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GarryVolchara [31]
3 years ago
12

The equation A = x(x - 7) describes the area, A, of a rectangular flower garden, where x is the width in feet.

Mathematics
1 answer:
mamaluj [8]3 years ago
8 0

x(x-7)=8

x^2-7x-8=0

(x-8)(x+1)=0

so x=8 (and -1 but width can’t be negative)

So your answer is C. 8

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A fair die is rolled once. Let A be the event of rolling an even number, and let B be the event of rolling a number greater than
IRISSAK [1]

Answer:

A ∩ B = {4, 6}

Step-by-step explanation:

A die had 6 faces

S = {1, 2, 3, 4, 5, 6}

If A  be the event of rolling an even number, then;

A = {2, 4, 6}

If B be the event of rolling a number greater than 3, then;

B = {4, 5, 6}

A ∩ B are the values that are common to both sets

A ∩ B = {4, 6}

3 0
3 years ago
Help!! (You will get 100 points! & Brainliest if you are correct)
Dafna1 [17]

Answer: The first number that appears in both sequences is 28.

Step-by-step explanation:

Let's write down numbers from each of the sequences

Sequence 1) We need to start from 7 and multiply 4

7x4=28, 28x4=112

The sequence is 7,28,112...

Sequence 2) We need to start from 8 and add 5

5+8=13, 13+5=18, 18+5=23, 23+5=28

The sequence is 8,13,18,23,28...

They both have 28

5 0
2 years ago
Read 2 more answers
What is the length of the darkened arc?​
Stells [14]
3/4 would have to be a answer
7 0
4 years ago
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Obtain the general solution of<br> y ln x ln y dx + dy = 0
oksano4ka [1.4K]
Dy/dx = -y inx iny

(y in x in y) dx + dy = 0

(y in x in y ) dx = - dy

(in x )dx =  - (1/ y in y) dy

you just need to integrate both sides

Hope this helps
4 0
4 years ago
Find an equation of the sphere with center (2, −10, 3) and radius 5. $$25=(x−2)2+(y−(−10))2+(z−3)2 use an equation to describe i
lana66690 [7]

In the x,y plane, we have z=0 everywhere. So in the equation of the sphere, we have

25=(x-2)^2+(y+10)^2+(-3)^2\implies(x-2)^2+(y+10)^2=16=4^2

which is a circle centered at (2, -10, 0) of radius 4.

In the x,z plane, we have y=0, which gives

25=(x-2)^2+10^2+(z-3)^2\implies(x-2)^2+(z-3)^2=-75

But any squared real quantity is positive, so there is no intersection between the sphere and this plane.

In the y,z plane, x=0, so

25=(-2)^2+(y+10)^2+(z-3)^2\implies(y+10)^2+(z-3)^2=21=(\sqrt{21})^2

which is a circle centered at (0, -10, 3) of radius \sqrt{21}.

3 0
4 years ago
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