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3 years ago

Can y’all help me what do I do here please I’ll give you brainliest

Mathematics

1 answer:

exis [7]3 years ago

3 0

Answer:

1, 2, -3, 51

Step-by-step explanation:

0 ÷ 2 = 0

0 + 1 = 1

2 ÷ 2 = 1

1 + 1 = 2

-8 ÷ 2 = -4

-4 + 1 = -3

100 ÷ 2 = 50

50 + 1 = 51

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Please Help!!!!!<br> Prove that polynomials are not closed under division.

AveGali [126]

Polynomials are not closed under division. When you divide polynomials it is possible to get quotients with negative exponents or with fractions that have exponents in the denominator, and neither of these could be included in polynomials.

5 0

3 years ago

Solve the linear equation <br> 5x-10y=-5

IceJOKER [234]

ANSWER WITH STEPS:

Let's solve for x.

5x−10y=−5

Step 1: Add 10y to both sides.

5x−10y+10y=−5+10y

5x=10y−5

Step 2: Divide both sides by 5.

10y−5

x=2y−1

6 0

2 years ago

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

3 years ago

Seven less than one-fourth of a number is 2. What is the number?

insens350 [35]

Based on the equation given, it can be deduced that the value of the number will be 36.

Let the number be represented by x

Therefore, seven less than one-fourth of a number will be:

=(1/4 × x) - 7

= 0.25x - 7

Now, seven less than one-fourth of a number is 2 will be written as:

0.25x - 7 = 2

0.25x = 2 + 7

0.25x = 9

x = 9/0.25

x = 36

Learn more about equation on:

brainly.com/question/13802812

6 0

2 years ago

some of the steps in the derivation of the quadratic formula are shown. step 3: –c b^2/4a=a(x^2 b/ax b^2/4a62) step 4a: –c b^2/4

wel

-converting to a common denominator

The terms -c of the left side was converted to -4ac/4a to have the same denominator of b^2/4a.

3 0

3 years ago

Read 2 more answers