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3 years ago

Solve for b: 2b = 20

Mathematics

2 answers:

JulsSmile [24]3 years ago

7 0

To solve for "b", you need to isolate/get the variable "b" by itself in the equation:

2b = 20 Divide 2 on both sides to get "b" by itself

$\frac{2b}{2}=\frac{20}{2}$

b = 10

iVinArrow [24]3 years ago

6 0

Answer:

b=10

Step-by-step explanation:

2b=20

divide both sides by 2

b=10

You might be interested in

Examine the diagram and information to answer the question. Rectangle EFGH has vertices at E(7,7), F(1,5), G(2,2), and H(8,4). H

schepotkina [342]

Answer:

Perimeter of rectangle = 6√10 units (the attachment isnt clear. It seems the answer is in option B)

Step-by-step explanation:

First we find the distance of the length and width of the rectangle.

Distance between E and F = Distance EF

Distance EF = √(∆y²-∆x²)

= √[(5-7)² + (1-7)²] = √[(-2)² + (-6)²]

= √(4+36) = √40

Distance EF = 2√10

Distance between F and G = Distance FG

Distance FG = √(∆y²-∆x²)

= √[(2-5)² + (2-1)²] = √[(-3)² + (1)²]

= √(9+1) = √10

Distance FG = √10

Length= distance EF = distance HG

And width = distance FG = distance EH

Perimeter of rectangle = 2(length + width)

Perimeter of rectangle = 2(2√10 + √10)

= 2(3√10)

Perimeter of rectangle = 6√10 units (the attachment isnt clear. It seems the answer is in option B)

7 0

4 years ago

AP Calculus redo! I know the answer but can't figure out exactly how to get there. Thank you! I want to work through the steps.

Monica [59]

You're approximating

$\displaystyle\int_1^5 x^2\,\mathrm dx$

with a Riemann sum, which comes in the form

$\displaystyle\int_a^b f(x)\,\mathrm dx=\lim_{n\to\infty}\sum_{i=1}^nf(x_i)\Delta x_i$

where $x_i$ are sample points chosen according to some decided-upon rule, and $\Delta x_i$ is the distance between adjacent sample points in the interval.

The simplest way of approximating the definite integral is by partitioning the interval into equally-spaced subintervals, in which case $\Delta x=\dfrac{b-a}n$ , and since $[a,b]=[1,5]$ , we have

$\Delta x=\dfrac{5-1}n=\dfrac4n$

Using the right-endpoint method, we approximate the area under $f(x)$ with rectangles whose heights are determined by their right endpoints. These endpoints are chosen by successively adding the subinterval length to the starting point of the interval of integration.

So if we had $n=4$ subintervals, we'd split up the interval of integration as

$[1,5]=[1,2]\cup[2,3]\cup[3,4]\cup[4,5]$

Note that the right endpoints follow a precise pattern of

$2=1+\dfrac44$
$3=1+\dfrac84$
$4=1+\dfrac{12}4$
$5=1+\dfrac{16}4$

The height of each rectangle is then given by the values above getting squared (since $f(x)=x^2$ ). So continuing with the example of $n=4$ , the Riemann sum would be

$\displaystyle\sum_{i=1}^4\left(1+\dfrac{4i}4\right)^2\dfrac44$

For $n=5$ ,

$\displaystyle\sum_{i=1}^5\left(1+\dfrac{4i}5\right)^2\dfrac45$

and so on, so that the definite integral is given exactly by the infinite sum

$\displaystyle\lim_{n\to\infty}\sum_{i=1}^n\left(1+\dfrac{4i}n\right)^2\dfrac4n$