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4 years ago

15.7 is it a rational,natural,whole, integer etc?

Mathematics

1 answer:

taurus [48]4 years ago

8 0

Answer:

Rational because it is a decimal that can be seen as a fraction.

Step-by-step explanation:

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Which are steps in the process of completing the square used to solve the equation 3 – 4x = 5x2 – 14x? Check all that apply.

Bond [772]

Answer:

1. 3 = 5x2 – 10x

2. 8 = 5(x2 – 2x + 1)

3. StartFraction 8 Over 5 EndFraction = (x – 1)2

Step-by-step explanation:

3-4x=5x^2-14x

3=5x^2-14x+4x

3=5x^2-10x

5x^2-10x-3=0

1. 3 = 5x2 – 10x

2. 8 = 5(x2 – 2x + 1)

8=5x^2-10x+5

8-5=5x^2-10x

3=5x^2-10x

3. StartFraction 8 Over 5 EndFraction = (x – 1)2

8/5=x^2-2x+1

Cross product

8=5(x^2-2x+1)

8=5x^2-10x+5

8-5=5x^2-10x

3=5x^2-10x

7 0

3 years ago

Read 2 more answers

California license plates used to consist of a system of a number then letter followed by a number then a letter and a number th

zhenek [66]

Answer:

for second and third letters, we again have 26 combinations available and thus 26×26×26=17576 combinations for letters. But digits are from 0 to 9 i.e. 10 combinations for each place and tolal 10×10=100 combinations. Hence for LLLDD , we have 1,757,600 combinations available for license plates.

Step-by-step explanation:

4 0

3 years ago

Y=9x-2 & 3x-2y=7 what is the value of x of the given system of equation

andre [41]

Y=9x-2
3x-2y=7
3x-18x+4=7
-15x=3
x=-1/5
y= -19/5

6 0

4 years ago

Harold uses the binomial theorem to expand the binomial (3x^5 - 1/9y^3)^4

riadik2000 [5.3K]

<h3>The complete question:</h3>

Harold uses the binomial theorem to expand the binomial $(3x^5 -\dfrac{1}{9}y^3)^4$

(a) What is the sum in summation notation that he uses to express the expansion?

(b) Write the simplified terms of the expansion.

Answer:

(a). $(3x^5 -\dfrac{1}{9}y^3)^4=$$\sum_{k=0}^{n} \binom{4}{k}(3x^5)^{4-k}( -\dfrac{1}{9}y^3)^k $$$

(b). $(3x^5 -\dfrac{1}{9}y^3)^4=81x^{20}-12x^{15}y^3+\dfrac{2x^{10}y^6}{3}-\dfrac{4x^5y^9}{243}+\frac{y^{12}}{6561}$

Step-by-step explanation:

(a).

The binomial theorem says

$(x+y)^n=$$\sum_{k=0}^{n} \binom{n}{k}x^{n-k}y^k $$$

For our binomial this gives

$\boxed{(3x^5 -\dfrac{1}{9}y^3)^4=$$\sum_{k=0}^{n} \binom{4}{k}x^{4-k}y^k $$}$

(b).

We simplify the terms of the expansion and get:

$$$\sum_{k=0}^{n} \binom{4}{k}(3x^5)^{4-k}y^k $$= \binom{4}{0}(3x^5)^{4-0}(-\dfrac{1}{9}y^3 )^0+\binom{4}{1}(3x^5)^{4-1}(-\dfrac{1}{9}y^3 )^1+\\\\\binom{4}{2}(3x^5)^{4-2}(-\dfrac{1}{9}y^3 )^2+\binom{4}{3}(3x^5)^{4-3}(-\dfrac{1}{9}y^3 )^3+\binom{4}{4}(3x^5)^{4-4}(-\dfrac{1}{9}y^3 )^4$

$$$\sum_{k=0}^{n} \binom{4}{k}(3x^5)^{4-k}(-\frac{1}{9}y^3 )^k $$= (3x^5)^{4}+4(3x^5)^{3}(-\frac{1}{9}y^3 )+6(3x^5)^{2}(-\frac{1}{9}y^3 )^2+\\\\4(3x^5)(-\frac{1}{9}y^3 )^3+(-\frac{1}{9}y^3 )^4$

$\boxed{(3x^5 -\dfrac{1}{9}y^3)^4=81x^{20}-12x^{15}y^3+\dfrac{2x^{10}y^6}{3}-\dfrac{4x^5y^9}{243}+\frac{y^{12}}{6561} }$

3 0

3 years ago

Fran graphs the equations y = 2x2 – 2 and y = –0.5x + 4. Her graph is shown below.

Firdavs [7]

The approximate solution you require is the x coordinate of the point of intersection of the 2 graphs.

8 0

4 years ago

Read 2 more answers

15.7 is it a rational,natural,whole, integer etc?​

15.7 is it a rational,natural,whole, integer etc?