Ten times the sum of four and three

3(x+4) - (x-1)<br> x=3 y=2

Ksenya-84 [330]

Answer: 19

Step-by-step explanation:

7 0

2 years ago

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22. Find the perimeter and area.

zmey [24]

Answer:

Part 22) The area is $A=15a^3b^6\ units^2($ and the perimeter is $P=10a^2b^4+6ab^2\ unit$

Part 24) The area is $A=16m^3n\ units^2$ and the perimeter is $P=24mn\ units$

Part 26) The area is equal to $A=9\pi x^6y^{2}\ units^2$

Step-by-step explanation:

Part 22) Find the perimeter and area

step 1

The area of a rectangle is equal to

$A=LW$

we have

$L=5(a^2)(b^4)\ units$

$W=3(a)(b^2)\ units$

Remember that

When multiply exponents with the same base, adds the exponents and maintain the base

substitute in the formula

$A=(5(a^2)(b^4))(3(a)(b^2))$

$A=15a^3b^6\ units^2$

step 2

The perimeter of a rectangle is equal to

$P=2(L+W)$

we have

$L=5(a^2)(b^4)\ units$

$W=3(a)(b^2)\ units$

substitute in the formula

$P=2(5(a^2)(b^4)+3(a)(b^2))$

$P=10a^2b^4+6ab^2\ unit$

Part 24) Find the perimeter and area

step 1

The area of triangle is equal to

$A=\frac{1}{2}bh$

where

$b=8mn\ units$

$h=4m^2\ units$

Remember that

When multiply exponents with the same base, adds the exponents and maintain the base

substitute the given values

$A=\frac{1}{2}(8mn)(4m^2)$

$A=16m^3n\ units^2$

step 2

Find the perimeter

I will assume that is an equilateral triangle (has three equal length sides)

The perimeter of an equilateral triangle is

$P=3b$

where

$b=8mn\ units$

substitute

$P=3(8mn)$

$P=24mn\ units$

Part 26) Find the area

The area of a circle is equal to

$A=\pi r^{2}$

where

$r=3x^3y\ units$

Remember the property

$(a^{m})^{n}=a^{m*n}$

substitute in the formula the given value

$A=\pi (3x^3y)^{2}$

$A=9\pi x^6y^{2}\ units^2$

6 0

3 years ago

Liam is a tyre fitter.

maks197457 [2]

Answer:

3 tires

Step-by-step explanation:

56 minutes divided by 4 is 14. 42 minutes divided by 14 is 3 :)

7 0

2 years ago

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The can of tomatoes is three times as tall but one-third as wide as the can of tuna. What is true about the areas of their label

yulyashka [42]

Problem One
Call the radius of the second can = r
Call the height of the second can = h

Then the radius of the first can = 1/3 r
The height of the first can = 3*h

A1 / A2 = (2*pi*(1/3r)*(3h)] / [2*pi * r * h]

Here's what will cancel. The twos on the right will cancel. The 3 and 1/3 will multiply to one. The 2 r's will cancel. The h's will cancel. Finally, the pis will cancel

Result A1 / A2 = 1/1
The labels will be shaped differently, but they will occupy the same area.

Problem Two
It seems like the writer of the problem put some lids on the new solid that were not implied by the question.

If I understand the problem correctly, looking at it from the top you are sweeping out a circle for the lid on top and bottom, plus the center core of the cylinder.

One lid would be pi r^2 = pi w^2 and so 2 of them would be 2 pi w^2
The region between the lids would be 2 pi r h for the surface area which is 2pi w h

Put the 2 regions together and you get
Area = 2 pi w^2 + 2 pi w h

Answer: Upper left corner <<<<< Answer

8 0

3 years ago

Angles C and E are complementary angles. Angle C measures (4× + 50). Angle E measures 28 degrees. Solve for x.

julsineya [31]

Answer:

x = 3

Step-by-step explanation:

4x + 50 + 28 = 90

4x + 78 = 90

4x = 12

x = 3

7 0

3 years ago

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