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Anestetic [448]
3 years ago
11

$627 is 95% of what amount ?

Mathematics
2 answers:
marusya05 [52]3 years ago
8 0
To find the amount you just need to divide 627 by .95 and then add that to 627 to get 1287. 
Hope I helped

Scrat [10]3 years ago
6 0
595.65 because u take 627 by 95 percent

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The answer is b or the second choice available
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Guess my age and ill make you brianlest 2007 8 14
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Which of the following inequalities would be graphed with an open circle?
icang [17]

3(x - 2) < 18 and h + 6 > -3h + 12 would be graphed with an open circle.

Open circle is a term used for numbers that are less than (<) or greater than (>). On the other hand, Closed circle is used for numbers that are less than or equal to (≤ ) and greater than or equal to (≥).

Here, we are given four inequalities, let us check them one by one-

A. h + 6 > -3h + 12

⇒ 4h > 6

or h > 3/2

Here, we see that there is no equality sign, this means that 3/2 won't be included and hence h + 6 > -3h + 12 would be an open circle inequality.

B. 7 ≥ d/3

⇒ d ≤ 21

Here, we can clearly see the equality sign which means that 21 will be included. Hence, 7 ≥ d/3 would not be graphed with an open circle.

C. 3(x - 2) < 18

x - 2 < 6

x < 8

Here, we see that there is no equality sign, this means that 8 won't be included and hence 3(x - 2) < 18 would be graphed with an open circle.

D. 2y ≤ 14

y ≤ 7

Here, we can clearly see the equality sign which means that 7 will be included. Hence, 2y ≤ 14 would not be graphed with an open circle.

Thus, 3(x - 2) < 18 and h + 6 > -3h + 12 would be graphed with an open circle.

Learn more about inequalities here-

brainly.com/question/24372553

#SPJ9

3 0
2 years ago
What is an interior obtuse angle​
jeka57 [31]

Answer:

An obtuse-angled triangle is a triangle in which one of the interior angles measures more than 90° degrees. In an obtuse triangle, if one angle measures more than 90°, then the sum of the remaining two angles is less than 90°.

6 0
3 years ago
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Suppose \nabla f (x,y) = 3 y \sin(xy) \vec{i} + 3 x \sin(xy)\vec{j}, \vec{f} = \nabla f(x,y), and c is the segment of the parabo
Anna11 [10]

I'll assume you're supposed to compute the line integral of \nabla f over the given path C. By the fundamental theorem of calculus,

\displaystyle\int_C\nabla f(x,y)\cdot\mathrm d\vec r=f(4,48)-f(1,3)

so evaluating the integral is as simple as evaluting f at the endpoints of C. But first we need to determine f given its gradient.

We have

\dfrac{\partial f}{\partial x}=3y\sin(xy)\implies f(x,y)=-3\cos(xy)+g(y)

Differentiating with respect to y gives

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f(x,y)=-3\cos(xy)+C

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8 0
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