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3 years ago

8

What is the solution to the following 21+x=13

1 answer:

Hitman42 [59]3 years ago

7 0

Answer:x=-8

Step-by-step explanation:subtract 21 from both sides then simplify 31-21 to -8

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Find angle B if triangle ABC below is isosceles.

crimeas [40]

2x-20 = x+8

x-20 =8

x = 28

28+8 =36

angle B = 36 degrees

answer is b

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3 years ago

351 students in 27 classes _____ students in each class

77julia77 [94]

Answer:

13

Step-by-step explanation:

351 divided by 27 = 13

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2 years ago

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2 years ago

Consider the sequence 8, 14, 20, 26, . . .. (a) What is the next term in the sequence? 2.2 Exercises 98 (b) Find a formula for t

omeli [17]

Answer:

Step-by-step explanation:

Each term of the given arithmetic sequence is 6 more than the previous one. Thus, the next term is 26+6, or 32.

a(n) = 8 + 6(n-1)

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3 years ago

A figure is composed of a rectangle and a right triangle. What is the area of the figure?

strojnjashka [21]

QUESTION 1

The figure is a tra-pezoid.

The area of a trapezoid is given by the formula;

$A=\frac{1}{2}(Sum\:of\:parallel\:sides)\times h$

We substitute the values to obtain;

$Area=\frac{1}{2}(13.2cm+8.4cm)\times 6cm$

$Area=\frac{1}{2}(21.6cm)\times 6cm$

$Area=64.8cm^2$

QUESTION 2

The approximate length of the ribbon is equal to the circumference of the circular table cloth.

We can find this by using the formula for calculating the circumference of a circle.

$C=2\pi r$

where $r=3.5ft$ is the radius of the circular table cloth.

We substitute this value and $\pi=3.14$ into the formula to get;

$C=2(3.14)(3.5)ft$

$C=21.98ft$

The approximate length of the ribbon is 22.0ft to the nearest tenth.

QUESTION 3

Type of quadrilateral:Rectangle

Explanation: A rectangle has 4 angles that are right angles.

The two pairs of opposite sides of a rectangle are also congruent.

QUESTION 4.

The circumference of a semi circle is calculated using the formula;

$C=\pi r$

where $r=12.5cm$ is the radius of the semicircle.

$C=3.14\times 12.5cm$

$C=39.25cm$

QUESTION 5

The area of the entire figure is the area of the semicircle plus the area of the isosceles triangle.

$Area=\frac{1}{2}\pi r^2+\frac{1}{2}bh$

$Area=\frac{1}{2}\times3.14\times12.5^2+\frac{1}{2}\times25\times 24$

$Area=345.3125+300$

$Area=645.3125cm^2$

$Area\approx645cm^2$

QUESTION 6

The area of the parallelogram is given by the formula;

$A=bh$

From the diagram, $b=y\:in.,h=3in.$ .

Given, A=23.7 inches squared.

We substitute the values to obtain;

$23.7=3y$

$y=\frac{23.7}{3}$

$y=7.9in.$

QUESTION 7

The area of a rectangle is given by the formula;

$Area=l\times b$

The area of the bigger rectangle $=9\times15=135in^2$

The area of the smaller rectangle $=8\times 13=104in^2$

The area of the shaded region $=135-104=31in^2$

5 0

3 years ago