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2 years ago

If y =-5 when x =12.5 find x when y =15

Mathematics

1 answer:

Naily [24]2 years ago

6 0

Answer:

x = -37.5

Step-by-step explanation:

Set up a proportion so -5/12.5 = 15/x

Solve for x so -5x = 15(12.5) so -5x = 187.5 so x = -37.5

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Find the solution to the system of equations given below using substitution. PLEASE ANSWER QUICKLY

irinina [24]

Answer:

option D

D. x = 5, y = 2

Step-by-step explanation:

Given in the question two equation,

Equation 1

5.3x + y = 28.5

Equation 2

4.2x + 3.1y = 27.2

rearrange equation 1 in terms of y

y = 28.5 - 5.3x

Substitute the value of y in equation 2

4.2x + 88.35 - 16.43x = 27.2

4.2x - 16.43x = 27.2 - 88.35

-12.23x = -61.15

x = 61.15/12.23

x = 5

put value of x in any of the equation

y = 28.5 - 26.5

y = 2

5 0

2 years ago

A circle has a circumference of 907.46 What is the diameter of the circle?

mafiozo [28]

The diameter would be d≈288.85!

I hope this helped! Brainliest? :) -Raven❤️

7 0

2 years ago

Please help me with 168÷45

velikii [3]

Answer:

3.73 ( The three is infinite)

Step-by-step explanation:

7 0

2 years ago

Given:

diamong [38]

Answer:

10-5 $\sqrt{2}$

Step-by-step explanation:

As per the attached figure, right angled $\triangle MDL$ has an inscribed circle whose center is $I$ .

We have joined the incenter $I$ to the vertices of the $\triangle MDL$ .

Sides MD and DL are equal because we are given that $\angle M = \angle L = 45 ^\circ.$

Formula for <em>area</em> of a $\triangle = \dfrac{1}{2} \times base \times height$

As per the figure attached, we are given that side <em>a = 10.</em>

Using pythagoras theorem, we can easily calculate that side ML = 10 $\sqrt{2}$

Points P,Q and R are at $90 ^\circ$ on the sides ML, MD and DL respectively so IQ, IR and IP are heights of $\triangle$ MIL, $\triangle$ MID and $\triangle$ DIL.

Also,

$\text {Area of } \triangle MDL = \text {Area of } \triangle MIL +\text {Area of } \triangle MID+ \text {Area of } \triangle DIL$

$\dfrac{1}{2} \times 10 \times 10 = \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10\sqrt2\\\Rightarrow r = \dfrac {10}{2+\sqrt2} \\\Rightarrow r = \dfrac{5\sqrt2}{\sqrt2+1}\\\text{Multiplying and divinding by }(\sqrt2 +1)\\\Rightarrow r = 10-5\sqrt2$

So, radius of circle = $10-5\sqrt2$

8 0

2 years ago

EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 9z on the curve of intersection of the plane x − y + z =

geniusboy [140]

The Lagrangian,

$L(x,y,z,\lambda,\mu)=x+2y+9z-\lambda(x-y+z-1)-\mu(x^2+y^2-1)$

has critical points where its partial derivatives vanish:

$L_x=1-\lambda-2\mu x=0$

$L_y=2+\lambda-2\mu y=0$

$L_z=9-\lambda=0$

$L_\lambda=x-y+z-1=0$

$L_\mu=x^2+y^2-1=0$

$L_z=0$ tells us $\lambda=9$ , so that

$L_x=0\implies-8-2\mu x=0\implies x=-\dfrac4\mu$

$L_y=0\implies11-2\mu y=0\implies y=\dfrac{11}{2\mu}$

Then with $L_\mu=0$ , we get

$x^2+y^2=\dfrac{16}{\mu^2}+\dfrac{121}{4\mu^2}=1\implies\mu=\pm\dfrac{\sqrt{185}}2$

and $L_\lambda=0$ tells us

$x-y+z=-\dfrac4\mu-\dfrac{11}{2\mu}+z=1\implies z=1+\dfrac{19}{2\mu}$

Then there are two critical points, $\left(\pm\frac8{\sqrt{185}},\mp\frac{11}{\sqrt{185}},1\pm\frac{19}{\sqrt{185}}\right)$ . The critical point with the negative $x$ -coordinates gives the maximum value, $9+\sqrt{185}$ .

8 0

2 years ago