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3 years ago

Find the area of each figure. Round to the nearest tenth if necessary.

Mathematics

1 answer:

Flauer [41]3 years ago

7 0

Answer:

Hi there!

The correct answer is: 220.5 cm²

Step-by-step explanation:

This is an example of a composite figure where you have to break up the figure into different shapes. In this case this figure, can break up into a semicircle and a triangle.

Triangle: A= 1/2bh A=1/2(16)(15) = 120 cm²

Semicircle: another way of saying half a circle so A= 1/2 πr²

A=(1/2)(π)(8)² = 100.5 cm²

And then you just add the two areas together and you get 220.53 cm²

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Als 3. Determine ir 12:16 and d 72:96 are equivalent ratios.

Gnoma [55]

Answer:

YES

Step-by-step explanation:

I'll start with 72/12

= 6

6 is does not have any decimals,

Therefore YES

5 0

2 years ago

Solve the quadratic equation: 2(x-3)^2-24-0

Keith_Richards [23]

Answer:

x=3+2×sqrt(3)

x=3-2×sqrt(3)

Step-by-step explanation:

2(x-3)²-24=0

2(x-3)²=24

(x-3)²=12

Case 1:

x-3=sqrt(12)=2×sqrt(3)

x=3+2×sqrt(3)

Case 2:

x-3=-sqrt(12)=-2×sqrt(3)

x=3-2×sqrt(3)

7 0

3 years ago

The function f(x) = -5x2 + 13x + 6 represents the height, in meters, of a

AnnZ [28]

Answer:

Step-by-step explanation:

Differentiate the function and set it to zero:

f'(x) = -10x + 13

0 = -10x + 13

x = 1.3 seconds

Plug the x value into f(x):

f(1.3) = -5(1.3)^2 + 13(1.3) + 6 = 14.45 meters

8 0

3 years ago

The scatter plot shows a correlation between the years and the rainfall in centimeters in Tennessee.

nataly862011 [7]

Answer:

Step-by-step explanation:

The actual values are shown on the given graph as blue points.

The line of regression is shown on the given graph as the red line.

From inspection of the graph, in the year 2000 the actual rainfall was 43 cm, shown by point (2000, 43). It appears that the regression line is at y = 50 when x is the year 2000.

⇒ Difference = 50 - 43 = 7 cm

In 2000, the actual rainfall was 7 centimeters below what the model predicts.

From inspection of the graph, in the year 2003 the actual rainfall was 44 cm, shown by point (2003, 40). It appears that the regression line is at y = 40 when x is the year 2003.

⇒ Difference = 44 - 40 = 4 cm

In 2003, the actual rainfall was 4 centimeters above what the model predicts.

3 0

2 years ago

Y=3xy-1, Solve for y

Ket [755]

Answer:

y = $\frac{1}{3x-1}$

Step-by-step explanation:

Given

y = 3xy - 1

Collect terms in y on the left side by subtracting 3xy from both sides

y - 3xy = - 1 ← factor out y on the left side

y(1 - 3x) = - 1 ( divide both sides by 1 - 3x

y = $\frac{-1}{1-3x}$ ← multiply numerator/denominator by - 1

y = $\frac{1}{3x-1}$

3 0

4 years ago