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3 years ago

Someone help me!!!!!

Mathematics

1 answer:

monitta3 years ago

7 0

Answer: 116

Step-by-step explanation:

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Need help with these 2 problems on my homework.

poizon [28]

4. 55 + 0.65*55 = $90.75

5. 90.75 * 0.25*90.75 = $113.44

6. 113.44 - 0.33*113.44 = $76.00

4 0

3 years ago

1/4(1-2/3)^2+1/3 Simplify

MrMuchimi

Simplifying the expression gives 13/ 36

<h3>How to solve the expression</h3>

Given;

1/4(1-2/3)^2+1/3

Represented as;

$\frac{1}{4} * ( 1 - \frac{2}3} )^2 + \frac{1}{3}$

Take the LCM of the bracket

$\frac{1}{4} * ( \frac{1}{3} )^2 + \frac{1}{3}$

Find the square

$\frac{1}{4} (\frac{1}{9} )+ \frac{1}{3}$

Find the LCM

$\frac{1}{36} + \frac{1}{3}$

1 + 12/36

13/ 36

Thus, simplifying the expression gives 13/ 36

Learn more about fractions here:

brainly.com/question/1622425

#SPJ1

6 0

1 year ago

Read 2 more answers

The school play was last night and a total of 100 tickets were sold.

maria [59]

Answer:

The answer will be 55 student tickets

Step-by-step explanation:

8 0

3 years ago

Sketch the region enclosed by y=3x and y=x^2 . Then find the area of the region.

kolezko [41]

Answer:

$A=4.5u^2$

Step-by-step explanation:

The integral of a function gives you the area under the curve, the subtraction of one of the areas from the other will give you the area in between.

The limits of integration are the points where the curves intersect each other(take the curves has a system of equation and solve for x and y):

$y=3x, y=x^2\\3x=x^2\\x^2-3x=0\\x(x-3)=0\\x_1=0\\x-3=0\\x_2=3$

$y_1=3x_1=3(0)=0\\y_2=3x_2=3(3)=9$

The integral will be the subtraction of the curve $y=3x$ and $y=x^2$ (In the graph you can see y=3x is the upper curve):

$\int\limits^3_0 {3x-x^2} \, dx =\left\frac{3x^2}{2}-\frac{x^3}{3}\right|^3_0=\frac{3(3)^2}{2}-\frac{(3)^3}{3}-\left(\frac{3(0)^2}{2}-\frac{(0)^3}{3}\right)\\\int\limits^3_0 {3x-x^2} \, dx =\frac{27}{2}-\frac{27}{3} =13.5-9=4.5u^2$

6 0

3 years ago

Me pueden ayudar ????

Julli [10]

La área va a ser 16ft^2

3 0

3 years ago

Someone help me!!!!!​

Someone help me!!!!!