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FromTheMoon [43]

3 years ago

12

3/4 of Ana's books are equal to two-thirds of Jake's books find the ratio of the number of Ana's books it to the number of Jake'

s books draw models to help you. Please show your work!!!!!!!

2 answers:

dimaraw [331]3 years ago

8 0

First you want to change the fractions to the same denominator
3/4 = 9/12 and 2/3 = 8/12
then you put it in ratio form...
the answer is 9:8

Paraphin [41]3 years ago

4 0

The answer would be 9:8

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What is the length of AA' , rounded to the nearest hundredth? Type a numerical answer is the space provided. Do not type spaces

inna [77]

Coordinate of A are (0,0)
Coordinates of A' are (5,2)

We can find the distance from A to A' using the distance formula:

$AA'= \sqrt{(5-0)^{2}+ (2-0)^{2} } \\ \\ 
AA'= \sqrt{25+4} \\ \\ 
AA'= \sqrt{29} \\ \\ 
AA'=5.39$

Thus, rounded to nearest hundredth, AA' is equal to 5.39

3 0

3 years ago

a group of 50 people were asked their gender and if they liked cats. the data from the survey are shown in the ben diagram. dete

sasho [114]

Answer:

A=13, b= 28, c= 6 d=22 e=31

Step-by-step explanation:

E = 15 +16 = 31

6 males like cats ( the middle section) so C = 6

d = 6 +16 = 22

a = 19-6 = 13

b = 50-22 = 28

so in order: a = 13, b = 28, c = 6, d = 22, e = 31

6 0

3 years ago

find the area of the trapezium whose parallel sides are 25 cm and 13 cm The Other sides of a Trapezium are 15 cm and 15 CM

Snezhnost [94]

$\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}$

Given - <u>A </u><u>trapezium</u><u> </u><u>ABCD </u><u>with </u><u>non </u><u>parallel </u><u>sides </u><u>of </u><u>measure </u><u>1</u><u>5</u><u> </u><u>cm </u><u>each </u><u>!</u><u> </u><u>along </u><u>,</u><u> </u><u>the </u><u>parallel </u><u>sides </u><u>are </u><u>of </u><u>measure </u><u>1</u><u>3</u><u> </u><u>cm </u><u>and </u><u>2</u><u>5</u><u> </u><u>cm</u>

To find - <u>Area </u><u>of </u><u>trapezium</u>

Refer the figure attached ~

In the given figure ,

AB = 25 cm

BC = AD = 15 cm

CD = 13 cm

<u>Construction</u><u> </u><u>-</u>

$draw \: CE \: \parallel \: AD \: \\ and \: CD \: \perp \: AE$

Now , we can clearly see that AECD is a parallelogram !

$\therefore$ AE = CD = 13 cm

Now ,

$AB = AE + BE \\\implies \: BE =AB - AE \\ \implies \: BE = 25 - 13 \\ \implies \: BE = 12 \: cm$

Now , In ∆ BCE ,

$semi \: perimeter \: (s) = \frac{15 + 15 + 12}{2} \\ \\ \implies \: s = \frac{42}{2} = 21 \: cm$

Now , by Heron's formula

$area \: of \: \triangle \: BCE = \sqrt{s(s - a)(s - b)(s - c)} \\ \implies \sqrt{21(21 - 15)(21 - 15)(21 - 12)} \\ \implies \: 21 \times 6 \times 6 \times 9 \\ \implies \: 12 \sqrt{21} \: cm {}^{2}$

Also ,

$area \: of \: \triangle \: = \frac{1}{2} \times base \times height \\ \\\implies 18 \sqrt{21} = \: \frac{1}{\cancel2} \times \cancel12 \times height \\ \\ \implies \: 18 \sqrt{21} = 6 \times height \\ \\ \implies \: height = \frac{\cancel{18} \sqrt{21} }{ \cancel 6} \\ \\ \implies \: height = 3 \sqrt{21} \: cm {}^{2}$

<u>Since </u><u>we've </u><u>obtained </u><u>the </u><u>height </u><u>now </u><u>,</u><u> </u><u>we </u><u>can </u><u>easily </u><u>find </u><u>out </u><u>the </u><u>area </u><u>of </u><u>trapezium </u><u>!</u>

$Area \: of \: trapezium = \frac{1}{2} \times(sum \: of \:parallel \: sides) \times height \\ \\ \implies \: \frac{1}{2} \times (25 + 13) \times 3 \sqrt{21} \\ \\ \implies \: \frac{1}{\cancel2} \times \cancel{38 }\times 3 \sqrt{21} \\ \\ \implies \: 19 \times 3 \sqrt{21} \: cm {}^{2} \\ \\ \implies \: 57 \sqrt{21} \: cm {}^{2}$

hope helpful :D

6 0

2 years ago

Write 2 fraction problem with any operation!!plz help!!!!!write them down and solve them!!!!

Deffense [45]

John wanted to go on a train. the equation y=9/5 x describes the number of miles y that a train travels in x minutes. what is the constant speed of the train in terms of hours?

The constant speed of the train is 108 miles per hour

4 0

3 years ago

A quadratic equation is shown below:

Valentin [98]

Part A)

The given equation is:

$25 x^{2} +10x+1=0$

The radicand or discriminant(d) of the equation will be:

$d=(10)^{2}-4(25)(1) \\ \\ 
d=100-100 \\ \\ 
d=0$

Since the discriminant is equal to 0, the given quadratic equation has only 1 root. In other words we can say the the given equation is a perfect square.

Part B)

The given equation is:

$4 x^{2} -4x+1=0$

We can solve this expression by factorization. Factors of middle term are to be made in such a way that their product equals the product of first and third term and sum is equal to the middle term i.e. product should be 4x² and sum should be -4x.
So the two such terms are -2x and -2x. Using the factors and simplifying the equation by taking common we get:

$4 x^{2} -2x-2x+1=0 \\ \\ 
2x(2x-1)-1(2x-1)=0 \\ \\ 
(2x-1)(2x-1)=0 \\ \\ 
(2x-1) ^{2}=0 \\ \\ 
2x-1=0 \\ \\ 
x= \frac{1}{2}$

4 0

3 years ago