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4 years ago

Single variable linear equation that has one solution

Mathematics

1 answer:

Tcecarenko [31]4 years ago

3 0

Is the variable segment negative or positive determining for the answer

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Helppppppppppppppppppp I’ll mark you brainlist don’t respond if you’re going to put something random I will report you :)

Savatey [412]

Y=5/2x-2
Count up 5 and 2 to the left and you bring at -2

8 0

3 years ago

Read 2 more answers

A box contains 240 lumps of sugar. Fie lumps are fitted across the box and the were three layers. How many lumps are fitted alon

lutik1710 [3]

Answer:

v240

Step-by-step explanation:

5 0

2 years ago

Jamie wants to build a rectangular fence around a new garden. He can afford at most 70

KonstantinChe [14]

Option B is correct. Yes, (9, 22) is a solution to the inequalities and the measurements will fit in the space.

The formula for calculating the perimeter of the rectangular fence is expressed as:

A = 2(L + W) where:

L is the length

W is the width

If Jamie can afford at most 70feet to build a rectangular fence is expressed as:

2L + 2W ≤ 70

We are to check if the garden measure 9 feet by 22 feet. To do this we are to substitute L = 9 and W = 22 into the formula to check if the result will be less than 70

On substituting:

= 2(9) + 2(22)

= 18 + 44

= 62 feet

Since 63 feet is less than 70 feet, hence we can conclude that Yes, (9, 22) is a solution to the inequalities and the measurements will fit in the space.

Learn more here: brainly.com/question/17229451

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%20%5Crm%5Csum%20%5Climits_%7Bn%20%3D%200%7D%5E%7B%20%5Cinfty%20%7D%20%20%5Carcsin%20%5Clar

sineoko [7]

Recall that over an appropriate domain,

$\arcsin(x) \pm \arcsin(y) = \arcsin\left(x \sqrt{1-y^2} \pm y \sqrt{1-y^2}\right)$

Let $x=\frac1{n+1}$ and $y=\frac1{n+2}$ (these belong to the "appropriate domain", so the identity holds). We have

$\dfrac{\sqrt{n+3}}{(n+2)\sqrt{n+1}} = \dfrac1{n+1} \sqrt{\dfrac{(n+3)(n+1)}{(n+2)^2}} = \dfrac1{n+1} \sqrt{1 - \dfrac1{(n+2)^2}} = x \sqrt{1-y^2}$

and

$\dfrac{\sqrt n}{(n+1)\sqrt{n+2}} = \dfrac1{n+2} \sqrt{\dfrac{n(n+2)}{(n+1)^2}} = \dfrac1{n+2} \sqrt{1 - \dfrac1{(n+1)^2}} = y \sqrt{1-x^2}$

Then the sum telescopes, as

$\displaystyle \sum_{n=0}^\infty \arcsin\left(x \sqrt{1-y^2} - y \sqrt{1-x^2}\right) = \sum_{n=0}^\infty \left( \arcsin(x) - \arcsin(y) \right) \\\\ = \left(\arcsin(1) - \arcsin\left(\frac12\right)\right) + \left(\arcsin\left(\frac12\right) - \arcsin\left(\frac13\right)\right) + \cdots \\\\ = \arcsin(1) = \boxed{\frac\pi2}$

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2 years ago

Which of the following are not polynomials?

VladimirAG [237]

Letter b and letter c

8 0

3 years ago