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stepan [7]
3 years ago
12

Suppose that a number written as a decimal has an infinite number of no -repeating digits after the decimal point. What is this

number called?

Mathematics
1 answer:
bekas [8.4K]3 years ago
3 0

Answer:

irrational

are the numbers like 7.55782902..where the numbers aren't repeated

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Solve the system of equations. 4y + 11x – 67 = 0 2y + 5x - 19 = 0 x= y=​
katen-ka-za [31]

Answer:

x = 29, y = -63

Step-by-step explanation:

Solve the following system:

{4 y + 11 x - 67 = 0 | (equation 1)

{2 y + 5 x - 19 = 0 | (equation 2)

Express the system in standard form:

{11 x + 4 y = 67 | (equation 1)

v5 x + 2 y = 19 | (equation 2)

Subtract 5/11 × (equation 1) from equation 2:

{11 x + 4 y = 67 | (equation 1)

{0 x+(2 y)/11 = -126/11 | (equation 2)

Multiply equation 2 by 11/2:

{11 x + 4 y = 67 | (equation 1)

{0 x+y = -63 | (equation 2)

Subtract 4 × (equation 2) from equation 1:

{11 x+0 y = 319 | (equation 1)

{0 x+y = -63 | (equation 2)

Divide equation 1 by 11:

{x+0 y = 29 | (equation 1)

{0 x+y = -63 | (equation 2)

Collect results:

Answer: {x = 29, y = -63

3 0
3 years ago
Read 2 more answers
Evaluate 8(7x--3) when x=9
Schach [20]

I will assume that by "Evaluate 8(7x--3) when x=9" you meant:

"Evaluate 8(7x-3) when x=9."

First, substitute 9 for x in 7x-3: 7(9)-3 = 60. Rewrite the problem as:

8(60). Multiply. Answer: 480

5 0
3 years ago
Complete. Express the product in simplest form.<br><br> 2/5 of 7/10 =
tester [92]
Of means multiply in math (most of the time). 
So multiply. 
2/5 x 7/10 = 14/50 divided by 2/2 = 7/25 
3 0
3 years ago
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The question is what statement is true about these triangles if you cant see :))) ty for helping
Anuta_ua [19.1K]

Answer:

its type hard to see ngl

Step-by-step explanation:

8 0
3 years ago
A geometric sequence is defined by the general term tn = 75(5n), where n ∈N and n ≥ 1. What is the recursive formula of the sequ
andreyandreev [35.5K]
The correct answer is C) t₁ = 375, t_n=5t_{n-1}.

From the general form,
t_n=75(5)^n, we must work backward to find t₁.

The general form is derived from the explicit form, which is
t_n=t_1(r)^{n-1}.  We can see that r = 5; 5 has the exponent, so that is what is multiplied by every time. This gives us

t_n=t_1(5)^{n-1}

Using the products of exponents, we can "split up" the exponent:
t_n=t_1(5)^n(5)^{-1}

We know that 5⁻¹ = 1/5, so this gives us
t_n=t_1(\frac{1}{5})(5)^n&#10;\\&#10;\\=\frac{t_1}{5}(5)^n

Comparing this to our general form, we see that
\frac{t_1}{5}=75

Multiplying by 5 on both sides, we get that
t₁ = 75*5 = 375

The recursive formula for a geometric sequence is given by
t_n=t_{n-1}(r), while we must state what t₁ is; this gives us

t_1=375; t_n=t_{n-1}(5)

3 0
4 years ago
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