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sesenic [268]
3 years ago
11

A point in quadrant IV is reflected across the x-axis. The new point is located in Quadrant..?

Mathematics
2 answers:
olga_2 [115]3 years ago
8 0
A point in quadrant IV is reflected across the x-axis. The new point is located in Quadrant..?
A) I
B) II
C) III
D) IV

It’s C) III
suter [353]3 years ago
4 0

Answer

It's D) IV

Step-by-step explanation:

Just look at a graph.

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Please help me to figure out this question. Thanks
Marta_Voda [28]

Answer:

a) 5 : 7

b) 25 black keys

c) 100 black keys will appear.

Step-by-step explanation:

a) From the picture,

There are 5 black keys and 7 white keys.

Therefore, the ratio of black keys to white keys are = 5 : 7 = 5/7

b) If the pattern continues, 25 black keys will be appeared on a portable full-sized keyboard.

Explanation -

From part A, we get the ratio of black keys to white keys = 5 : 7. Therefore,

If there are 7 white keys, there are 5 black keys

If there are 1 white keys, there are 5/7 black keys

If there are 35 white keys, there are (5*35)/7 black keys

= 25 black keys.

c) If there are 240 keys, the black keys will appear -

We get, the ratio of black keys to white keys = 5 : 7

The summation of the black keys and white keys = (5 + 7) = 12.

Therefore, if there are 240 keys, the black keys will be

= (5/12) x 240

= (5 x 240)/12

= 100

Therefore, there are 100 black keys.

7 0
3 years ago
Solve the following and show all work: √8x+1=x+2
BlackZzzverrR [31]

Answer:

x= 3, 1

Step-by-step explanation:

7 0
2 years ago
How do I solve this
Montano1993 [528]
Use arccos...cos^-1 so

cos^-1 (0.8195)

it equals 34.96522689°
6 0
2 years ago
How many gallons of punch will sadie make?
antoniya [11.8K]
She will make 20,000000. Gallons a punch.
5 0
3 years ago
How many integers between 10000 and 99999, inclusive, are divisible by 3 or<br> 5 or 7?
Yuki888 [10]

Answer: Hence, there are approximately 48884 integers are divisible by 3 or 5 or 7.

Step-by-step explanation:

Since we have given that

Integers between 10000 and 99999 = 99999-10000+1=90000

n( divisible by 3) = \dfrac{90000}{3}=30000

n( divisible by 5) = \dfrac{90000}{5}=18000

n( divisible by 7) = \dfrac{90000}{7}=12857.14

n( divisible by 3 and 5) = n(3∩5)=\dfrac{90000}{15}=6000

n( divisible by 5 and 7) = n(5∩7) = \dfrac{90000}{35}=2571.42

n( divisible by 3 and 7) = n(3∩7) = \dfrac{90000}{21}=4285.71

n( divisible by 3,5 and 7) = n(3∩5∩7) = \dfrac{90000}{105}=857.14

As we know the formula,

n(3∪5∪7)=n(3)+n(5)+n(7)-n(3∩5)-n(5∩7)-n(3∩7)+n(3∩5∩7)

=30000+18000+12857.14-6000-2571.42-4258.71+857.14\\\\=48884.15

Hence, there are approximately 48884 integers are divisible by 3 or 5 or 7.

5 0
3 years ago
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