

OPTION C is correct.

3 0

2 years ago

100 POINTS AND BRAINLIYEST!!!

Helga [31]

See explanations below

Step-by-step explanation:

Chanel

(i) 90° counterclockwise and then dilation by a factor of 1/2

Trapezoid PQRS is given as :

P(6,-4) -------P'(4,6)

Q(2,-2)------Q'(2,2)

R(2,-6)-------R'(6,2)

S(6,-6)-------S'(6,6)

The transformation switches the position of (x,y) to and adds a -ve sign to the original y value

B. The new coordinates are : P'(4,6), Q'(2,2),R'(6,2) and S'(6,6)

The rule for rotation 90° counterclockwise about the origin is (x,y)⇒(-y,x)

The rule for dilation with scale factor of 1/2 is (x,y)------(1/2x,1/2y)

P'(4,6) -----P''(2,3)

Q'(2,2)-----Q''(1,1)

R'(6,2)-----R''(3,1)

S'( 6,6)-----S''(3,3)

The dilation reduces the coordinates by 1/2

Preston

E. (i)Reflection over x-axis where the rule is : (x,y)----(x,-y)

(ii)Dilation by scale factor of 1/2 the rule is : (x,y)---(1/2x, 1/2y)

Applying reflection over x-axis rule;

P(6,-4) -----P'(6,4)

Q(2,-2)-----Q'(2,2)

R(2,-6)------R'(2,6)

S(6,-6)------S'(6,6)

The coordinates have changed in that a negative sign is added to the y-axis coordinate.

F. P'(6,4), Q'(2,2), R'(2,6) and S'(6,6)

G. Applying the dilation by scale factor of 1/2

P'(6,4)-------P''(3,2)

Q'(2,2)------Q''(1,1)

R'(2,6)------R''(1,3)

S'(6,6)------S''(3,3)

The coordinates of P'Q'R'S are divide by 2 to give coordinates of P''Q''R''S''

Comparing the answers to the final figure given

K(3,2)

L(1,1)

M(1,3)

N(3,3)

Preston is correct because his sequence proves trapezoid PQRS is similar to trapezoid KLMN.

Learn More

Reflection and dilation :brainly.com/question/13408902

Keywords :trapezoid, maps, rotate, counterclockwise, dilate,reflect,scale factor,shape, coordinate

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2 years ago

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