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asambeis [7]
3 years ago
14

Write a word phrase for 12 + a

Mathematics
2 answers:
Shkiper50 [21]3 years ago
6 0
It is the sum of twelve plus a hope this helps
skelet666 [1.2K]3 years ago
3 0
The sum of twelve and a number a.
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Your friend is graphing the point (2,5) 2 , 5 . Her first step is to start at the origin and go up 2 2 units. Is her method corr
Alchen [17]
No. You need to first move along the x-axis.
3 0
3 years ago
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Evaluate.<br><br> 2 • {[5 • (60 – 14 ÷ 7)] + 25} • 5
boyakko [2]

Answer:

3,150.

Step-by-step explanation:

  1. 2 · {[ 5 · ( 60 - 2 )] + 25 } · 5.
  2. 2 · {[ 5 · ( 58 )] + 25 } · 5.
  3. 2 · { 290 + 25 } · 5.
  4. 2 · { 315 } · 5.
  5. 630 · 5.
  6. 3,150.

These are all of the steps to completely and correctly solve this question.

Hope this helps.

Kyle.

5 0
3 years ago
Can someone help me with this
Phantasy [73]
I pretty sure the answer would be B  :)

8 0
3 years ago
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Given △ABC, use a dilation with the center at the origin to make a similar triangle with side lengths three times as large. What
Artist 52 [7]
The image of the dilation is shown below, with the centre of dilation (0,0) and scale factor of 3

The coordinate of C' is (6, -3) which is three times of the coordinate of C(2, -1)

7 0
3 years ago
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Select the functions that have a value of -1. sin180° cos180° tan180° csc180° sec180° cot180°
kupik [55]

We have to break each degree in terms of 90

A) sin180^\circ=sin(90\times2+0)

Which is in third quadrant, therefore sine is negative hence

sin(90\times2+0)= -sin0 ^\circ = 0


B) cos180^\circ =cos(90\times2+0)

Which is in third quadrant, therefore cosine is negative hence

cos(90\times2+0)= -cos0^\circ  = -1


C) tan180^\circ=tan(90\times2+0)

Which is in third quadrant, therefore tangent is positive hence

tan(90\times2+0)= tan0^\circ  = 0


D) csc180^\circ=csc(90\times2+0)

Which is in third quadrant, therefore cosec is negative hence

cosec(90\times2+0)= -csc0^\circ  =not defined


E)sec180^\circ=sec(90\times2+0)

Which is in third quadrant, therefore secant is negative hence

sec(90\times2+0)= -sec0^\circ  = -1


F) cot180^\circ=cot(90\times2+0)

Which is in third quadrant, therefore tangent is positive hence

cot(90\times2+0)= cot0^\circ = not defined


Hence only cos 180^\circ and

sec180^\circ have value -1

Hope this will help

7 0
3 years ago
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