All categories

3 years ago

Which set is closed under subtraction?

Mathematics

2 answers:

ss7ja [257]3 years ago

8 0

The set of rational numbers.

Gemiola [76]3 years ago

7 0

the answer is the set of rational numbers

You might be interested in

LM=9, NR=16, SR=8. Find the perimeter of △SMP.

Amanda [17]

Answer:

<h2>perimeter of △SMP = 25</h2>

Step-by-step explanation:

The perimeter of the triangle △SMP is the sum of al the sides of the triangle.

Perimeter of △SMP = ||MS|| + ||MP|| + ||SP||

Note that the triangle △LRN, △LSM, △MPN and △SRP are all scalene triangles showing that their sides are different.

Given LM=9, NR=16 and SR=8

NR = NP+PR

Since NP = PR

NR = NP+NP

NR =2NP

NP = NR/2 = 16/2

NP = 8

From △LSM, NP = PR = <u>MS</u><u> = 8</u>

Also since LM = MN, MN = 9

From △SRP, SR = RP = <u>PS = 9</u>

Also SR =<u> MP = 8</u>

From the equation above, perimeter of △SMP = ||MS|| + ||MP|| + ||SP||

perimeter of △SMP = 8+8+9

perimeter of △SMP = 25

5 0

3 years ago

HELP! Which triangles are congruent?

kati45 [8]

9514 1404 393

Answer:

4. A, B, C, and D

Step-by-step explanation:

Each right triangle has a short leg of 3 units and a long leg of 4 units. They are all congruent by the SAS congruence postulate.

8 0

3 years ago

A circle has the equation 2x²+12x+2y²−16y−150=0.

KonstantinChe [14]

Answer: B. The coordinates of the center are (-3,4), and the length of the radius is 10 units.

Step-by-step explanation:

The equation of a circle in the center-radius form is:

$(x-h)^{2} +(y-k)^{2}=r^{2}$ (1)

Where $(h,k)$ are the coordinates of the center and $r$ is the radius.

Now, we are given the equation of this circle as follows:

$2x^{2}+12x+2y^{2}-16y-150=0$ (2)

And we have to write it in the format of equation (1). So, let's begin by applying common factor 2 in the left side of the equation:

$2(x^{2}+6x+y^{2}-8y-75)=0$ (3)

Rearranging the equation:

$x^{2}+6x+y^{2}-8y=75$ (4)

$(x^{2}+6x)+(y^{2}-8y)=75$ (5)

Now we have to complete the square in both parenthesis, in order to have a perfect square trinomial in the form of $(a\pm b)^{2}=a^{2}\pm+2ab+b^{2}$ :

<u>For the first parenthesis:</u>

$x^{2}+6x+b^{2}$

We can rewrite this as:

$x^{2}+2(3)x+b^{2}$

Hence in this case $b=3$ and $b^{2}=9$ :

$x^{2}+2(3)x+3^{2}=x^{2}+6x+9=(x+3)^{2}$

<u>For the second parenthesis:</u>

$y^{2}-8y+b^{2}$

We can rewrite this as:

$y^{2}-2(4)y+b^{2}$

Hence in this case $b=-3$ and $b^{2}=9$ :

$y^{2}-2(4)y+4^{2}=y^{2}-8y+16=(y-4)^{2}$

Then, equation (5) is rewritten as follows:

$(x^{2}+6x+9)+(y^{2}-8y+16)=75+9+16$ (6)

<u>Note we are adding 9 and 16 in both sides of the equation in order to keep the equality.</u>

Rearranging:

$(x-3)^{2}+(y-4)^{2}=100$ (7)

At this point we have the circle equation in the center radius form $(x-h)^{2} +(y-k)^{2}=r^{2}$

Hence:

$h=-3$

$k=4$

$r=\sqrt{100}=10$

8 0

3 years ago

WRTH 50 PNTS!

andrew-mc [135]

Answer:

So, Michael completed the proof incorrectly because ∠AKG ≅ ∠AKL are not congruent because they are not vertically opposite angles.

Step-by-step explanation:

Hope this helps! (:

4 0

2 years ago

Read 2 more answers

What is the value of (4-2)^3-3X4

Mazyrski [523]

<h2>Answer:</h2>

$3 x^{4}-8=0$

<h2>Given: </h2>

$(4-2)^{3}-3 x^{4}$

<h2>Step-by-step explanation:</h2>

In this problem, we need to find the value of the given expression for which we need to simplify the given expression.

$\Rightarrow(4-2)^{3}-3 x^{4}=0$

We need to solve the numbers within the parentheses.

$\Rightarrow(2)^{3}-3 x^{4}=0$

We can bring the constant one side and variables another side.

$\Rightarrow 8=3 x^{4}$

Taking fourth root for ‘8’ and ‘3’ will give numbers which are very tough to solve further.

Since, we cannot solve the above equation further. The value of the given equation remains,

$\therefore 3 x^{4}-8=0$

6 0

3 years ago