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kolezko [41]
4 years ago
14

What’s a divided by a% of a

Mathematics
1 answer:
SSSSS [86.1K]4 years ago
5 0
A ÷ a% = a ÷a/100 = a x 100/a = 100

Any number divided by its percentage = 100

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What is the solution of √-4x=100?<br> A) x = –2500<br> B) x = –50<br> C) x = –2.5<br> D) no solution
yawa3891 [41]
First we raise both members of the equality to the square:
 (√-4x) ^ 2 = (100) ^ 2
 Then, we solve:
 -4x = 10000
 We cleared x in the last instance
 x = 10000 / (- 4)
 x = -2500
 answer:
 the solution of √-4x=100 is 
 A) x = –2500
4 0
3 years ago
GIVING LOTS OF POINTS!!!!
Orlov [11]

Answer:

80 cubes

Step-by-step explanation:

volume of one small cube = 1 /2 × 1 / 2 × 1 / 2

= 1 / 8cm^3

let x be the number of cubes needed

1 / 8 × x = 10

x = 10 × 8

x = 80 cubes

6 0
3 years ago
Select each expression that is equivalent to 18x + 3
emmasim [6.3K]
What’s x...............
4 0
3 years ago
And electrician needs 2 rolls of electrical wire to wire each room in a house how many rooms can he wire with 3/4 of a roll of w
Natasha_Volkova [10]

Answer:

3/8 room

Step-by-step explanation:

Given data

We are told that 2 rolls of wire are needed to wire 1 room

Hence if 1 room needs 2 rolls

               x rooms will need 3/4 rolls

cross multiply we have

2x= 3/4

Multiply  both sides by 1/2

x= 3/4*1/2

x= 3/8 room

8 0
3 years ago
Find the point on the plane ax + by + cz = d at minimum distance from the origin using the method of lagrange multipliers.
kupik [55]
The distance between some point (x,y,z) and the origin is given by

f(x,y,z)=\sqrt{x^2+y^2+z^2}

so this is the function we're trying to minimize. But notice that f(x,y,z) and f(x,y,z)^2 attain their critical points at the same (x,y,z), so we can solve the same problem by minimizing x^2+y^2+z^2 instead.

So let's take the Lagrangian to be

L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(ax+by+cz-d)

with partial derivatives (set equal to 0)

L_x=2x+a\lambda=0
L_y=2y+b\lambda=0
L_z=2z+c\lambda=0
L_\lambda=ax+by+cz-d=0

Now, notice that

aL_x+bL_y+cL_z=2(ax+by+cz)+(a^2+b^c+c^2)\lambda=0
\implies\lambda=-\dfrac{2d}{a^2+b^2+c^2}

and we can use this to solve for x,y,z. We get

x=\dfrac{ad}{a^2+b^2+c^2}
y=\dfrac{bd}{a^2+b^2+c^2}
z=\dfrac{cd}{a^2+b^2+c^2}

At this critical point, we get a minimum distance of

\sqrt{\left(\dfrac{ad}{a^2+b^2+c^2}\right)^2+\left(\dfrac{bd}{a^2+b^2+c^2}\right)^2+\left(\dfrac{cd}{a^2+b^2+c^2}\right)^2}=\sqrt{\dfrac{d^2}{a^2+b^2+c^2}}
8 0
3 years ago
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