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3 years ago

Is 23/40 a rational or irrational number? I'm not sure how to solve this one.

Mathematics

1 answer:

vfiekz [6]3 years ago

5 0

Answer:

23/40 is a rational number

Step-by-step explanation:

an irrational number cannot be written as a fraction - with a finite end. An irrational number would be a repeating decimal

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Simplify to create an equivalent expression.<br> 8(a+2)+2(2+3a)=

Alchen [17]

Answer:

14a + 20

Step-by-step explanation:

8(a + 2) + 2(2 + 3a)

expand the bracket

8a + (8*2) + (2*2) + (2*3a)

8a + 16 + 4 + 6a

bring like terms together

8a + 6a + 16 + 4

14a + 20

3 0

4 years ago

What is the area of the trapezoid if a = 10 b= 18 and h = 6 inches?

yanalaym [24]

[ Answer ]

$\boxed{\bold{84 \ In.^{2} }}$

[ Explanation ]

Formula For Area Of Trapezoid: $\boxed{\bold{\frac{a \ + \ b}{2}Height }}$

-----------------------------------------

A = 10

B = 18

H = 6

Plug Numbers Into Equation: $\boxed{\bold{\frac{10 \ + \ 18}{2}Height }}$

Solve

10 + 18 = 28

28 ÷ 2 = 14

14 · 6 = 84

Final Answer

$\boxed{\bold{[] \ Eclipsed \ []}}$

4 0

3 years ago

Solve it by substitution method:- 5x - 6y=-9 and 3x + 4y= 25

user100 [1]

$\left\{\begin{array}{ccc}-5x-6y=-9&/\cdot2\\3x+4y=25&/\cdot3\end{array}\right\\+\left\{\begin{array}{ccc}-10x-12y=-18\\9x+12y=75\end{array}\right\\------------\\.\ \ \ \ \ \ \ -x=57\\.\ \ \ \ \ \ \ \ \ x=-57\\\\3\cdot(-57)+4y=25\\-171+4y=25\\4y=25+171\\4y=196\ \ \ \ /:4\\y=49\\\\Solution:x=-57;\ y=49.$

8 0

3 years ago

Write each of these in scientific notation <br> 1) 0.00000402<br> 2) 1,900,000

Mkey [24]

Answer:

See below in bold.

Step-by-step explanation:

1. 0.00000402

= 4.02 * 10^-6 (Counting the digits after the decimal point until we get to the 4 gives us the -6).

2. 1,900,000

= 1.9 * 10^6 ( counting the number of digits after the 1 gives us 6).

3 0

3 years ago

What are the solutions to the equation (2x – 5)(3x – 1) = 0?

user100 [1]

Answer: $x=\frac{5}{2}, x=\frac{1}{3}$

Explanation:

The equation is:

$(2x-5)(3x-1)=0$

The term on the left consists of a product of two different factors: therefore, this product can be zero if either the first term (2x-5) or the second term (3x-1) is equal to zero.

This means that we can solve separately for the two terms:

$2x-5=0\\3x-1=0$

Solving the first equation:

$2x-5=0\\2x=5\\x=\frac{5}{2}$

Solving the second equation:

$3x-1=0\\3x=1\\x=\frac{1}{3}$

8 0

3 years ago

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