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3 years ago

. What is the solution of 53x – 900 ? Round your answer to the nearest hundredth.

Mathematics

1 answer:

katrin [286]3 years ago

4 0

Answer:

I Believe Its 01.24 :))

Step-by-step explanation:

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Question in picture above ^

Andrei [34K]

[ 4(4) x 4(3) ] / 4(5)

= [4(4+3)] / 4(5)

= 4(7) / 4(5)

= 4(7-5)

= 4(2)

5 0

3 years ago

What is the sum of -3 and -12

goblinko [34]

Answer:

-15

Step-by-step explanation:

-3 + -12 = -15

4 0

3 years ago

Janie earned $15.50 at her part time job on Saturday

mina [271]

Answer:48 dollars

Step-by-step explanation:Just add 32.50 and 15.50 to get 48

8 0

3 years ago

Read 2 more answers

Someone help with this

elena-14-01-66 [18.8K]

The quadrants are I, II, III, and IV, (meaning 1, 2, 3, and 4). The first quadrant is in the upper right, which has both positive x and positive y values. The second quadrant is the upper left, which has negative x and positive y values. The third quadrant is the lower left, which has both negative x and y values. The fourth quadrant is the lower right, which has positive x and negative y values. Using this knowledge and our positive and negative signs, the following are the answers to questions 1-6.

1) (-4, -2) - Quadrant III, both x and y are negative

2) (0, -7) - This point is actually on the y-axis. The x value is 0 and the y value is -7, so the graph is 7 units down from the origin on the y-axis.

3) (0,0) - This point is the origin, or where the x and y axes cross (the middle of the graph).

4) (6, -9) - This point is in Quadrant IV, because it has a positive x value and a negative y value

5) (3,5) - This point is in Quadrant I, because both the x and y values are positive.

6) (8,0) - This point is on the x-axis. The y-value is zero, so this point is 8 units to the right of the origin between Quadrant I and Quadrant IV.

Using the knowledge presented above, to graph the points given to you in the second part of the problem, first you can figure out what quadrant or part of the graph the point is on. Then, you can count the number of units (squares on the graph) in the right direction (remember that up is positive on the y-axis and down is negative, and to the right is positive on the x-axis and to the left is negative) in order to plot the points. Then, you must connect the points that correspond to the same figure in order to create the figures.

Please comment if you have any questions!

Hope this helps!

7 0

3 years ago

Can a sequence be both arithmetic and geometric?

artcher [175]

An aritmetic sequence is like this
$a_n=a_1+d(n-1)$ where a1=first term and d=common difference

geometric is $a_n=a_1(r)^{n-1}$ where a1=first term and r=common ratio

can it be both aritmetic and geometric
hmm, that means that the starting terms should be the same

therfor we need to solve $d(n-1)=(r)^{n-1}$
what values of d and r make all natural numbers of n true?
are there values that make all natural numbers for n true?

when n=1, then d(1-1)=0 and r^(1-1)=1, so already they are not equal

the answer is no, a sequence cannot be both aritmetic and geometric

4 0

3 years ago