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2 years ago

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Mathematics

1 answer:

Elanso [62]2 years ago

5 0

Answer:

I don't see a yellow point

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Giving away 100 points :)<br> Hope you have a nice day

Verizon [17]

Answer:

THANKS BRO

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

The bottom of Ignacio's desktop is 74.5cm from the floor. Ignacio's sits in his adjustable chair, and the tops of his legs are 4

dsp73

Answer:

Ignacio can make 5 clockwise rotations.

Step-by-step explanation:

Given that the Ignacio's legs are already at a height of 49.3cm and each rotation of the chair knob raises his legs another 4.8cm, we can set up an inequality to determine the number of rotations Ignacio could make without his legs touching the desk, which is at a height of 74.5cm:

4.8r + 49.3 < 74.5 where 'r' is equal to the number of rotations

The sum of the Ignacio's original leg height plus the amount of height increased from the rotations of the know must be less than 74.5 in order for his legs not to touch. Now, solve for 'r':

Subtract 49.3 from both sides: 4.8r + 49.3 - 49.3 < 74.5 - 49.3 or 4.8r < 25.2

Divide 4.8 from both sides: 4.8r/4.8 < 25.2/4.8 or r < 5.25

Since the number of rotations must be less than 5.25, he can make 5 complete rotations.

6 0

2 years ago

I will give brainliest away if you solve this for me please.

Elis [28]

Turn both equations into slope-intercept form [ y = mx + b ].

x + 3y = 3

~Subtract x to both sides

3y = 3 - x

~Divide 3 to everything

y = 1 - x/3

~Reorder

y = -1/3x + 1

4x + 3y = -6

~Subtract 4x to both sides

3y = -6 - 4x

~Divide 3 to everything

y = -2 - 4x/3

~Reorder

y = -4/3x - 2

Graph of the equations will be shown below. Note that the solution of graphing two equations will be where both equations intersect. Both lines intersect at (-3, 2), hence making that the solution.

Best of Luck!

8 0

2 years ago

Read 2 more answers

Given f(x)=2x-8, find f(3)

IgorLugansk [536]

F(3)=2*3-8=-2.
Hope that helps.

6 0

2 years ago

We are interested in the dimensions of a certain square a rectangle had length triple the side of this square and width two unit

Sphinxa [80]

Let s be the side length of the square. The dimensions of the rectangle are three times the side of the square (i.e. 3s), and two less than the side of the square (i.e. s-2).

So, the area of this rectangle is

$3s(s-2)=3s^2-6s$

The area of the square is $s^2$ , and we know that the two areas are the same, so we have

$3s^2-6s=s^2 \iff 2s^2-6s=0 \iff 2s(s-3)=0 \iff s=0 \lor s=3$

The solution s=0 would lead to the extreme case where the rectangle and the square are actually a point, so we accept the solution s=3.

3 0

2 years ago