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3 years ago

Solve 3.8 = 2x - 11.2. X=__

Mathematics

1 answer:

olga55 [171]3 years ago

8 0

Answer:

x= 15/2 im pretty sure

Step-by-step explanation:

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A local hamburger shop sold a combined total of 500 hamburgers and cheeseburgers on Wednesday. There were 50 fewer cheeseburgers

Virty [35]

Hey there! I'm happy to help!

Let's call the hamburgers h and the cheeseburgers c.

h+c=500

c=h-50

Let's plug this value for c into the first equation to solve for h.

h+h-50=500

2h-50=500

Add 50 to both sides.

2h=550

Divide both sides by 2.

h=275

Therefore, 275 hamburgers were sold on Wednesday.

Have a wonderful day! :D

6 0

3 years ago

Solve for x. 0.6x = 1.2x + 6

hichkok12 [17]

Step-by-step explanation:

0.6x- 1.2x = 6

-0.6x = 6

x = 6/0.6

x = -10

8 0

2 years ago

Which expression are equivalent to 4+4x+2y+x-y)?

In-s [12.5K]

Answer:

The answer to that is 5x+y+4 but the equivalent is

4(3x + y) and 12x + 4y

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Use implicit differentiation to find the points where the parabola defined by x2−2xy+y2+4x−8y+20=0 has horizontal and vertical t

Komok [63]

Answer:

The parabola has a horizontal tangent line at the point (2,4)

The parabola has a vertical tangent line at the point (1,5)

Step-by-step explanation:

Ir order to perform the implicit differentiation, you have to differentiate with respect to x. Then, you have to use the conditions for horizontal and vertical tangent lines.

-To obtain horizontal tangent lines, the condition is:

$\frac{dy}{dx}=0$ (The slope is zero)

--To obtain vertical tangent lines, the condition is:

$\frac{dy}{dx}=\frac{1}{0}$ (The slope is undefined, therefore the denominator is set to zero)

Derivating respect to x:

$\frac{d(x^{2}-2xy+y^{2}+4x-8y+20)}{dx} = \frac{d(x^{2})}{dx}-2\frac{d(xy)}{dx}+\frac{d(y^{2})}{dx}+4\frac{dx}{dx}-8\frac{dy}{dx}+\frac{d(20)}{dx}$ = $2x -2(y+x\frac{dy}{dx})+2y\frac{dy}{dx}+4-8\frac{dy}{dx}= 0$

Solving for dy/dx:

$\frac{dy}{dx}(-2x+2y-8)=-2x+2y-4\\\frac{dy}{dx}=\frac{2y-2x-4}{2y-2x-8}$

Applying the first conditon (slope is zero)

$\frac{2y-2x-4}{2y-2x-8}=0\\2y-2x-4=0$

Solving for y (Adding 2x+4, dividing by 2)

y=x+2 (I)

Replacing (I) in the given equation:

$x^{2}-2x(x+2)+(x+2)^{2}+4x-8(x+2)+20=0\\x^{2}-2x^{2}-4x+x^{2} +4x+4+4x-8x-16+20=0\\-4x+8=0\\x=2$

Replacing it in (I)

y=(2)+2

y=4

Therefore, the parabola has a horizontal tangent line at the point (2,4)

Applying the second condition (slope is undefined where denominator is zero)

2y-2x-8=0

Adding 2x+8 both sides and dividing by 2:

y=x+4(II)

Replacing (II) in the given equation:

$x^{2}-2x(x+4)+(x+4)^{2}+4x-8(x+4)+20=0\\x^{2}-2x^{2}-8x+x^{2}+8x+16+4x-8x-32+20=0\\-4x+4=0\\x=1$

Replacing it in (II)

y=1+4

y=5

The parabola has vertical tangent lines at the point (1,5)

4 0

3 years ago

Find two numbers whose sum is 64 and whose difference is 42

vovangra [49]

These numbers are "a" and "b".
Sum is 64, so a+b=64
Difference is 42 so a-b=42
So you've got:
$+\begin{cases} a+b=64 \\ a-b=42\end{cases} \\ \overline{, \quad2a \quad \ = \ 106} \qquad /:2 \\ a=53$
When a+b=64 and a=53 you've got:
53+b=64
b=11

So these numbers is 53 and 11.

4 0

3 years ago