Prove 1 and 3 are supplementary

Mathematics

1 answer:

vovikov84 [41]2 years ago

6 0

Answer:

Also question is, can 3 angles be supplementary? Remember, only a pair of angles can be supplementary. Sure, the three angles in a triangle may add up to 180 degrees, but there are three angles in a triangle, so they are not supplementary!

Step-by-step explanation:

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A shere has a diameter of 24 units what is it’s volume in cubic units

Vika [28.1K]

Answer:

7238 rounded to the nearest whole number.

Step-by-step explanation:

To find the volume of the sphere, we use the formula.

$v = \frac{4}{3} \pi {r}^{3}$

The radius will be 24 units ÷ 2 = 12 units.

Hence, the volume is

$\frac{4}{3} \pi \times {12}^{3}$

When entered in a calculator you'll get a result of 7238.23 cubic units in or 7238 rounded to the nearest whole number.

4 0

2 years ago

◆ Quadratic Equations ◆<br>Please help !

Mila [183]

I'm sure there's an easier way of solving it than the way I did, but I'm not sure what it could be. Never dealt with a problem like this before.

Anyway, I just plugged in and tested. Chose random values for a, b, c, and d, which follow the rule 0 < a < b < c < d:

a = 1
b = 2
c = 3
d = 4

$\sf ax^2+(1-a(b+c))x+abc-d)$

$\sf 1x^2+(1-1(2+3))x+(1)(2)(3)-(4))$

Simplify into standard form:

$\sf x^2+(1-1(5))x+6-4$

$\sf x^2+(1-5)x+2$

$\sf x^2-4x+2$

Use the quadratic formula to solve:

$\sf x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

For functions in the form of $\sf ax^2+bx+c$ . So in this case:

a = 1
b = -4
c = 2

Plug them in:

$\sf x=\dfrac{4\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}$

Solve for 'x':

$\sf x=\dfrac{4\pm\sqrt{16-8}}{2}$

$\sf x=\dfrac{4\pm\sqrt{8}}{2}$

$\sf x\approx\dfrac{4\pm 2.83}{2}$

$\sf x\approx 0.59,3.41$

So the answer would be A.

3 0

3 years ago

Let ABCD and EFGH be two quadrilaterals such that ABCD ~ EFGH. If AB = 15 cm, EF = 18 cm and the perimeter of ABCD is 40 cm, fin

Andrews [41]

Step-by-step explanation:

this photo is correct answer of this question.