20=v+9-16 20=v+9+ ( ) 20=v- =v

PLEASE HELP ASAP WILL MARK THE BRAINLEST

statuscvo [17]

Answer:

similar triangles

Step-by-step explanation:

First of all, what are similar shapes? Well, two shapes are similar if you can turn one into the other by moving, rotating, flipping, or scaling. That means you can make one shape bigger or smaller. In this case, we know that triangles ABC and DEF are mathematically similar. The area of triangles ABC is , so we need to know the area of triangle DEF.

From math, let's call the scaling factor, so we know that for any similar figures, the ratio of the areas of any are in proportion to . In other words, if is the area of triangle ABC, and is the area of triangle DEF, then we can write the following relationship:

4 0

2 years ago

Evaluate csc(-315°).

Maru [420]

Answer:

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

Step-by-step explanation:

Notice from the attached image that -315 degrees is the same as 45 degrees.

Recall as well that the csc function is defined as the reciprocal of the sine function: $csc(x)=\frac{1}{sin(x)}$

Then, you need to evaluate $\frac{1}{sin(45)} = 1/\frac{\sqrt{2} }{2} = \frac{2}{\sqrt{2} } =\sqrt{2}$

8 0

2 years ago

Can someone help me with this question?

Crazy boy [7]

Answer:

m<S' = 115

Step-by-step explanation:

A translation is when one shifts the image over on a coordinate plane. A translation doesn't affect the image itself. Hence m<S is the same as m<S'. Therefore m<S' = 115

4 0

3 years ago

Read 2 more answers

I need help asap!!!! Please It’s an Algebra 2 question

Elza [17]

Answer:

8) Real; Rational

9) Real; Rational; Integer; Whole; Natural

10) Real; Rational

Step-by-step explanation:

Real numbers consists of the subsets:

Irrational Numbers:

Irrational numbers is the subset complementary to rational numbers. It includes numbers such as √(2) or π. These numbers do not repeat nor terminate.

Rational Numbers:

Rational numbers are all the others. They include the integers as well as the repeating and terminating decimals.

Integers:

Integers include the entire set of whole numbers and the negative numbers. This subset does not include decimals nor fractions. Thus, the integers are: ...-3, -2, -1, 0, 1, 2, 3... etc.

Whole Numbers:

Whole numbers is the set of natural numbers that includes 0. So, whole numbers are: 0, 1, 2, 3... and so on.

Natural Numbers:

Natural numbers or the counting numbers are all the positive, non-decimal numbers. These are 1, 2, 3... and so on. This set does not include 0.

8) $1.\overline{3}=1.3333...$

The given number is not natural, whole, nor an integer because it is a decimal. It repeats, so it belongs to the rational numbers subset.

9) $\sqrt4=2$

The square root of 4 is simply 2. 2 is a natural number. Because of this, this means that 2 is also a whole number, integer, and a rational number.

10) $-3/4=-0.75$

The given number is a fraction and it does not reduce to a whole number, so it's not a natural number, whole number, nor an integer. It does terminate, so it belongs to the rational numbers subset.

7 0

3 years ago

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

2 years ago

20=v+9-16 20=v+9+ ( )20=v- =v