All categories

3 years ago

2⁴+(2.75+1.75)÷0.9 fast pls helppppp

Mathematics

1 answer:

alisha [4.7K]3 years ago

5 0

Answer:

Step-by-step explanation:

First, add the parenthesis

2^4 + 4.50 ÷ 0.9

Second, divide 4.50 by 0.9

2^4 + 5

Third, get rid of the ^4

(2x2x2x2)

16 + 5

You might be interested in

What is the length of a rectangle with width 12 in. and area 90 in^2?

Alex

Answer: 7.5

Step-by-step explanation:

All you have to do is divide the base/width by area.

6 0

3 years ago

Read 2 more answers

Ys is the perpendicular bisector of xz. What is the length of Xs if xz is 18 inches long?

exis [7]

Answer:

XS is 9 inches long.

Step-by-step explanation:

Given that

YS is the perpendicular bisector of XZ.

Length of XZ = 18 inches

To find:

Length of XS = ?

Solution:

First of all, let us learn about the perpendicular bisector.

Perpendicular bisector of a line AB is a line PQ, which divides the line AB in two equal parts and is at an angle of $90^\circ$ with the line AB.

If B is on the line PQ, then $BP = BQ = \frac{PQ}{2}$ and

$\angle ABP = \angle ABQ = 90^\circ$

Applying the above property in our given question.

Kindly refer to the attached image for the given dimensions.

S is on the line XZ.

$XS = SZ = \frac{XY}{2} = \dfrac{18}{2} \\\Rightarrow \bold{XS = 9\ inches}$

So, the answer is XS is 9 inches long.

8 0

3 years ago

Given the funtion f(x)=-4-x: A. Evaluate: f(-2) B. Solve for x when f(x) =-8

NNADVOKAT [17]

f(x)=-4-x

A. f(-2)=-4-(-2)=-4+2=-2

f(-2)=-2

B. f(x)=8

-4-x=8 /+4

-x=12 /*(-1)

x=-12

3 0

3 years ago

What Is Least Common Multiple Of 5 and 9

muminat

Answer: 45

Step-by-step explanation: just multiply the two number together

6 0

3 years ago

Read 2 more answers

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