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3 years ago

Move the dot to the number that equals 7 X

Mathematics

1 answer:

balu736 [363]3 years ago

8 0

Answer:

the answer is 21

Step-by-step explanation:

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Y= -x +3 and y = x - 1

inysia [295]

Answer:

x=2 and y=1

Step-by-step explanation:

Rewrite equations:

y=−x+3;y=x−1

Step: Solve

y=−x+3

Step: Substitute−x+3foryiny=x−1:

y=x−1

−x+3=x−1

−x+3+−x=x−1+−x(Add -x to both sides)

−2x+3=−1

−2x+3+−3=−1+−3(Add -3 to both sides)

−2x=−4−2x−2=−4−2

(Divide both sides by -2)

x=2

Step: Substitute2forxiny=−x+3:

y=−x+3

y=−2+3

y=1(Simplify both sides of the equation)

Hope this Helps you :)

3 0

3 years ago

and a list of numbers the pattern increases by 0.001 as you move to the right if the third number list is 0.0 64 what is the fir

Ksivusya [100]

Answer:

0.062

Step-by-step explanation:

The numbers will decrease by 0.001 as you move the the left, so the list of numbers can be found as ...

3rd number: 0.064

2nd number: 0.064 -0.001 = 0.063

1st number: 0.063 -0.001 = 0.062

7 0

3 years ago

Use the substitution method to solve each linear system. 4x - 3y = -13,-2x + y = 4

alisha [4.7K]

Answer:

Step-by-step explanation:

(X,Y)=(-55/2,-38

6 0

3 years ago

Find the domain of the Bessel function of order 0 defined by [infinity]J0(x) = Σ (−1)^nx^2n/ 2^2n(n!)^2 n = 0

Snowcat [4.5K]

Answer:

Following are the given series for all x:

Step-by-step explanation:

Given equation:

$\bold{J_0(x)=\sum_{n=0}^{\infty}\frac{((-1)^{n}(x^{2n}))}{(2^{2n})(n!)^2}}\\$

Let the value a so, the value of $a_n$ and the value of $a_(n+1)$ is:

$\to a_n=\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}$

$\to a_{(n+1)}=\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}$

To calculates its series we divide the above value:

$\left | \frac{a_(n+1)}{a_n}\right |= \frac{\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}}{\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}}\\\\$

$= \left | \frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2} \cdot \frac {2^{2n}(n!)^2}{(-1)^2n x^{2n}} \right |$

$= \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)!^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |$

$= \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\= \left | \frac{x^{2n}\cdot x^2}{2^{2n} \cdot 2^2(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\$

$= \frac{x^2}{2^2(n+1)^2}\longrightarrow 0$ for all x

The final value of the converges series for all x.

8 0

4 years ago

Donna bought 5 bags of dog treats for $13.10. What is the cost per bag of dog treats?

ch4aika [34]

It is answer C because 13.10 divided by 5 is 2.62 Pls mark brainliest

3 0

3 years ago