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4 years ago

I dont get this problem

Mathematics

1 answer:

lions [1.4K]4 years ago

6 0

6 servings.

3/4 pounds can be converted into 6/8 by multiplying the numerator and denominator by 2. So 3x2=6 and 4x2= 8, giving you the fraction of 6/8, then you divide that by 1/8 which gives you six servings

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The circumference of a circle is 100.48 inches. What is the circle's diameter? Use 3.14 for .

krok68 [10]

Answer:

d = 32 inches

Step-by-step explanation:

C = 2 × 3.14 × r

100.48 = 2 × 3.14 × r

Now divide 100.48 by 2 and 3.14 to get the radius:

r = 16 inches

The radius is half of the diameter, so...

d = 32 inches

7 0

3 years ago

Which similarity statements are true

Sliva [168]

The first and last one are similar

3 0

3 years ago

PLEASE HELP ASAP In this task, you will practice finding the area under a nonlinear function by using rectangles. You will use g

mrs_skeptik [129]

Answer:

a) $1280 u^{2}$

b) $1320 u^{2}$

c) $\frac{4000}{3} u^{2}$

Step-by-step explanation:

In order to solve this problem we must start by sketching the graph of the function. This will help us visualize the problem better. (See attached picture)

You can sketch the graph of the function by plotting as many points as you can from x=0 to x=20 or by finding the vertex form of the quadratic equation by completing the square. You can also do so by using a graphing device, you decide which method suits better for you.

So we are interested in finding the area under the curve, so we divide it into 5 rectangles taking a right hand approximation. This is, the right upper corner of each rectangle will touch the graph. (see attached picture).

In order to figure the width of each rectangle we can use the following formula:

$\Delta x=\frac{b-a}{n}$

in this case a=0, b=20 and n=5 so we get:

$\Delta x=\frac{20-0}{5}=\frac{20}{5}=4$

so each rectangle must have a width of 4 units.

We can now calculate the hight of each rectangle. So we figure the y-value of each corner of the rectangles. We get the following heights:

h1=64

h2=96

h3=96

h4= 64

h5=0

so now we can use the following formula to find the area under the graph. Basically what the formula does is add the areas of the rectangles:

$A=\sum^{n}_{i=1} f(x_{i}) \Delta x$

which can be rewritten as:

$A=\Delta x \sum^{n}_{i=1} f(x_{i})$

So we go ahead and solve it:

$A=(4)(64+96+96+64+0)$

so:

$A= 1280 u^{2}$

B) The same procedure is used to solve part B, just that this time we divide the area in 10 rectangles.

In order to figure the width of each rectangle we can use the following formula:

$\Delta x=\frac{b-a}{n}$

in this case a=0, b=20 and n=10 so we get:

$\Delta x=\frac{20-0}{10}=\frac{20}{10}=2$

so each rectangle must have a width of 2 units.

We can now calculate the hight of each rectangle. So we figure the y-value of each corner of the rectangles. We get the following heights:

h1=36

h2=64

h3=84

h4= 96

h5=100

h6=96

h7=84

h8=64

h9=36

h10=0

so now we can use the following formula to find the area under the graph. Basically what the formula does is add the areas of the rectangles:

$A=\sum^{n}_{i=1} f(x_{i}) \Delta x$

which can be rewritten as:

$A=\Delta x \sum^{n}_{i=1} f(x_{i})$

So we go ahead and solve it:

$A=(2)(36+64+84+96+100+96+84+64+36+0)$

so:

$A= 1320 u^{2}$

In order to find part c, we calculate the area by using limits, the limit will look like this:

$\lim_{n \to \infty} \sum^{n}_{i=1} f(x^{*}_{i}) \Delta x$

so we start by finding the change of x so we get:

$\Delta x =\frac{b-a}{n}$

$\Delta x =\frac{20-0}{n}$

$\Delta x =\frac{20}{n}$

next we find $x^{*}_{i}$

$x^{*}_{i}=a+\Delta x i$

so:

$x^{*}_{i}=0+\frac{20}{n} i=\frac{20}{n} i$

and we find $f(x^{*}_{i})$

$f(x^{*}_{i})=f(\frac{20}{n} i)=-(\frac{20}{n} i)^{2}+20(\frac{20}{n} i)$

cand we do some algebra to simplify it.

$f(x^{*}_{i})=-\frac{400}{n^{2}}i^{2}+\frac{400}{n}i$

we do some factorization:

$f(x^{*}_{i})=-\frac{400}{n}(\frac{i^{2}}{n}-i)$

and plug it into our formula:

$\lim_{n \to \infty} \sum^{n}_{i=1}-\frac{400}{n}(\frac{i^{2}}{n}-i) (\frac{20}{n})$

And simplify:

$\lim_{n \to \infty} \sum^{n}_{i=1}-\frac{8000}{n^{2}}(\frac{i^{2}}{n}-i)$

$\lim_{n \to \infty} -\frac{8000}{n^{2}} \sum^{n}_{i=1}(\frac{i^{2}}{n}-i)$

And now we use summation formulas:

$\lim_{n \to \infty} -\frac{8000}{n^{2}} (\frac{n(n+1)(2n+1)}{6n}-\frac{n(n+1)}{2})$

$\lim_{n \to \infty} -\frac{8000}{n^{2}} (\frac{2n^{2}+3n+1}{6}-\frac{n^{2}}{2}-\frac{n}{2})$

and simplify:

$\lim_{n \to \infty} -\frac{8000}{n^{2}} (-\frac{n^{2}}{6}+\frac{1}{6})$

$\lim_{n \to \infty} \frac{4000}{3}+\frac{4000}{3n^{2}}$

and solve the limit

$\frac{4000}{3}u^{2}$

4 0

3 years ago

Find the volume and the total surface area of a cuboide whose dimension 2m, 80cm and 60cm

DENIUS [597]

Answer:

Volume= 0.96m^3

TSA= 6.56m^2

Step-by-step explanation:

Given data

Length = 2m

Width= 80cm= 0.8m

Height= 60cm= 0.6m

Volume= L*W*H

Volume= 2*0.8*0.6

Volume= 0.96m^3

TSA=2lw+2lh+2hw

TSA= 2*2*0.8+2*2*0.6+2*0.6*0.8

TSA= 3.2+ 2.4+ 0.96

TSA= 6.56m^2

5 0

3 years ago

Select the factors that can be divided out to simplify the multiplication problem: 4/26 x 13/20. Select all that apply. A. 2 B.

Tcecarenko [31]

$\dfrac{4}{26} \times \dfrac{13}{20}$

-------------------------------------
Divide by factor of 4.
-------------------------------------
*Or divide by the factor of 2 twice

$\dfrac{1 \times 4}{26} \times \dfrac{13}{5 \times 4}$

$\dfrac{1}{26} \times \dfrac{13}{5}$

-------------------------------------
Divide by factor of 13.
-------------------------------------

$\dfrac{1}{2 \times 13} \times \ \dfrac{1 \times 13}{5}$

$\dfrac{1}{2} \times \dfrac{1 }{5}$

-------------------------------------
Combine into single fraction.
-------------------------------------
$\dfrac{1}{10}$

--------------------------------------------------------------------------
Answer: The factors are 2, 4 and 13
--------------------------------------------------------------------------

3 0

3 years ago

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