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Kitty [74]
3 years ago
12

What is the value of 20+6•5/-2

Mathematics
1 answer:
scoray [572]3 years ago
7 0

Answer:

The answer is 25

Step-by-step explanation:

6•5=30

30+20=50

50/-2=25

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I need help on 4 and 6 please help me. Don’t put a link.
Arte-miy333 [17]

Answer:

I don't know about 4 but of no. 6, it is A

Step-by-step explanation:

3 0
3 years ago
A Fulcrum of a Pyramid is 6cm Square at the
Len [333]

Answer:

288 cubic cm

Step-by-step explanation:

Volume of the pyramid = 1/3 * base area * height

Base area = 12 * 12

Base area - 144 sq. cm

Height = 6cm

Volume of the pyramid = 144 * 6/3

Volume of the pyramid = 144 * 2

Volume of the pyramid  = 288 cubic cm

4 0
3 years ago
What is the sum of the roots of the quadratic equation x²+6x-14=0?​
aivan3 [116]

Answer:

<h3>Sum of Roots =<u> (-8)</u><u> </u><u> </u></h3><h3>Product of Roots = <u> </u><u> </u><u>(-84)</u><u> </u><u> </u></h3>

Step-by-step explanation:

{x}^{2}  + 6x - 14 = 0

Roots :- 6 and -14

Sum of the roots :

[6 + (-14)]

= (-8)//

Product of the roots :

[(6)(-14)]

= (-84)//

8 0
2 years ago
What is the next sequence 1/2, 1/2, 3/8, 1/4, 5/32
NemiM [27]
1/8 is tha answer............
5 0
3 years ago
I don't know if this is right... please someone help mee
worty [1.4K]
For the first circle, let's use the pythagorean theorem

\bf \textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;c=\sqrt{8^2+15^2}\implies c=\sqrt{289}\implies c=17

now, it just so happen that the hypotenuse on that triangle, is actually 17, but we used the pythagorean theorem to find it, and the pythagorean theorem only works for right-triangles.

 so if the hypotenuse is actually 17, that means that triangle there is actually a right-triangle, meaning that the radius there, and the outside line there, are both meeting at a right-angle.

when an outside line touches the radius line, and they form a right-angle, the outside line is indeed a tangent line, since the point of tangency is always a right-angle with the radius.



now, let's check for second circle

\bf \textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;c=\sqrt{11^2+14^2}\implies c=\sqrt{317}\implies c\approx 17.8044938

well, low and behold, we didn't get our hypotenuse as 16 after all, meaning, that triangle is NOT a right-triangle, and that outside line is not touching the radius at a right-angle, therefore is NOT a tangent line.



let's check the third circle

\bf \textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;c=\sqrt{33^2+56^2}\implies c=\sqrt{4225}\implies c=\stackrel{33+32}{65}

this time, we did get our hypotenuse to 65, the triangle is a right-triangle, so the outside line is indeed a tangent line.
6 0
3 years ago
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