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3 years ago

Can someone help me figure this out ..I have 7 more questions like this and I don’t get it ugh

Mathematics

1 answer:

lord [1]3 years ago

3 0

It would be answer A because something in the negatives can not be greater than something in the negatives

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I need help with this <br>7f + 2f - 3b + b<br><br>I can't figure it out

Leni [432]

Answer:

9f-4b

Step-by-step explanation:

you add the two fs together and you add the two bs together to get the answer

5 0

3 years ago

For the noon meal today she ate 1/2 of a 3-ounce serving of meatloaf, 3/4 of her 3-ounce serving of mashed potatoes, and 1/3 of

Leya [2.2K]

4.41 ounces. 1.5 +2.25 + .66=4.41. You just multiply the faction by the number and add it all up.

3 0

3 years ago

24 is what percent of 96?

Andrei [34K]

Answer: 25%

Step-by-step explanation:

make it into a fraction

24/96 then simplify

decimal answer is .25

move the decimal to the right 2 times

7 0

3 years ago

Please help with this table

Lyrx [107]

Answer:

Step-by-step explanation:

Mean = 8+12+10+6+14 = 50

50 divided by 5 = 10

(Divided by 5 because there are 5 numbers)

7 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