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HACTEHA [7]
3 years ago
9

In which quadrant are all coordinates positive? I II III IV

Mathematics
2 answers:
Vlad1618 [11]3 years ago
8 0
<span>In which quadrant are all coordinates positive?

answer is I (first choice)</span>
erastovalidia [21]3 years ago
7 0
The answer is quadrant I
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Write the slope intercept form of the equation of the line through the given point with the given slope through (-2,5),slope-3/2
hjlf
Answer: y= -3/2x + 2
y= -3/2 + b
5 = -3/2 * -2 + b
5 = 3 + b
2 = b
6 0
3 years ago
Lenora buys 3.7 pounds of potatoes that cost $0.29 per pound. How much does Lenora pay for the potatoes. Round to the nearest ce
kap26 [50]
If it is 29.cents the awnser should be 30 cants
7 0
3 years ago
Read 2 more answers
Solve these simultaneous equations:
ivolga24 [154]

Answer:

1) We have the system:

5*x - 3*y = 15

4*x + 3*y = 6

To solve this, we first need to isolate one of the variables in one of the equations, let's isolate x in the first equation:

x = 15/5 + (3/5)*y = 3 + (3/5)*y

Now we can replace this in the other equation to get:

4*( 3 + (3/5)*y) + 3*y = 6

and solve this for y.

12 + (12/5)*y  + 3*y = 6

(12/5 + 3)*y = 6 - 12 = -6

(12/5 + 15/5)*y = -6

(27/5)*y = -6

y = -6*(5/27) = 1.11

Now we can replace this in the equation:

x = 3 + (3/5)*y

To get the value of x.

x = 3 + (3/5)*1.11 = 3.67

Then the solution of this system is the point (3.67, 1.11)

2) Now we have the system:

2*x + 5*y = 26

4*x + 3*y = 24

The solution method is the same as before:

x = 26/2 - (5/2)*y = 13 - (5/2)*y

Now we replace this in the other equation:

4*( 13 - (5/2)*y) + 3*y = 24

52 - 10*y + 3*y = 24

52 - 7*y = 24

52 - 24 = 7*y

28 = 7*y

28/7 = y

4 = y

now we replace this in the equatio:

x = 13 - (5/2)*y

x = 13 - (5/2)*4 = 13 - 10 = 3

The solution of this sytem is (3, 4)

3) Now we have the system:

3*x + 3*y = 39

2*x - 3*y = -2

first we isolate x in the first equation:

x = 39/3 - 3*y/3 = 13 - y

Now we can replace this in the other equation:

2*(13 - y) - 3*y = -2

26 - 2*y - 3*y = -2

26 - 5*y = -2

26 + 2 = 5*y

28 = 5*y

28/5 = y = 5.6

Now we can replace this in the equation:

x = 13 - y

To get the x-value

x = 13 - 5.6 = 7.4

Then the solution for this system is (7.4, 5.6)

4 0
2 years ago
At 9 a.m. a car (A) began a journey from a point, traveling at 40 mph. At 10 a.m.
Kisachek [45]
At three hours as car a has already traveled a hour which making it three therefore making car b equal 2 hours or in a math equation 40×3=car b and 60×2=car a
5 0
3 years ago
If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t
sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, 6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296

This implies that 6^{4} exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, 6^{3}=216 and 216\times 5=1080, 216\times 4=864. This implies that 216\times 5 exceeds 1052 and thus there can be at most 4 groups of 6^{3}.

Then,

1052-4\times6^{3}

1052-4\times216

1052-864

188

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, 6^{2}=36 and 36\times 5=180, 36\times 6=216. This implies that 36\times 6 exceeds 188 and thus there can be at most 5 groups of 6^{2}.

Then,

188-5\times6^{2}

188-5\times36

188-180

8

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, 6^{1}=6 and 6\times 1=6, 6\times 2=12. This implies that 6\times 2 exceeds 8 and thus there can be at most 1 group of 6^{1}.

Then,

8-1\times6^{1}

8-1\times6

8-6

2

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, 6^{0}=1 and 1\times 2=2. This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

Then,

1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}

So, the number 1052, written as a base 6 number is 4512.

Learn more about change of base of numbers here:

brainly.com/question/14291917

6 0
2 years ago
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