The work of a student to solve a set of equations is shown:

1 answer:

6 0

Error in step 2

<span>Step 2: –7y = 16 – 2z [Equation A in Step 1 is simplified.]
(forgot to multiply -7 on the right side)

should be
</span><span>Step 2: –7y = -112 + 14z [Equation A in Step 1 is simplified.]</span>

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Find the point on the line <br> y = 2x + 3<br> that is closest to the origin.

Ludmilka [50]

I believe the answer is (-1,1) . The origin is the same as (0, 0) so I drew the line on a quick graph and if you are talking about distance, then the answer should be correct.
Hope this helped :)

6 0

3 years ago

Can someone help me asapp !!!!

Doss [256]

Answer:

he has 10 type of each coin

Step-by-step explanation:

Total number of cents in possession is calculated as;

= 25 + 10 + 5

= 40 cent

Total amount in possession = $4

= 400 cent

Since he has equal number of each cent, let this equal number be "y";

40cent (y) = 400cent

40y = 400

divide both sides by 40

$\frac{40y}{40} = \frac{400}{40} \\\\y = 10$

Therefore, he has 10 type of each coin

4 0

2 years ago

The area of a right triangle is sixty square centimetres. Its base is one centimetre less than twice its height. If the base and

NemiM [27]

Answer:

5. None of the above

Step-by-step explanation:

We have this information:

A = 60 cm2

b = 2*h - 1

Let's find out b and h:

The equation of the are of a triangle is A = (b*h)/2, if we replace it with our information, we have

$60 = ((2*h - 1) * h)/2\\ 120 = 2*h^2 - h\\ 0 = 2*h^2 - h - 120$

Let's find h with the quadratic formula:

$\fbox {Quadratic formula:} \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$h = \frac{-(-1)\pm\sqrt{(-1)^2-4*2*(-120)}}{2*2} = \frac{1\pm\sqrt{961}}{4} = \frac{1\pm\ 31}{4}$

h = 8 or h = -7.5

But h represents the height of the triangle, so it has to be a positive number, that's h = 8.

If we replace this in the equation we had for b, we have that b = 2*8 - 1 = 15.

Now we can calculate the hypotenuse with the Pythagorean equation

$\fbox {Pythagorean equation:}$ <em>The square of the length of the hypotenuse (the side opposite the right angle) of a right triangle is equal to the sum of the squares of the two legs (the two sides that meet at a right angle). </em>