All categories

3 years ago

According to exponent rules, when we multiply the expressions we __ the exponents

Mathematics

1 answer:

serious [3.7K]3 years ago

6 0

Answer:

Add

Step-by-step explanation:

You might be interested in

Which is bigger 2/5 or 3/4

sp2606 [1]

3/4 = 0.75
2/5 = 0.4

0.4 < 0.75

Therefore,

2/5 < 3/4

4 0

3 years ago

a recipe calls for 2 2/4 cups of Raisins but Julie has only a 1/4 cup. how many 1/4 cups does she need to measure out?

schepotkina [342]

1/4+1/4+1/4+1/4+1/4+1/4+1/4+1/4+1/4+1/4=2 2/4
or: 1/4X9=2 2/4=2 1/2

4 0

4 years ago

Read 2 more answers

QUICK SOLVE 2y=x+6 show work !!!

marusya05 [52]

Answer:

y=0.5x+3

Step-by-step explanation:

Divide both sides by 2 which will give you y=0.5x+3.

8 0

3 years ago

Read 2 more answers

PLEASE HELP!!! I don't understand it

Len [333]

Answer:

Step-by-step explanation:

This is exponential decay; the height of the ball is decreasing exponentially with each successive drop. It's not going down at a steady rate. If it was, this would be linear. But gravity doesn't work on things that way. If the ball was thrown up into the air, it would be parabolic; if the ball is dropped, the bounces are exponentially dropping in height. The form of this equation is

$y=a(b)^x$ , or in our case:

$A(n)=a(b)^{n-1}$ , where

a is the initial height of the ball and

b is the decimal amount the bounce decreases each time. For us:

a = 1.5 and

b = .74

Filling in,

$A(n)=1.5(.74)^{n-1}$

If ww want the height of the 6th bounce, n = 6. Filling that into the equation we already wrote for our model:

$A(6)=1.5(.74)^{6-1}$ which of course simplifies to

$A(6)=1.5(.74)^5$ which simplifies to

$A(6)=1.5(.22190066)$

So the height of the ball is that product.

A(6) = .33 cm

A is your answer

3 0

4 years ago

Find the integrate of 2x/sqrt(1-x^2) dx from 0 to 1/2

Ludmilka [50]

Compute the definite integral:

$\large\begin{array}{l}\mathsf{\displaystyle\int_{0}^{\frac{1}{2}}\frac{2x}{\sqrt{1-x^2}}\,dx}\\\\ =\mathsf{\displaystyle\int_{0}^{\frac{1}{2}}\frac{-1}{\sqrt{1-x^2}}\cdot (-2x)\,dx\qquad\quad(i)} \end{array}$

$\large\begin{array}{l} \textsf{Substitute}\\\\ \mathsf{1-x^2=u~~\Rightarrow~~-2x\,dx=du}\\\\\\ \textsf{Finding the new limits of integration:}\\\\ \begin{array}{lcl} \textsf{When }\mathsf{x=0}&~\Rightarrow~&\mathsf{u=1-0^2}\\\\ &&\mathsf{u=1}\\\\\\ \textsf{When }\mathsf{x=\dfrac{1}{2}}&~\Rightarrow~&\mathsf{u=1-\left(\dfrac{1}{2}\right)^2}\\\\ &&\mathsf{u=1-\dfrac{1}{4}}\\\\ &&\mathsf{u=\dfrac{3}{4}} \end{array} \end{array}$

$\large\begin{array}{l} \textsf{Then (i) becomes}\\\\ =\mathsf{\displaystyle\int_{1}^{\frac{3}{4}}\frac{-1}{\sqrt{u}}\,du}\\\\ =\mathsf{\displaystyle\int_{1}^{\frac{3}{4}}\frac{-1}{u^{\frac{1}{2}}}\,du}\\\\ =\mathsf{\displaystyle\int_{1}^{\frac{3}{4}} (-1)\cdot u^{-\frac{1}{2}}\,du}\\\\ =\mathsf{(-1)\cdot \dfrac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\bigg|_{1}^{\frac{3}{4}}} \end{array}$

$\large\begin{array}{l} =\mathsf{(-1)\cdot \dfrac{~~u^{\frac{1}{2}}~~}{\frac{1}{2}}\bigg|_{1}^{\frac{3}{4}}}\\\\ =\mathsf{(-1)\cdot 2u^{\frac{1}{2}}\Big|_{1}^{\frac{3}{4}}}\\\\ =\mathsf{-2\cdot \sqrt{u}\Big|_{1}^{\frac{3}{4}}}\\\\ =\mathsf{-2\cdot \left(\sqrt{\dfrac{3}{4}}-\sqrt{1}\right)} \end{array}$

$\large\begin{array}{l} =\mathsf{-2\cdot \left(\dfrac{\sqrt{3}}{2}-1\right)}\\\\ =\mathsf{-\diagup\!\!\!\! 2\cdot \dfrac{\sqrt{3}}{\diagup\!\!\!\! 2}+(-2)\cdot (-1)}\\\\ =\mathsf{-\sqrt{3}+2}\\\\ =\mathsf{2-\sqrt{3}} \end{array}$

$\large\begin{array}{l} \boxed{\begin{array}{c}\mathsf{\displaystyle\int_{0}^{\frac{1}{2}}\frac{2x}{\sqrt{1-x^2}}\,dx=2-\sqrt{3}} \end{array}}\qquad\checkmark \end{array}$

If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2153950

$\large\textsf{I hope it helps.}$

Tags: definite integral integrate limits function irrational square root sqrt fraction composite substitution integral calculus

4 0

4 years ago