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2 years ago

Simplify. −5x = 50 A. 10 B. −10 C. 250 D. −250

Mathematics

2 answers:

Anika [276]2 years ago

7 0

The answer to your question is “B”

Xelga [282]2 years ago

6 0

The answer is B because -5 times -10 makes a positive 50.

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Complete the statement to describe the expression a b c + d e f abc+defa, b, c, plus, d, e, f. the expression consists of terms,

DedPeter [7]

The statement of expressions is two terms and three factors.

<h3>The statement and expressions:</h3>

According to the question, the statement that teaches us about expression is made up of two terms, each of which has three factors.

The ABC and DEF expressions define the expression of two terms, each of which comprises two elements.

As a result, the answer is two terms and three factors.

The terms of the expressions are connected by mathematical signs like addition and subtraction.
Variables are combined using multiplication signs to form the factor.
The three components are found in each word.
Looking at the first term ABC, we can see that it has three variables, a, b, and c, which is multiplied together.

Therefore, the answer is two terms and three factors.

Know more about an expression here:

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1 year ago

Mike had some candy to give to his four

zimovet [89]

10 pieces there was five people each got two pieces

6 0

2 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=prove%20that%5C%20%20%5Ctextless%20%5C%20br%20%2F%5C%20%20%5Ctextgreater%20%5C%20%5Cfrac%20%7B

inysia [295]

$\large \bigstar \frak{ } \large\underline{\sf{Solution-}}$

Consider, LHS

$\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered} \\ \\ \text{So, using this identity, we get} \\ \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered} \\$

So, using this identity, we get

$\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

can be rewritten as

$\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}$

<h2>Hence,</h2>

$\begin{gathered} \\ \rm\implies \:\boxed{\sf{ \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}$

$\rule{190pt}{2pt}$

5 0

2 years ago

How do you find the complete square of a math problem

dolphi86 [110]

Step-by-step explanation:

I am not sure what exactly you mean.

do you mean the complete square of an expression or

term ?

if so, then by multiplying this term by itself, and that means in general, every part is multiplied by every part and the part results are added considering the signs involved.

e.g.

squaring a+b

(a+b)(a+b) = a×a + a×b + b×a + b×b = a² + 2ab + b²

remember that multiplication and addition are commutative (you can flip the right and left sides with each other and still get the same result : a+b = b+a, a×b = b×a).

squaring a-b

(a-b)(a-b) = a×a + a×-b + -b×a + -b×-b = a² - 2ab + b²

remember that

+×- = -×+ = -

-×- = +

+×+ = +

a more complex example ?

squaring a-b+c

(a-b+c)(a-b+c) =

= a×a + a×-b + a×c + -b×a + -b×-b + -b×c + c×a + c×-b + c×c =

= a² - 2ab - 2bc + 2ac + b² + c²

6 0

2 years ago

Tehtävä 44<br><br>b) 12,2,10,4, ,

Tpy6a [65]

I am not sure it is 2i think

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2 years ago