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Viefleur [7K]
2 years ago
13

Point A is located in which quadrant? IV III I II

Mathematics
2 answers:
Ipatiy [6.2K]2 years ago
6 0
The correct answer would be Quadrant ll
Romashka-Z-Leto [24]2 years ago
5 0

Answer:

Quadrant II

Step-by-step explanation:

This is because the top right is I, the top left

is II, the bottom left is III, and the bottom right is IV.

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The domain of y = cosx is the set of real numbers. True False
artcher [175]

Answer:

true

Step-by-step explanation:

3 0
2 years ago
At the fair 4 employees were paid $8.75 an hour to work at a funnel cake stand. They worked 8 hours at regular pay then for 5 ho
nordsb [41]

Answer:

$125.25

Step-by-step explanation:

8.75*h+11.25*e

h is hours regular pay and e is extra pay for overtime. Since its an extra $2.50 for overtime, add that to the regular pay. Put in the hours:

8.75*8+11.25*5

Simplify by multiplying 8.75 by 8

70+11.25*5

Simplify by multiplying 11.25 by 5

70+56.25

Add

70+56.25=126.25

Since it says how much did each earn, as in individually, leave it as it is (unless regular pay was $2.18 an hour, which would be a rip-off). I'm pretty sure including the number of employees was meant to throw you off.

7 0
3 years ago
Brainliest if it its right
Anika [276]

Answer:

A = <u>40</u> mm^{2}

Step-by-step explanation:

The equation for finding the area of a trapezoid is A= \frac{a + b}{2}h.

In this equation, a=4, b=6, h=8.

A = \frac{4 + 6}{2}8.

4+6=10
10/2=5
5x8=40

Hope this helps!

8 0
2 years ago
Read 2 more answers
Help me pleaseee.....
Lana71 [14]

Answer:

titutex=cos\alp,\alp∈[0:;π]

\displaystyle Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1)\Leftright |cos\alp +sin\alp |=\sqrt{2}(2cos^2\alp -1)Then∣x+

1−x

2

∣=

2

(2x

2

−1)\Leftright∣cos\alp+sin\alp∣=

2

(2cos

2

\alp−1)

\displaystyle |\N {\sqrt{2}}cos(\alp-\frac{\pi}{4})|=\N {\sqrt{2}}cos(2\alp )\Right \alp\in[0\: ;\: \frac{\pi}{4}]\cup [\frac{3\pi}{4}\: ;\: \pi]∣N

2

cos(\alp−

4

π

)∣=N

2

cos(2\alp)\Right\alp∈[0;

4

π

]∪[

4

3π

;π]

1) \displaystyle \alp \in [0\: ;\: \frac{\pi}{4}]\alp∈[0;

4

π

]

\displaystyle cos(\alp -\frac{\pi}{4})=cos(2\alp )\dotscos(\alp−

4

π

)=cos(2\alp)…

2. \displaystyle \alp\in [\frac{3\pi}{4}\: ;\: \pi]\alp∈[

4

3π

;π]

\displaystyle -cos(\alp -\frac{\pi}{4})=cos(2\alp )\dots−cos(\alp−

4

π

)=cos(2\alp)…

1

Top

Display

6 0
3 years ago
Can someone help me with this please?
AfilCa [17]

Answer:

87

Step-by-step explanation:

14 = Minimum

22 = Lower quartile

32 = Median

87 = Upper quartile

95 = Maximum

8 0
3 years ago
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