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Flura [38]
2 years ago
14

When using a vertical number line do the negative numbers go on the bottom or top?

Mathematics
1 answer:
pshichka [43]2 years ago
8 0
Bottom Like They Go on the left horizontally
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9(x+3)=4x-3 I need to use the distributive property to solve
algol [13]

Answer:

x= 6

Step-by-step explanation:

8 0
3 years ago
Which is the solution to the inequality?
makvit [3.9K]

Answer:

b>3.133333333

Step-by-step explanation:

13/5<b-8/15

13/5+8/15<b

39+8/15<b

47/15<b

3.13333333<b

b>3.13333333

6 0
3 years ago
Pls only do this if yk it because i’m giving correct answer brainliest! :))
Leno4ka [110]

Answer:

x=2+\frac{1}{2}\sqrt[]{21}

or

x=2-\frac{1}{2}\sqrt{{21}

Step-by-step explanation:

4x^2-16x-26=-21

Add 21 on both sides.

4x^2-16x-26+21=-21+21

4x^2-16x-5=0

a=4

b=-16

c=-5

x=\frac{-b\frac{+}{}\sqrt[]{b^2-4ac}  }{2a}

x=\frac{-(-16)\frac{+}{}\sqrt[]{(-16)^2-4(4)(-5)}  }{2(4)}

x=\frac{16\frac{+}{}\sqrt[]{256+80}  }{8}

x=\frac{16\frac{+}{}\sqrt[]{336}  }{8}

x=\frac{16\frac{+}{}\sqrt[]{2^2*2^2*21}  }{8}

x=\frac{16\frac{+}{}2*2\sqrt[]{21}  }{8}

x=\frac{16\frac{+}{}4\sqrt[]{21}  }{8}

---------------------------------------------------------------------------

x=\frac{16}{8}+\frac{4\sqrt[]{21}}{8}

x=2+\frac{1}{2}\sqrt[]{21}

---------------------------------------------------------------------------

x=\frac{16}{8}-\frac{4\sqrt[]{21}}{8}\\x=2-\frac{1}{2}\sqrt{{21}

6 0
2 years ago
Read 2 more answers
The scale on a map is 1: 75 000. (a) Calculate the ACTUAL distance, in km, between two places which are 16 cm apart on the map.
finlep [7]

Answer:

a. 1 200 000 km

b. 0.000 773 cm

c. 67 500 000 000 km²

d. 0.000 000 016 cm²

Step-by-step explanation:

a. Calculate the ACTUAL distance, in km, between two places which are 16 cm apart on the map.

Since the scale is 1 : 75000 that is 1 cm : 75000 km, we want to find the actual distance of 16 cm on the map. Let x be the actual distance in km.

So, 1 cm : 75000 km = 16 cm : x km

So, 1 cm/75000 km = 16 cm/x km

Thus x km = 16 cm × 75000 km/1 cm = 1 200 000 km

(b) Calculate length of a road on the map which is 0.58 km.

Let y be the length of road on the map. So, using our scale

1 cm : 75000 km = y cm : 0.58 km

So, 1 cm/75000 km = y cm/0.58 km

Thus y cm = 0.58 km × 1 cm/75000 km = 0.000 773 cm

(c) Calculate, in km², the ACTUAL area of a lake which is 12 cm2 on the map.

Since the scale is 1 : 75000 , the scale for the area would be the square of this. So, 1² : 75000² = 1 : 5 625 000 000 which is 1 cm² : 5 625 000 000 km²

Let z be the actual area of the lake.

So,  1 cm² : 5 625 000 000 km² = 12 cm² : z km²

So, 1 cm²/5 625 000 000 km² = 12 cm²/z km²

Thus z km² = 12 cm² × 5 625 000 000 km²/1 cm² = 67 500 000 000 km²

(d) Calculate the area of a play field on the map which has has an actual area of 90 km².

Let a represent the area of the play field on the map. Using our scale for the area,

1 cm² : 5 625 000 000 km² = a cm² : 90 km²

So, a cm² = 90 km² × 1/5 625 000 000 km² = 0.000 000 016 cm²

4 0
2 years ago
Which point satisfies the system of equations y = 3x − 2 and y = -2x + 3?
Sedbober [7]

Answer:

(1,1)

Step-by-step explanation:

The choices are not visible, however are unnecessary to solve this problem.

Given y = 3x - 2 and y = -2x + 3

[Set them equal to each other]

3x - 2 = -2x + 3

[Find x]

3x + 2x - 2 = 3

3x + 2x = 5

5x = 5

x = 1

[Plug x = 1 to both equations to check and find y]

y = 3x - 2

y = 3(1) - 2 = 3 - 2 = 1

y = -2x + 3

y = -2(1) + 3 = -2 + 3 = 1

Both are equal to y = 1 when x = 1

The point is then: (1,1)

3 0
3 years ago
Read 2 more answers
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