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zhenek [66]
3 years ago
14

Please include how to solve this!! Thanks you!

Mathematics
1 answer:
Anna007 [38]3 years ago
6 0

Here is how to do the problem:


First, set each of the shapes equal to a variable. So here's how I am going to do it:

circle=y

square=x

triangle=z

x+y=53

2y=36

z+18=45


This means that y=18.

Then, let's plug in y=18 into x+y=53.

x=35


Then, finally, we subtract 18 from 45.

z=27


The circle=18

The square=35

The triangle=27

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Prove With Diagram Which fraction is larger, 5/8 or 3/4 Brainliest
4vir4ik [10]

Answer:

3/4

Step-by-step explanation:

Convert the fractions into decimals.

5/8 = 0.625

3/4 = 0.75

0.75 is greater than 0.625.

8 0
3 years ago
If x and y are integers greater than 1, is x a multiple of y ? (1) \small 3y^{2}+7y=x (2) \small x^{2}-x is a multiple of y. Sta
Nata [24]

Answer:

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Step-by-step explanation:

A multiple of a number is obtained after multiplying the number by an integer.

Here,

x, y are any two integers greater than 1,

(1) We have,

\small 3y^{2}+7y=x

\implies y(3y+7) = x

∵ y is an integer ⇒ 3y + 7 is also an integer,

⇒ y × an integer = x

That is, when we multiply y by a number we obtain x,

∴ x is a multiple of y.

Thus, statement (1) ALONE is sufficient.

(2),

\small x^{2}-x\text{ is a multiple of y}

I.e.

y\times a = x^2-x, where a is an integer,

\implies y\times a = x(x-1)

∵ x and x - 1 are disjoint numbers,

There are three possible cases,

Case 1 : x is multiple of y

Case 2 : (x-1) is a multiple of y,

Case 3 : neither x nor x - 1 are multiple of y but their product is multiple of y,

Thus, statement (2) is not sufficient.

8 0
3 years ago
What is the sum of the geometric series
lbvjy [14]
If S denotes the sum of the first n terms of a geometric series with first term a and common ratio r, then

S=a+ar+ar^2+\cdots+ar^{n-3}+ar^{n-2}+ar^{n-1}
rS=ar+ar^2+ar^3+\cdots+ar^{n-2}+ar^{n-1}+ar^n
\implies S-rS=a+(ar-ar)+(ar^2-ar^2)+\cdots+(ar^{n-1}-ar^{n-1})-ar^n
\implies (1-r)S=a(1-r^n)
\implies S=a=\dfrac{1-r^n}{1-r}

Using summation notation, you have

S=\displaystyle\sum_{x=1}^nar^{x-1}=\sum_{x=0}^{n-1}ar^x=a\dfrac{1-r^n}{1-r}

In this case, you have a=2, r=\dfrac12, and n=16. So the value of the sum is

\displaystyle\sum_{x=0}^{15}2\left(\frac14\right)^x=2\dfrac{1-\left(\frac14\right)^{16}}{1-\frac14}\approx2.67

Rounded to the nearest whole number, the answer would be 3.
4 0
3 years ago
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olganol [36]
The answer is 1/2. The reasoning is 1/2 x 1/2=1/4 1/4 x 1/2=1/8 fractions multiply on numerator and denominator then simplify
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3 years ago
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denpristay [2]
5.01 is greater

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7 0
2 years ago
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