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3 years ago

Which equation can be used to calculate the area of the shaded triangle in the figure below?

Mathematics

1 answer:

s2008m [1.1K]3 years ago

8 0

Answer:D

Since the bottom half the triangle is shaded the width becomes 4 and the length stays the same. 4x14= 56, or 1/2(14x8)

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I will give lot of points please help

zheka24 [161]

Answer:

V = 43323 mm³

(second answer is already correct)

Step-by-step explanation:

these pennies make up the shape of a cylinder, so, we can solve this question by solving for the volume of a cylinder.

the volume of a cylinder can be found with the formula:

V = πr²h

(V = volume ; π = pi ; r = radius ; h = height)

We know the diameter of this cylinder is 19.05 mm. Diameter is two times the length of the radius, (diameter is all the way across, radius is from center to outer / halfway across) meaning that we can find the radius by dividing 19.05 by 2.

diameter = 19.05

radius = 19.05 / 2 = 9.525

so, r = 9.525

we also are given the height of this cylinder, 38 mm

so, h = 38

π = 3.14159 (π is a constant)

plugging our measurements into the volume formula:

V = πr²h

V = (3.14159)(9.525²)(38)

V = (3.14159)(362.9025)(38)

V = 43323.4894628

[rounded to V = 43323 mm³]

And yes, the volume is not affected by a change in shape. (you are correct)

hope this helps!!

4 0

2 years ago

Read 2 more answers

I JUST NEED #13 please and thnx

emmainna [20.7K]

$\bf \cfrac{18}{32}\implies \cfrac{2\cdot 3\cdot 3}{2\cdot 2\cdot 2\cdot 2\cdot 2}\implies \cfrac{3\cdot 3}{2\cdot 2\cdot 2\cdot 2}\implies \boxed{\cfrac{9}{16}}\qquad \checkmark \\\\[-0.35em] ~\dotfill\\\\ \cfrac{27}{48}\implies \cfrac{3\cdot 3\cdot 3}{2\cdot 2\cdot 2\cdot 2\cdot 3}\implies \cfrac{3\cdot 3}{2\cdot 2\cdot 2\cdot 2}\implies \boxed{\cfrac{9}{16}}\qquad \checkmark$

3 0

2 years ago

How many zeroes do we write when we write all the integers 1 to 243 in base 3?

Monica [59]

Answer:

289 numbers

Step-by-step explanation:

Above you will find the list of integers from 1 to 243 in base 3:

(1, 2, 10, 11, 12, 20, 21, 22, 100, 101, 102, 110, 111, 112, 120, 121, 122, 200, 201, 202, 210, 211, 212, 220, 221, 222, 1000, 1001, 1002, 1010, 1011, 1012, 1020, 1021, 1022, 1100, 1101, 1102, 1110, 1111, 1112, 1120, 1121, 1122, 1200, 1201, 1202, 1210, 1211, 1212, 1220, 1221, 1222, 2000, 2001, 2002, 2010, 2011, 2012, 2020, 2021, 2022, 2100, 2101, 2102, 2110, 2111, 2112, 2120, 2121, 2122, 2200, 2201, 2202, 2210, 2211, 2212, 2220, 2221, 2222, 10000, 10001, 10002, 10010, 10011, 10012, 10020, 10021, 10022, 10100, 10101, 10102, 10110, 10111, 10112, 10120, 10121, 10122, 10200, 10201, 10202, 10210, 10211, 10212, 10220, 10221, 10222, 11000, 11001, 11002, 11010, 11011, 11012, 11020, 11021, 11022, 11100, 11101, 11102, 11110, 11111, 11112, 11120, 11121, 11122, 11200, 11201, 11202, 11210, 11211, 11212, 11220, 11221, 11222, 12000, 12001, 12002, 12010, 12011, 12012, 12020, 12021, 12022, 12100, 12101, 12102, 12110, 12111, 12112, 12120, 12121, 12122, 12200, 12201, 12202, 12210, 12211, 12212, 12220, 12221, 12222, 20000, 20001, 20002, 20010, 20011, 20012, 20020, 20021, 20022, 20100, 20101, 20102, 20110, 20111, 20112, 20120, 20121, 20122, 20200, 20201, 20202, 20210, 20211, 20212, 20220, 20221, 20222, 21000, 21001, 21002, 21010, 21011, 21012, 21020, 21021, 21022, 21100, 21101, 21102, 21110, 21111, 21112, 21120, 21121, 21122, 21200, 21201, 21202, 21210, 21211, 21212, 21220, 21221, 21222, 22000, 22001, 22002, 22010, 22011, 22012, 22020, 22021, 22022, 22100, 22101, 22102, 22110, 22111, 22112, 22120, 22121, 22122, 22200, 22201, 22202, 22210, 22211, 22212, 22220, 22221, 22222, 100000)

If you count them, you will find that there are 289 numbers in total!

8 0

3 years ago

Tommy has 6 more than 3 times the amount of money in his bank account than Judy. Judy has $7500 more than a number in her accoun

Ad libitum [116K]

Answer:

Amount of money Judy has in her account is $\$(7500+n)$ .

Amount of money Tommy has in his account is $\$(22506+3n)$

Step-by-step explanation:

To find : Amount of money in Judy account and amount of money in Tommy account:

Solution:

Given:

Judy has $7500 more than a number in her account.

Let the number be 'n'.

So we can say that;

Amount of money in Judy account is equal to 7500 plus number.

framing in equation form we get;

Amount of money in Judy account = $\$(7500+n)$

Hence Amount of money Judy has in her account is $\$(7500+n)$ .

Now Given:

Tommy has 6 more than 3 times the amount of money in his bank account than Judy.

So we can say that;

Amount of money in Tommy account is equal to 3 multiplied by Amount of money in Judy account plus 6.

framing in equation form we get;

Amount of money in Tommy account = $3(7500+n)+6 =22500+3n+6=\$(22506+3n)$

Hence Amount of money Tommy has in his account is $\$(22506+3n)$ .

8 0

2 years ago

Approximate area under the curve f(x) =-x^2+2x+4 from x=0 to x=3 by using summation notation with six rectangles and use the the

bekas [8.4K]

Answer:

Summation notation:

$\frac{1}{2}\sum_{k=1}^6f((.5k))$

or after using your function part:

$\frac{1}{2}\sum_{k=1}^6(-(.5k)^2+2(.5k)+4)$

After evaluating you get 11.125 square units.

Step-by-step explanation:

The width of each rectangle is the same so we want to take the distance from x=0 to x=3 and divide by 6 since we want 6 equal base lengths for our rectangles.

The distance between x=0 and x=3 is (3-0)=3.

We want to divide that length of 3 units by 6 which gives a length of a half per each base length.

We are doing right endpoint value so I'm going to stat at x=3. The first rectangle will be drawn to the height of f(3).

The next right endpoint is x=3-1/2=5/2=2.5, and the second rectangle will have a height of f(2.5).

The next will be at x=2.5-.5=2, and the third rectangle will have a height of f(2).

The fourth rectangle will have a height of f(2-.5)=f(1.5).

The fifth one will have a height of f(1.5-.5)=f(1).

The last one because it is the sixth one will have a height of f(1-.5)=f(.5).

So to find the area of a rectangle you do base*time.

So we just need to evaluate:

$\frac{1}{2}f(3)+\frac{1}{2}f(2.5)+\frac{1}{2}f(2)+\frac{1}{2}f(1.5)+\frac{1}{2}f(1)+\frac{1}{2}f(.5)$

or by factoring out the 1/2 part:

$\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))$

To find f(3) replace x in -x^2+2x+4 with 3:

-3^2+2(3)+4

-9+6+4

To find f(2.5) replace x in -x^2+2x+4 with 2.5:

-2.5^2+2(2.5)+4

-6.25+5+4

2.75

To find f(2) replace x in -x^2+2x+4 with 2:

-2^2+2(2)+4

-4+4+4

To find (1.5) replace x in -x^2+2x+4 with 1.5:

-1.5^2+2(1.5)+4

-2.25+3+4

4.75

To find f(1) replace x in -x^2+2x+4 with 1:

-1^2+2(1)+4

-1+2+4

To find f(.5) replace x in -x^2+2x+4 with .5:

-.5^2+2(.5)+4

-.25+1+4

4.75

Now let's add those heights. After we obtain this sum we multiply by 1/2 and we have our approximate area:

$\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))$

$\frac{1}{2}(1+2.75+4+4.75+5+4.75)$

$\frac{1}{2}(22.25)$

$11.125$

Okay now if you wanted the summation notation for:

$\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))$

is it

$\frac{1}{2}\sum_{k=1}^{6}(f(.5+.5(k-1)))$

or after simplifying a bit:

$\frac{1}{2}\sum_{k=1}^6 f((.5+.5k-.5))$

$\frac{1}{2}\sum_{k=1}^6f((.5k))$

If you are wondering how I obtain the .5+.5(k-1):

I realize that 3,2.5,2,1.5,1,.5 is an arithmetic sequence with first term .5 if you the sequence from right to left (instead of left to right) and it is going up by .5 (reading from right to left.)

6 0

3 years ago