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3241004551 [841]

2 years ago

8

What is the value of the expression below? 2[32-(4-1)^3]

1 answer:

butalik [34]2 years ago

8 0

The value of the expression is 10.

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55÷11 7×(2 14) show work pls

BlackZzzverrR [31]

55/11 7*(2 14) 2 times 14 is 28
=55/11 7*28 55/11=5 7 times 28 = 196
=5*196=980

7 0

2 years ago

ASAP 30 + Brainliest <br><br> Please only solve 2 - 5

hichkok12 [17]

<u>QUESTION 2a</u>

We want to find the area of the given right angle triangle.

We use the formula

$Area=\frac{1}{2}\times base\times height$

The height of the triangle is $=a cm$ .

The base is $12cm$ .

We substitute the given values to obtain,

$Area=\frac{1}{2}\times 12\times a cm^2$ .

This simplifies to get an expression for the area to be

$Area=6a cm^2$ .

<u>QUESTION 2b</u>

The given diagram is a rectangle.

The area of a rectangle is given by the formula

$Area=length \times width$

The length of the rectangle is $l=7cm$ and the width of the rectangle is $w=ycm$ .

We substitute the values to obtain the area to be

$Area=7 \times y$

The expression for the area is

$Area=7y$

<u>QUESTION 2c.</u>

The given diagram is a rectangle.

The area of a rectangle is given by the formula

$Area=length \times width$

The length of the rectangle is $l=2x cm$ and the width of the rectangle is $w=4 cm$ .

We substitute the values to obtain the area to be

$Area=2x \times 4$

The expression for the area is

$Area=8x$

<u>QUESTION 2d</u>

The given diagram is a square.

The area of a square is given by,

$Area=l^2$ .

where $l=b m$ is the length of one side.

The expression for the area is

$Area=b^2 m^2$

<u>QUESTION 2e</u>

The given diagram is an isosceles triangle.

The area of this triangle can be found using the formula,

$Area=\frac{1}{2}\times base\times height$ .

The height of the triangle is $4cm$ .

The base of the triangle is $6a cm$ .

The expression for the area is

$Area=\frac{1}{2}\times 6a \times 4cm^2$

$Area=12a cm^2$

<u>QUESTION 3a</u>

Perimeter is the distance around the figure.

Let P be the perimeter, then

$P=x+x+x+x$

The expression for the perimeter is

$P=4x mm$

<u>QUESTION 3b</u>

The given figure is a rectangle.

Let P, be the perimeter of the given figure.

$P=L+B+L+B$

This simplifies to

$P=2L+2B$

Or

$P=2(L+B)$

<u>QUESTION 3c</u>

The given figure is a parallelogram.

Perimeter is the distance around the parallelogram

$Perimeter=3q+P+3q+P$

This simplifies to,

$Perimeter=6q+2P$

Or

$Perimeter=2(3q+P)$

<u>QUESTION 3d</u>

The given figure is a rhombus.

The perimeter is the distance around the whole figure.

Let P be the perimeter. Then

$P=5b+5b+5b+5b$

This simplifies to,

$P=20b mm$

<u>QUESTION 3e</u>

The given figure is an equilateral triangle.

The perimeter is the distance around this triangle.

Let P be the perimeter, then,

$P=2x+2x+2x$

We simplify to get,

$P=6x mm$

QUESTION 3f

The figure is an isosceles triangle so two sides are equal.

We add all the distance around the triangle to find the perimeter.

This implies that,

$Perimeter=3m+5m+5m$

$Perimeter=13m mm$

<u>QUESTION 3g</u>

The given figure is a scalene triangle.

The perimeter is the distance around the given triangle.

Let P be the perimeter. Then

$P=(3x+1)+(2x-1)+(4x+5)$

This simplifies to give us,

$P=3x+2x+4x+5-1+1$

$P=9x+5$

<u>QUESTION 3h</u>

The given figure is a trapezium.

The perimeter is the distance around the whole trapezium.

Let P be the perimeter.

Then,

$P=m+(n-1)+(2m-3)+(n+3)$

We group like terms to get,

$P=m+2m+n+n-3+3-1$

We simplify to get,

$P=3m+2n-1$ mm

QUESTION 3i

The figure is an isosceles triangle.

We add all the distance around the figure to obtain the perimeter.

Let $P$ be the perimeter.

Then $P=(2a-b)+(a+2b)+(a+2b)$

We regroup the terms to get,

$P=2a+a+a-b+2b+2b$

This will simplify to give us the expression for the perimeter to be

$P=4a+3b$ mm.

QUESTION 4a

The given figure is a square.

The area of a square is given by the formula;

$Area=l^2$

where $l=2m$ is the length of one side of the square.

We substitute this value to obtain;

$Area=(2m)^2$

This simplifies to give the expression of the area to be,

$Area=4m^2$

QUESTION 4b

The given figure is a rectangle.

The formula for finding the area of a rectangle is

$Area=l\times w$ .

where $l=5a cm$ is the length of the rectangle and $w=6cm$ is the width of the rectangle.

We substitute the values into the formula to get,

$Area =5a \times 6$

$Area =30a cm^2$

QUESTION 4c

The given figure is a rectangle.

The formula for finding the area of a rectangle is

$Area=l\times w$ .

where $l=7y cm$ is the length of the rectangle and $w=2x cm$ is the width of the rectangle.

We substitute the values into the formula to get,

$Area =7y \times 2x$

The expression for the area is

$Area =14xy cm^2$

QUESTION 4d

The given figure is a rectangle.

The formula for finding the area of a rectangle is

$Area=l\times w$ .

where $l=3p cm$ is the length of the rectangle and $w=p cm$ is the width of the rectangle.

We substitute the values into the formula to get,

$Area =3p \times p$

The expression for the area is

$Area =3p^2 cm^2$

See attachment for the continuation

6 0

2 years ago

Read 2 more answers

Miguel is making an obstacle course for field day. At every 1/6 of the course there is a tire. At every 1/3 of the course there

Y_Kistochka [10]

Locations can be seen in the attachment below. Answers: 1/3, 1/2, 2/3, 1.

4 0

2 years ago

when the solutions to each of the two equations below are graphed in the xy-coordinate plane, the graphs intersect at two places

Neporo4naja [7]

The answer is -2 and 6

8 0

2 years ago

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ASAP Can someone help with this problems just I only need which formula do I used I just need for 4,5 and 6 just tell me the for

Sveta_85 [38]

Explanation:

The formula for law of cosines is:

$a^{2} = b^{2} + c^{2} - 2bc(cosA)$

This is written at the top of your paper.

The angle is substituted where the capital A is. The side opposite to the angle is the angle name's lowercase. The two other sides are b and c.

Substitute all the information you know, then isolate the variable that you do not know.

When your formula is rearranged to look like this:

$\frac{a^{2} - b^{2} - c^{2}}{- 2bc} = (cosA)$

Solve the left side and punch into your calculator:

cos⁻¹(left side) to find the angle A.

3 0

2 years ago