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mylen [45]
3 years ago
15

PLS HELP! DUE SOON, ANSWER ASAP

Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
4 0

Answer: y-1= 1/2(x-2)

Or

Y-3=1/2(x-6)

Step-by-step explanation:

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Solve for x.<br> Your answer must be simplified.<br><br> x+5≤−4
iren2701 [21]

Answer:

x ≤ -9

Step-by-step explanation:

to solve this equation you must subtract 5 from both sides

you get

x ≤ -9

that is your answer

4 0
2 years ago
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A group of friends went to lunch. The bill, before sales tax and tip was $35.55. A sale tax of 6% was added. The group then tipp
Mashcka [7]
I did the math and I think it is $62.20??
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2 years ago
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What is 3x3-45 divided by (4+1)
Lorico [155]

Answer:

3x3-45÷(4+1) is 0.

Step-by-step explanation:

We have to do order of operations so we follow PEMDAS:

P is for Parentheses so we do what's in parentheses first:

(4+1)=5

E is for Exponents but we don't have exponents.

Next is M and D for Multiplication and Division. We have division before multiplication so we should do division first:

45÷5=9

We now have multiplication:

3x3=9

Now we have A and S for addition and subtraction

We don't have addition so we'll do subtraction:

9-9=0

The final answer is 0.

Hope this helps! :)

<h3>CloutAnswers</h3>
5 0
3 years ago
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What is the length of the hypotenuse of the right triangle? Enter your answer in the box.
ki77a [65]

Answer:

10 ft

Step-by-step explanation:

Using Pythagoras' identity in the right triangle

let h be hypotenuse, then

h² = 6² + 8² = 36 + 64 = 100 ( take the square root of both sides )

h = \sqrt{100} = 10

The hypotenuse is 10 ft

5 0
3 years ago
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A circle is centered at J(3, 3) and has a radius of 12.
stealth61 [152]

Answer:

(-6,\, -5) is outside the circle of radius of 12 centered at (3,\, 3).

Step-by-step explanation:

Let J and r denote the center and the radius of this circle, respectively. Let F be a point in the plane.

Let d(J,\, F) denote the Euclidean distance between point J and point F.

In other words, if J is at (x_j,\, y_j) while F is at (x_f,\, y_f), then \displaystyle d(J,\, F) = \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}.

Point F would be inside this circle if d(J,\, F) < r. (In other words, the distance between F\! and the center of this circle is smaller than the radius of this circle.)

Point F would be on this circle if d(J,\, F) = r. (In other words, the distance between F\! and the center of this circle is exactly equal to the radius of this circle.)

Point F would be outside this circle if d(J,\, F) > r. (In other words, the distance between F\! and the center of this circle exceeds the radius of this circle.)

Calculate the actual distance between J and F:

\begin{aligned}d(J,\, F) &= \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}\\ &= \sqrt{(3 - (-6))^{2} + (3 - (-5))^{2}} \\ &= \sqrt{145}  \end{aligned}.

On the other hand, notice that the radius of this circle, r = 12 = \sqrt{144}, is smaller than d(J,\, F). Therefore, point F would be outside this circle.

5 0
3 years ago
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