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Gwar [14]
3 years ago
15

Show me a graph that best represents the solution set to the system of a linear inequalities; y-x>0. y-1>0

Mathematics
1 answer:
tangare [24]3 years ago
4 0
<h2>Answer:</h2>

The graph is shown in the attached image

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Kate is saving to take an SAT prep course that costs $250, so far , she has saved $150, and she adds $10to her savings each week
matrenka [14]

Answer:

It will be ten weeks

Step-by-step explanation:

First you would need  to find out what is 250 - 150 = 100 So they will need 100 dollars 100 divided by 10 is 10 so that will be 10 weeks

7 0
3 years ago
Question 7<br> #7) What value of m makes the equation true?<br> 3(8 – m) = 5m – 40
Anastaziya [24]

Answer:

8

Step-by-step explanation:

3(8-m)=5m-40

you first distribute 3 through the parenthesis

24-3m=5m-40

Move the variable to the left and change sign

24-3m-5m=-40

now move the constant to the right

-3m-5m=-40-24

collect like terms

-8m=-40-24

-8m=-64

Divide both sides by -8 and negative divide negative number becomes positive

so m = 8

7 0
3 years ago
Read 2 more answers
A cake has circumference of 25 1/7inches. What is the area of the cake? Use 22/7 to approximate π. Round to the nearest hundredt
bixtya [17]
We know that
area of the circle=pi*r²

circumference==2*pi*r----------> r=circumference/(2*pi)
circumference=25 1/7 in----> (25*7+1)/7----> 176/7 in
r=(176/7)/(2*22/7)----> r=176/44----> r=4 in

area of the circle is equal to the area of the cake
area=pi*r²---> (22/7)*4²-----> area=50.29 in²

the answer is
50.29 in²
5 0
3 years ago
;-; hehehehehehehehehe
iogann1982 [59]

Answer: hahahahahaha

Step-by-step explanation:

Hahahahaha mines more cringe ;)

8 0
3 years ago
Read 2 more answers
Find a and b such that gcd(a, b) = 14, a &gt; 2000, b &gt; 2000, and the only prime divisors of a and b are 2 and 7.
choli [55]

Answer:

a = 2^\alpha.14 \ ,\alpha\geq 8\\b = 7^\delta.14 \ , \delta\geq 3

Step-by-step explanation:

As 2  and 7 are the only prime divisors of both a and b we know that both can be written as:

a = 2^\alpha . 7^\beta \ , \alpha ,\beta\ \in \mathbb{N}_{0}\\b = 2^\gamma . 7^\delta \ , \gamma ,\delta\ \in \mathbb{N}_{0}\\

Where \mathbb{N}_{0} is the set of all natural numbers adding the zero (careful because this part is important as I'll explain next).

We also know that 14 divides both numbers and that is actually the greatest common divisor between them. So we can rewrite a and b as follows:

a = 2^\alpha . 7^\beta.14 \ , \alpha ,\beta\ \in \mathbb{N}_{0}\\b = 2^\gamma . 7^\delta.14 \ , \gamma ,\delta\ \in \mathbb{N}_{0}\\

Why do I write them like this? Because this way is easier to observe that if \alpha and \gamma were both greater than zero, then 28 would divide both hence 14 wouldn't be their g.c.d.. Likewise, if \beta and \delta were both greater than zero, then 98 would divide both and once again, 14 wouldn't be their g.c.d.

So either of them has to be equal to zero. And then we have that

a = 2^\alpha.14 \\b = 7^\delta.14

All we have left to do is find the possible values for \alpha and \delta so that a>2000 \ , b>2000 and that only happens if \alpha\geq 8 and \delta\geq 3

8 0
3 years ago
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