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grandymaker [24]
3 years ago
10

Please HELP HELP JUST CHOSE THE RIGHT ANSWER AND I WILL MARK BRAINLEST

Mathematics
1 answer:
Zepler [3.9K]3 years ago
7 0
I think the answer is A
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In what form is the following linear equation y-3=2/3(x-1)
AveGali [126]
Y=2/3x-7/3
Hope this helps
8 0
3 years ago
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Pleaseeeee help I’m begging someone I PROMISE. I will mark brainlest
Lerok [7]

Answer:


Step-by-step explanation:

a) x>0 → items can not be negative

b) c(2,000) = 5,000+1.3(2,000) = $7,600

c)5,000+1.3x<10,000

5,000-5,000+1.3x<10,000-5,000

1.3x<5,000

x<3846 items


5 0
4 years ago
a circle has an area of 225 square inches. if the radius of the circle is multiples by 6, then what is the new area of the circl
Lesechka [4]

Answer:

D:  8,100π in²

Step-by-step explanation:

The formula for area of a circle is

A = πr²

The radius is squared, so if the radius or a new circle is 6 times larger (multiplied by 6), the difference in area will be 36 times greater.  So the area is

225π(36) = 8,100π

3 0
3 years ago
Which of the following is a factor of x2 − 8x + 15?
brilliants [131]

Answer:

(x−3)(x−5)

Step-by-step explanation:

Let's factor x2−8x+15

x2−8x+15

The middle number is -8 and the last number is 15.

Factoring means we want something like

(x+_)(x+_)

Which numbers go in the blanks?

We need two numbers that...

Add together to get -8

Multiply together to get 15

Can you think of the two numbers?

Try -3 and -5:

-3+-5 = -8

-3*-5 = 15

Fill in the blanks in

(x+_)(x+_)

with -3 and -5 to get...

(x-3)(x-5)

3 0
3 years ago
Read 2 more answers
Find the smallest positive $n$ such that \begin{align*} n &amp;\equiv 3 \pmod{4}, \\ n &amp;\equiv 2 \pmod{5}, \\ n &amp;\equiv
Alex777 [14]

4, 5, and 7 are mutually coprime, so you can use the Chinese remainder theorem right away.

We construct a number x such that taking it mod 4, 5, and 7 leaves the desired remainders:

x=3\cdot5\cdot7+4\cdot2\cdot7+4\cdot5\cdot6

  • Taken mod 4, the last two terms vanish and we have

x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod4

so we multiply the first term by 3.

  • Taken mod 5, the first and last terms vanish and we have

x\equiv4\cdot2\cdot7\equiv51\equiv1\pmod5

so we multiply the second term by 2.

  • Taken mod 7, the first two terms vanish and we have

x\equiv4\cdot5\cdot6\equiv120\equiv1\pmod7

so we multiply the last term by 7.

Now,

x=3^2\cdot5\cdot7+4\cdot2^2\cdot7+4\cdot5\cdot6^2=1147

By the CRT, the system of congruences has a general solution

n\equiv1147\pmod{4\cdot5\cdot7}\implies\boxed{n\equiv27\pmod{140}}

or all integers 27+140k, k\in\mathbb Z, the least (and positive) of which is 27.

3 0
3 years ago
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