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3 years ago

Convert to a decimal using long division. 3/16

Mathematics

1 answer:

sineoko [7]3 years ago

3 0

Answer: 0.1875

Step-by-step explanation:

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Simplify the expression by combining like terms. 4y+7x- 2y+4x

juin [17]

Hello!

Answer:

$\boxed{2y+11x}$

Step-by-step explanation:

Distributive property: a(b+c)=ab+ac

First, you switch sides. *Group like terms*

$4y-2y+7x+4x$

Then, you add the similar to elements. You can also add the numbers from left to right.

$4y-2y=2y$

$2y+7x+4x$

$7x+4x=11x$

$\boxed{2y+11x}\checkmark$ , is the correct answer.

Hope this helps!

Have a nice day! :)

4 0

3 years ago

Read 2 more answers

Please help me it’s really important to me

shepuryov [24]

1. b
2. e
3. a
4. c
5. d
6. f
7. g
8. h
hope this helps there is really no way of explaining you would have to study you theorems

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3 years ago

Which of the following will form the composite function G(F(x)) shown

photoshop1234 [79]

The answer is c F(x)= x2-5 and g(x)=c+4

6 0

3 years ago

9) Brenda spent $30 on eating utensils.

vredina [299]

She bought 4 spoons and 5 forks

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3 years ago

The x- and y-coordinates of the focus of a parabola are (1,3) and the directrix is y=-5 . A second parabola is a translation 1 u

Marianna [84]

Answer:

The equation of the second parabola is $y=\frac{1}{16} \left(x - 2\right)^{2} + 1$

Step-by-step explanation:

We know that the focus of the first parabola is (1,3) and the directrix is y = -5. We also know that the second parabola is a translation 1 unit right and 2 units up of the first parabola.

We can use the focus of the first parabola to find the focus of the second parabola (1+1, 3+2) = (2, 5) and the directrix of the second parabola is moved 2 units up. The equation of the directrix of the second parabola is y = -3.

To find the parabola equation we start by assuming a general point on the parabola (x,y).

Next, with the help of the distance formula we find that the distance between (x,y) and the focus.

$\sqrt{(x-2)^2+(y-5)^2}$

The distance between (x,y) and the directrix y = -3 is $\sqrt{(y+3)^2}$

On the parabola, these distances are equal:

$\sqrt{\left(y+3\right)^2}=\sqrt{\left(x-2\right)^2+\left(y-5\right)^2}\\\left(\sqrt{\left(y+3\right)^2}\right)^2=\left(\sqrt{\left(x-2\right)^2+\left(y-5\right)^2}\right)^2\\\\\left(y+3\right)^2=\left(x-2\right)^2+\left(y-5\right)^2\\\\\left(y+3\right)^2= x^2-4x+4+y^2-10y+25\\\left(y+3\right)^2=x^2-4x+y^2+29-10y\\\\y^2+6y+9=x^2-4x+y^2+29-10y\\16y=x^2-4x+20\\y=\frac{x^2-4x+20}{16}\\\\y=\frac{1}{16} \left(x - 2\right)^{2} + 1$

The equation of the second parabola is $y=\frac{1}{16} \left(x - 2\right)^{2} + 1$

7 0

4 years ago