1+4=5, 2+7=12 8+11=?

Identify the terms. Then identify the coefficients of the variable terms of the expression. (Enter your answers as a comma-separ

sleet_krkn [62]

X is a term and 6 and 7 are coefficients

5 0

3 years ago

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Find the surface area. Use 3.14 for pi

Tatiana [17]

Answer:

S=1444.4

Step-by-step explanation:

they already gave the formula so

13 is the height, 10 is the radius

plug it in to the formula (using 3.14 for pi)

S=2(3.14)(10)^2+2(3.14)(10)(13)

S=628+816.4

S=1444.4

5 0

2 years ago

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7.Find the lengths of the missing sides in the triangle. If your answer is not an integer, leave it in simplest radical form. Th

Fed [463]

Here a right angled triangle given. one angle with measure $45^o$ given. The three sides of the triangle given 4, x, y.

We have to find, the sides which is opposite, adjacent and hypotenuse here.

We know that the side opposite to right angle is always hypotenuse. So, hypotenuyse = y.

The side adjacent to the given angle $45^o$ is x. So, here adjacent = x.

The opposite side is opposite to the given angle. So, opposite = 4.

Now we will use SOHCAHTOA that is $sin(x) =\frac{Opposite}{Hypotenuse} , cos(x) =\frac{Adjacent}{Hypotenuse} , tan(x) =\frac{Opposite}{Adjacent}$ , where x is the angle given.

To get x, we will use tan. So we will get,

$tan(45^o) = \frac{4}{x}$

We know the value of $tan(45^o) = 1$ . By substituting the value we will get,

$1 =\frac{4}{x}$

To find x, we have to move x here to the left side by multiplying it to both sides. We will get,

$(1)(x) = (\frac{4}{x}) (x)$

$x = 4$

So we have got the value of x here.

Now to find y, we will use the trigonometric function sine.

$sin(45^o) =\frac{4}{y}$

we know the value of $sin(45^o) =\frac{\sqrt{2}}{2}$

By substituting the value we will get,

$\frac{\sqrt{2}}{2} = \frac{4}{y}$

By cross multiplying we will get,

$(\sqrt{2}) (y) = (4)(2)$

$\sqrt{2}y = 8$

We will get y by dividing both sides by $\sqrt{2}$ , we will get,

$\frac{\sqrt{2}y}{\sqrt{2}} =\frac{8}{\sqrt{2} }$

$y =\frac{8}{\sqrt{2} }$

Now we will rationalize the denominator by multiplying $\sqrt{2}$ to the top and bottom.

$y =\frac{8\sqrt{2}}{(\sqrt{2})(\sqrt{2})}$

$y =\frac{8\sqrt{2}}{2}$

$y = 4\sqrt{2}$

So we have got the required values of x and y.

8 0

3 years ago

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For the class of 2013's prom Norman's dress shop sold cheaper dresses for $90 each and more expensive dresses for $140 each. The

Tema [17]

Answer:

Number of cheaper dresses sold is 35

Number of expensive dresses sold is 15

Step-by-step explanation:

Given:

Cost of cheaper dresses = $90

Cost of expensive dresses = $140

Total cost of the dresses = $5250

To Find:

Number of cheaper dress = ?

Number of expensive dress = ?

Solution:

Let

The number of cheaper dresses be x

The number of expensive dresses be y

(Number of cheaper dresses X cost of cheap dress) + (Number of Expensive dresses X cost of expensive dress) = $5250

$x \times90 +y \times 140 = 5250$ = $5250

It is given that the 20 more of the cheaper dresses than the expensive dresses is sold

So,

number of cheaper dress = 20 + number of expensive dress

x = 20 + y---------------------------------------(1)

$(20+y) \times90 +y \times 140 = 5250 = 5250$

$(20 \times 90 +y\times 90) +y \times 140= 5250$

$1800 + 90y+ 140y = 5250$

$1800 + 230y = 5250$

$230y =5250 -1800$

$230y = 3450$

$y = \frac{3450}{230}$

y = 15

Substituting y in (1)

x = 20 +15

x= 35

5 0

3 years ago

AABC has vertices at A(12, 8), B(4,8), and C(4, 14).

borishaifa [10]

Answer:

<u>Triangle ABC and triangle MNO</u> are congruent. A <u>Rotation</u> is a single rigid transformation that maps the two congruent triangles.

Step-by-step explanation:

ΔABC has vertices at A(12, 8), B(4,8), and C(4, 14).

length of AB = √[(12-4)² + (8-8)²] = 8

length of AC = √[(12-4)² + (8-14)²] = 10

length of CB = √[(4-4)² + (8-14)²] = 6

ΔMNO has vertices at M(4, 16), N(4,8), and O(-2,8).

length of MN = √[(4-4)² + (16-8)²] = 8

length of MO = √[(4+2)² + (16-8)²] = 10

length of NO = √[(4+2)² + (8-8)²] = 6

Therefore:

AB ≅ MN

AC ≅ MO

CB ≅ NO

and ΔABC ≅ ΔMNO by SSS postulate.

In the picture attached, both triangles are shown. It can be seen that counterclockwise rotation of ΔABC around vertex B would map ΔABC into the ΔMNO.

3 0

3 years ago