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3 years ago

What is the area of the composite figure

Mathematics

1 answer:

lara [203]3 years ago

4 0

Answer:

Area of composite figure = 216 cm²

Hence, option A is correct.

Step-by-step explanation:

The composite figure consists of two figures.

1) Rectangle

2) Right-angled Triangle

We need to determine the area of the composite figure, so we need to find the area of an individual figure.

Determining the area of the rectangle:

Given

Length l = 14 cm

Width w = 12 cm

Using the formula to determine the area of the rectangle:

A = wl

substituting l = 14 and w = 12

A = (12)(14)

A = 168 cm²

Determining the area of the right-triangle:

Given

Base b = 8 cm

Height h = 12 cm

Using the formula to determine the area of the right-triangle:

A = 1/2 × b × h

A = 1/2 × 8 × 12

A = 4 × 12

A = 48 cm²

Thus, the area of the figure is:

Area of composite figure = Rectangle Area + Right-triangle Area

= 168 cm² + 48 cm²

= 216 cm²

Therefore,

Area of composite figure = 216 cm²

Hence, option A is correct.

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What is the 38th term of 459,450,441,..

galina1969 [7]

Answer:

The 38th term of 459,450,441,.. will be:

$a_{13}=351$

Step-by-step explanation:

Given the sequence

$459,450,441,..$

An arithmetic sequence has a constant difference 'd' and is defined by

$a_n=a_1+\left(n-1\right)d$

computing the differences of all the adjacent terms

$450-459=-9,\:\quad \:441-450=-9$

$d=-9$

The first element of the sequence is

$a_1=459$

so the nth term will be

$a_n=-9\left(n-1\right)+459$

$a_n=-9n+468$

Putting n=38 to find the 38th term

$a_n=-9n+468$

$a_{13}=-9\left(13\right)+468$

$a_{13}=-117+468$

$a_{13}=351$

Therefore, the 38th term of 459,450,441,.. will be:

$a_{13}=351$

3 0

3 years ago

line AB passes through points A(-6,6) and B(12,3) if the equation of the line is written in slope intercept form

Paul [167]

Y=(-1/6)x+5
m=(y2-y1)/(x2-x1)
m=(3-6)/(12-(-6))
m=(-3)/(18)
m=-1/6
y=mx+b
y=(-1/6)x+b
6=(-1/6)(-6)+b
6=1+b
b=5
y=(-1/6)x+5

7 0

3 years ago

2 dots plots with number lines going from 0 to 10. Plot A has 0 dots above 0, 1, and 2, 1 above 3, 2 above 4, 2 above 5, 2 above

alexandr402 [8]

Answer:

First one:

Both the mean and median are greater for Plot A than for Plot B

Step-by-step explanation:

Set A:

Mean:

[1×10 + 2×7 + 2×6 + 2×5 + 2×4 + 1×3]/10

= 5.7

Median:

Median position: (10+1)/2 = 5.5th value

(5+6)/2

Median = 5.5

Set B:

Mean:

[1×7 + 3×6 + 3×5 + 2×4 + 1×3]/10

= 5.1

Median:

Median position: (10+1)/2 = 5.5th value

(5+5)/2

Median = 5

Mean: A is greater

Median: A is greater

8 0

3 years ago

Please can someone help me ill mark brainliest

kramer

#5 is 4, #6 is graph A, and #7 is graph B. I hope this helped, Please mark as brainliest because I am 3 more away from the next level.

7 0

3 years ago

Show all work and reasoning

Natalija [7]

Split up the interval [2, 5] into $n$ equally spaced subintervals, then consider the value of $f(x)$ at the right endpoint of each subinterval.

The length of the interval is $5-2=3$ , so the length of each subinterval would be $\dfrac3n$ . This means the first rectangle's height would be taken to be $x^2$ when $x=2+\dfrac3n$ , so that the height is $\left(2+\dfrac3n\right)^2$ , and its base would have length $\dfrac{3k}n$ . So the area under $x^2$ over the first subinterval is $\left(2+\dfrac3n\right)^2\dfrac3n$ .

Continuing in this fashion, the area under $x^2$ over the $k$ th subinterval is approximated by $\left(2+\dfrac{3k}n\right)^2\dfrac{3k}n$ , and so the Riemann approximation to the definite integral is

$\displaystyle\sum_{k=1}^n\left(2+\frac{3k}n\right)^2\frac{3k}n$

and its value is given exactly by taking $n\to\infty$ . So the answer is D (and the value of the integral is exactly 39).

8 0

4 years ago