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3 years ago

If an answer for this question is not a whole number, enter it as a decimal.

Mathematics

2 answers:

Naily [24]3 years ago

7 0

Answer: The answer would be A. Or if that is what you are asking.

Step-by-step explanation:

When you multiply decimals, you multiply like how you would for any other set of numbers. (It doesn't matter if you line the decimal points up or not.)

The only difference between multiplying normal numbers and decimals is the decimal point. How many digits there are to the right of the decimal point is how many times you move the decimal point for the new number to the left.

In the decimals 5.15 and 0.3, there are three digits to the right of the decimal in all, so, when you work this problem and come out with 1545, you move the decimal point to the left three times.

1^545

So the decimal point goes right between the 1 and the 5.

Hope I made it clear enough

Please give me Brainliest

zhenek [66]3 years ago

4 0

Answer:

1. (A - B) + C = -233.84

2. C - (B + A) = -322.72

Step-by-step explanation:

A = 220.03 - (55.27 + 115.32)

A = 49.44

B = 111.45 + (198.22 - 76.76)

B = -10.01

C = 28.15 - 202.98 - 108.46

C = -283.29

To find : (A - B) + C =?

On substituting the above values:

$(49.44-(-10.01)) + (-283.29)$

$(49.44 +10.01) - 283.29 = -233.84$

(A - B) + C = -233.84

To find : C - (B + A) =?

On substituting the above values:

$(-283.29)- ((-10.01)+49.44)$

$-283.29 - (-10.01+49.44)$

$-283.29 - (39.43) = -322.72$

C - (B + A) = -322.72

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How do I solve (|x-3|+5|x|) ?

11111nata11111 [884]

Answer:

x≤0 ⇒|x-3|+5|x| = -6x + 3

0≤x≤3 ⇒ |x-3|+5|x| = 4x + 3

x≥3 ⇒ |x-3|+5|x| = 6x - 3

Step-by-step explanation:

Check the picture below.

6 0

3 years ago

If you are offered one slice from a round pizza (in other words, a sector of a circle) and the slice must have a perimeter of 24

VARVARA [1.3K]

Answer:

A 12 in diameter will reward you with the largest slice of pizza.

Step-by-step explanation:

Let $r$ be the radius and $\theta$ be the angle of a circle.

According with the graph, the area of the sector is given by

$A=\frac{1}{2}r^2\theta$

The arc lenght of a circle with radius $r$ and angle $\theta$ is $r \theta$

The perimeter of the pizza slice is composed of two straight pieces, each of length r inches, and an arc of the circle which you know has length s = rθ inches. Thus the perimeter has length

$2r+r\theta=24 \:in$

We need to express the area as a function of one variable, to do this we use the above equation and we solve for $\theta$

$2r+r\theta=24\\r\theta=24-2r\\\theta=\frac{24-2r}{r}$

Next, we substitute this equation into the area equation

$A=\frac{1}{2}r^2(\frac{24-2r}{r})\\A=\frac{1}{2}r(24-2r)\\A=12r-r^2$

The domain of the area is

$0$

To find the diameter of pizza that will reward you with the largest slice you need to find the derivative of the area and set it equal to zero to find the critical points.

$\frac{d}{dr} A=\frac{d}{dr}(12r-r^2)\\A'(r)=\frac{d}{dr}(12r)-\frac{d}{dr}(r^2)\\A'(r)=12-2r$

$12-2r=0\\-2r=-12\\\frac{-2r}{-2}=\frac{-12}{-2}\\r=6$

To check if $r=6$ is a maximum we use the Second Derivative test

if $f'(c)=0$ and $f''(c)$ , then f(x) has a local maximum at x = c.

The second derivative is

$\frac{d}{dr} A'(r)=\frac{d}{dr} (12-2r)\\A''(r)=-2$

Because $A''(r)=-2$ the largest slice is when r = 6 in.

The diameter of the pizza is given by

$D=2r=2\cdot 6=12 \:in$

A 12 in diameter will reward you with the largest slice of pizza.

3 0

2 years ago

Oh need to know alot and a lot

o-na [289]

Answer:

yeah so it is 74882828 it is pretty easy

5 0

1 year ago

Look at the table below. Plot and label points M and N on the coordinate

marta [7]

M = (1,4)
N = (5,2)

For M, you simply move to 1 on the x axis abs for 4 you move on the y axis. Same does for N, for 5 you move on the x axis and for 2 you move on the y axis.

6 0

3 years ago

PLZ ANSWER QUICK!!!

alexira [117]

Answer; c^2 = a^2 + b^2

c = 17 inches

a = 8 inches

b^2 = c^2 – a^2

b^2 = 17^2 – 8^2

b^2 = 289 – 64

b^2 = 225

b = square root of 225

b = 15 inches ≈ c.

5 0

3 years ago