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3 years ago

Write 1000 in words. TQVM pls help

Mathematics

2 answers:

marusya05 [52]3 years ago

7 0

Thousand

Please Mark As Brainliest

Elis [28]3 years ago

6 0

Answer:

One thousand

Step-by-step explanation:

You might be interested in

Find x: 3/(x-4)(x-7) + 6/(x-7)(x-13) + 15/(x-13)(x-28) - 1/x-28 = -1/20

Novay_Z [31]

The value of x<em> </em>in the polynomial fraction 3/((x-4)•(x-7)) + 6/((x-7)•(x-13)) + 15/((x-13)•(x-28)) - 1/(x-28) = -1/20 is <em>x </em>= 24

<h3>How can the polynomial with fractions be simplified to find<em> </em><em>x</em>?</h3>

The given equation is presented as follows;

$\frac{3}{(x - 4) \cdot (x - 7) } + \frac{6}{(x - 7) \cdot (x - 13) } + +\frac{15}{(x - 13) \cdot (x - 28) } - \frac{1}{(x - 28) } = - \frac{1}{20}$

Factoring the common denominator, we have;

$\frac{3\cdot(x - 13) \cdot(x - 28) + 6 \cdot(x - 4) \cdot(x - 28) + 15 \cdot(x - 4) \cdot(x - 7) - (x - 4) \cdot (x - 7)\cdot(x - 13)}{(x - 4) \cdot (x - 7)\cdot(x - 13) \cdot(x - 28)} + = - \frac{1}{20}$

Simplifying the numerator of the right hand side using a graphing calculator, we get;

By expanding and collecting, the terms of the numerator gives;

-(x³ - 48•x + 651•x - 2548)

Given that the terms of the numerator have several factors in common, we get;

-(x³ - 48•x + 651•x - 2548) = -(x-7)•(x-28)•(x-13)

Which gives;

$\frac{-(x - 7) \cdot(x - 28)\cdot (x - 13)}{(x - 4) \cdot (x - 7)\cdot(x - 13) \cdot(x - 28)} + = - \frac{1}{20}$

Which gives;

$\frac{-1}{(x - 4)} + = - \frac{1}{20}$

x - 4 = 20

Therefore;

x = 20 + 4 = 24

Learn more about polynomials with fractions here:

brainly.com/question/12262414

#SPJ1

5 0

2 years ago

Please help i need this done QUICK im timed ill give the best answer brainliest

seraphim [82]

Answer:

x=−32/5

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

1/2(1/4x−3/5)=1/4(2/5+3/4x)

(1/2)(1/4x)+(1/2)(−3/5)=(1/4)(2/5)+(1/4)(3/4x)(Distribute)(1/2)(1/4x)+(1/2)(−3/5)=(1/4)(2/5)+(1/4)(3/4x)(Distribute)

1/8x+−3/10=1/10+3/16x

1/8x+−3/10=3/16x+1/10

Step 2: Subtract 3/16x from both sides.

1/8x+−3/10−3/16x=3/16x+1/10−3/16x

−1/16x+−3/10=1/10

Step 3: Add 3/10 to both sides.

−1/16x+−3/10+3/10=1/10+3/10

−1/16x=2/5

Step 4: Multiply both sides by 16/(-1).

(16/−1)*(−1/16x)=(16/−1)*(2/5)

x=−3/25

7 0

3 years ago

Please help! 5 Points. Hurry!

larisa [96]

<h2>Answer:</h2>

<u>The correct option is 3x + 4y = 10</u>

<h2>Step-by-step explanation:</h2>

The standard form for any linear equations having two variables is written as Ax+By=C. For example, 3x+4y=10 is a linear equation in standard form. When an equation is given in this form, it becomes easy to find both intercepts (x and y). This form is also very useful when solving systems of two linear equations whose solution is required to find the point of intersection of given lines.

7 0

4 years ago

(5.6x - 4) + (16 - 2.1x)<br><br> What is the sum of this expression?

Damm [24]

$\huge\text{Hey there!}$

$\large\textsf{(5.6x - 4) + (16 - 2.1x)}$

$\large\text{DROP the PARENTHESES}$

$\large\textsf{5.6x - 4 + 16 - 2.1x}$

$\large\text{COMBINE the LIKE TERMS}$

$\large\textsf{5.6x - 2.1x - 4 + 16}$

$\large\textsf{5.6x - 2.1x = \bf 3.5x}$

$\large\textsf{-4 + 16 = \bf 12}$

$\large\textsf{= \bf 3.5x + 12 }$

$\boxed{\boxed{\large\text{Answer: \huge \bf 3.5x + 12}}}\huge\checkmark$

$\text{Good luck on your assignment and enjoy your day!}$

~ $\frak{Amphitrite1040:)}$

7 0

3 years ago

Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

Bezzdna [24]

Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

$a_2 = a_1 + 2$

$a_3 = a_2 + 2 = a_1 + 2\cdot2$

$a_4 = a_3+2 = a_1+3\cdot2$

$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$