Hi please help i’ll give brainliest thanks

(-5,-4) rotated 270° counterclockwise

stepladder [879]

Answer:

(4, - 5 )

Step-by-step explanation:

Under a counterclockwise rotation about the origin of 270°

a point (x, y ) → (- y, x ) , thus

(- 5, - 4 ) → (4, - 5 )

6 0

3 years ago

State the property that justifies each statement:1. If 80 mA, then me = 802. If RS - TU and TU = YP, then RS = YP2.3. If 7x 28,

BabaBlast [244]

Question 3

if 7x = 28

divide both sides by 7

7x/7 = 28/7

x = 4

This is division property of equality

Question 1

if angle m is 80 degrees the m = 80 degrees

this can be a congruent angles or alternate angles are equal

this is symmetric property of equality

Question 5

m1 = 30 and mz1 = m2, then m2 = 30

this property is transitive property of equality

Question 2

if RS - TU and TU =YP, then RS = YP

this means you can substitute RS with YP

then we have our new expression as

YP - YP

this is substitution property of equality, this is because we are substituting a function with another function

4 0

1 year ago

4x + 9 + = 24 matematicas

Harlamova29_29 [7]

Answer:

3.75

Step-by-step explanation:

4x+9=24

4x=24-9

4x=15

x=15/4

x=3.75

7 0

2 years ago

Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on

Nookie1986 [14]

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

If he is given with two kings.
If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by = $\frac{^{3}C_2 }{^{47}C_2}$ {∵ one king is already there}

= $\frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}$ {∵ $^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{3}{1081}$ = 0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by = $\frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}$

= $\frac{3!}{1! \times 2!}\times \frac{2!}{1! \times 1!} \times \frac{2! \times 45!}{47!}$

= $\frac{6}{1081}$ = 0.0055

Now, probability that he ends up with a full house = $\frac{3}{1081} + \frac{6}{1081}$

= $\frac{9}{1081}$ = <u>0.0083</u>.

3 0

3 years ago

Read 2 more answers

Apply Newton’s method to estimate the solution of x3 − x − 1 = 0 by taking x1 = 1 and finding the least n such that xn and xn +

ss7ja [257]

Solution :

Given

$f(x)=x^3-x-1, x_1=1$

$f'(x)=3x^2-1$

Let the initial approximation is $x_1 =1$

So by Newton's method, we get

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)},n=1,2,...$

$x_2=x_1-\frac{f(x_1)}{f'(x_1)}=1-\frac{1^3-1-1}{3(1)^2-1}=1.5$

$x_3=1.5-\frac{(1.5)^3-1.5-1}{3(1.5)^2-1}=1.34782608$

$x_4=1.34782608-\frac{(1.34782608)^3-1.34782608-1}{3(1.34782608)^2-1}=1.32520039$

$x_5=1.32520039-\frac{(1.32520039)^3-1.32520039-1}{3(1.32520039)^2-1}=1.32471817$

$x_6=1.32471817-\frac{(1.32471817)^3-1.32471817-1}{3(1.32471817)^2-1}=1.32471795$

$x_5 \approx x_6$ are identical up to eight decimal places.

The approximate real root is x ≈ 1.32471795

∴ x = 1.32471795

4 0

3 years ago