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2 years ago

Can someone help me on this math question pls I have 12 more questions I need help on too !

Mathematics

1 answer:

love history [14]2 years ago

5 0

A is the answer. Since they will both be divided by 4 due to the same amount of points, because Globrite has a greater range, it will have a greater mean. Glo brite also ranges bigger than Luminate

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Find the area of this circle please…Use 3 for pi

ZanzabumX [31]

Given:

The radius of circle is 11 inch

The value for pi is 3

The objective is to find the area of the circle.

The area of the circle is given by the formula:

$A=\pi\times r^2^{}$

Substituting ,r = 11

$\pi=3$

We get:

$\begin{gathered} A=\pi\times r^2 \\ =3\times(11)^2 \\ =3\times121 \\ =363 \end{gathered}$

Hence, the area of the circle is 363 square inches

5 0

1 year ago

Explain how you can use the distributive property and mental math to find 5 x 198

viktelen [127]

Answer is 990
I hope It is the right answer

7 0

2 years ago

The opening in a rectangular door frame measures 9 feet by 3 feet. Each of the four doors described in the table measures 9 feet

just olya [345]

Answer:

Door 2

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

What value should be added to both sides for completing the square in this equation?

Mila [183]

Hey there!

x^2 + 4x = 12

SUBTRACT 12 to BOTH SIDES

x^2 + 4x - 12 = 12 - 12

SIMPLIFY IT

x^2 + 4x - 12 = 0

FACTOR the LEFT SIDE of your EQUATION

(x - 2)(x + 6) = 0

• EQUATION #1: x - 2 = 0

• EQUATION #2: x + 6 = 0

SIMPLIFY IT

• EQUATION #1 answer: x = 2

• EQUATION #2 answer: x = -6

OVERALL ANSWER: x = 2 or x = -6

YOUR ANSWER: x = 2 (Option A.)

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

3 0

2 years ago

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$

$\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )$

$\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du$

$\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]$

Reverting the change of variable, we have:

$\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]$

Then, evaluating we have:

$\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797$

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0

3 years ago