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antoniya [11.8K]
1 year ago
13

Square MATH is shown at the right. Assume its vertices have integer coordinates. What are the coordinates of the midpoint of the

diagonal from H to A?​

Mathematics
1 answer:
Mariana [72]1 year ago
8 0

Answer: F(-0.5,-0.5)

Step-by-step explanation:

A(3,3)      H(-4,-4)

The coordinates of the midpoint of the diagonal AH F(x,y):

\displaystyle\\F(x,y)=(\frac{3+(-4)}{2} ,\frac{3+(-4)}{2} )\\\\F(x,y)=(\frac{3-4}{2} ,\frac{3-4}{2})\\\\F(x,y)=(\frac{-1}{2},\frac{-1}{2})\\\\F (x,y) =(-0.5,-0.5)

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What is area of each circle below?
KiRa [710]

Answer:

Step-by-step explanation: A= (Pi) x r^2

4 0
3 years ago
What’s the distance between the points (-11, -16) and (20, -16)
pishuonlain [190]

Answer:

Sorry no one has answered for so long, by now you probably already answered but anyway, the answer would be 31.

Step-by-step explanation:

Because the y axes values are the same, we just need to figure out how far they are on the x axes. Which is indeed 31.

5 0
3 years ago
Answers should be exact decimals (please do not leave as fractions or unfinished calculations).
aksik [14]

Answer:

P(rolling the die results in a 3 or a 6)=0.3333

P(rolling the die results in not getting a 6)=0.8333

P(rolling the die results in the number being either greater than 2 or less than 5)=0.3333

P(rolling the die results in an even number or a 5)=0.6667

P( rolling the die twice results in a 6 followed by a 4)=0.0278

P(rolling the die three times results in all 6's)=0.0046

Step-by-step explanation:

a)

As the all outcomes are equally likely, so each number has equal probability of occurring.

For six sided die the sample space is {1,2,3,4,5,6} and each outcome has equal probability of 1/6.

1)

P(rolling the die results in a 3 or a 6)

P(rolling the die results in a 3 or a 6)=P(3)+P(6)=1/6+1/6=2/6=1/3

P(rolling the die results in a 3 or a 6)=0.3333

P(rolling the die results in not getting a 6)

P(rolling the die results in not getting a 6)=1-P(6)=1-1/6=5/6

P(rolling the die results in not getting a 6)=0.8333

P(rolling the die results in the number being either greater than 2 or less than 5)

number either greater than 2 or less than 5={3,4}

P(rolling the die results in the number being either greater than 2 or less than 5)=2/6=1/3

P(rolling the die results in the number being either greater than 2 or less than 5)=0.3333

P(rolling the die results in an even number or a 5)

number is an even number or a 5={2,4,5,6}

P(rolling the die results in an even number or a 5)=4/6=2/3

P(rolling the die results in an even number or a 5)=0.6667

P( rolling the die twice results in a 6 followed by a 4)

For six sided die rolled twice the sample space is

S={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}.

n(S)=6²=36.

rolling the die twice results in a 6 followed by a 4={(6,4)}

P( rolling the die twice results in a 6 followed by a 4)=1/36.

P( rolling the die twice results in a 6 followed by a 4)=0.0278

P(rolling the die three times results in all 6's)

When three die are rolled the number of outcomes=n(S)=6³=216.

rolling the die three times results in all 6's={(6,6,6)}

P(rolling the die three times results in all 6's)=1/216

P(rolling the die three times results in all 6's)=0.0046

Note: All answers are rounded to four decimal places.

8 0
3 years ago
Unsure how to do this calculus, the book isn't explaining it well. Thanks
krok68 [10]

One way to capture the domain of integration is with the set

D = \left\{(x,y) \mid 0 \le x \le 1 \text{ and } -x \le y \le 0\right\}

Then we can write the double integral as the iterated integral

\displaystyle \iint_D \cos(y+x) \, dA = \int_0^1 \int_{-x}^0 \cos(y+x) \, dy \, dx

Compute the integral with respect to y.

\displaystyle \int_{-x}^0 \cos(y+x) \, dy = \sin(y+x)\bigg|_{y=-x}^{y=0} = \sin(0+x) - \sin(-x+x) = \sin(x)

Compute the remaining integral.

\displaystyle \int_0^1 \sin(x) \, dx = -\cos(x) \bigg|_{x=0}^{x=1} = -\cos(1) + \cos(0) = \boxed{1 - \cos(1)}

We could also swap the order of integration variables by writing

D = \left\{(x,y) \mid -1 \le y \le 0 \text{ and } -y \le x \le 1\right\}

and

\displaystyle \iint_D \cos(y+x) \, dA = \int_{-1}^0 \int_{-y}^1 \cos(y+x) \, dx\, dy

and this would have led to the same result.

\displaystyle \int_{-y}^1 \cos(y+x) \, dx = \sin(y+x)\bigg|_{x=-y}^{x=1} = \sin(y+1) - \sin(y-y) = \sin(y+1)

\displaystyle \int_{-1}^0 \sin(y+1) \, dy = -\cos(y+1)\bigg|_{y=-1}^{y=0} = -\cos(0+1) + \cos(-1+1) = 1 - \cos(1)

7 0
1 year ago
Mai bank account is overdrawn by $125 which means her balance is negative 125 she gets $85 for her birthday and deposit into her
stira [4]
The answer is -125+85= -40 because she is adding -125 by 85 which equals-40
4 0
3 years ago
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