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2 years ago

13

What is 5PX7P^3 simplified?

1 answer:

morpeh [17]2 years ago

5 0

$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

Here's the solution ~

$\qquad \sf \dashrightarrow \: 5P \times 7 {P}^{3}$

$\qquad \sf \dashrightarrow \: (5 \times 7)(P \times P {}^{3} )$

$\qquad \sf \dashrightarrow \: 35(P {}^{ 1 + 3} )$

$\qquad \sf \dashrightarrow \:35 P {}^{4}$

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A triangle has side lengths 11, 12, and 15. Is it acute, obtuse, or right?

Sedbober [7]

Answer:

right triangle

Step-by-step explanation:

because 15 would be the base so 11 and 12 would be the sides and fot the triangle to be acute or obtuse the total of the sides (11+12) has to be lass than the base (15), but it is more 23 so it has to be a right triangle.

8 0

2 years ago

Calculate the area of the shaded sector.<br><br> 11.73<br><br> 11.2<br><br> 11.4<br><br> 11.83

laiz [17]

The formula for the area of a circle is

A=πr²
plug in the data we know
A=π4²
A=16π

Since we only we need to find the area of the shaded section, we need to twerk this equation a bit. (remember that the shaded section is 84/360 of the original circle.)

A=84/360*16π
A=11.728
A=11.73 units²

answer: <span>A=11.73 units²</span>

4 0

2 years ago

What is 30 divided by 75 and show the work plz?

lions [1.4K]

Answer:

75/30= 2 with a remainder of 15

Step-by-step explanation:

Divide 75 by 30 to get 2

Multiply the quotient digit (2) by the divisor 30

subtract 60 from 75

the solution for long division of 75/ 30 is 2 remainder of 15

convert it to mixed number

2 15/30

4 0

3 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

2 years ago

work out the equation of the line which passes through the point (-1, 2) and is paralell to the line y = x + 4

notka56 [123]

Step-by-step explanation:

working g and everything is above

hope its of help

3 0

2 years ago