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2 years ago

With solution needed please!!

Mathematics

2 answers:

Alex Ar [27]2 years ago

6 0

Answer:

x < 2

Step-by-step explanation:

-x + 6 > x + 2

2x < 4

x < 2

Alika [10]2 years ago

3 0

Answer: 2 > x

Step-by-step explanation:

Add x on both sides to get 6 > 2x + 2

Minus 2 on both sides to get 4 > 2x

Divide 2 on both sides to get 2 > x

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Evaluate 3a for a = 5

zhannawk [14.2K]

Answer:

Hello There!!

Step-by-step explanation:

The answer is 15 as 3a=3×a so 3×5=15.

hope this helps,have a great day!!

~Pinky~

7 0

3 years ago

What value of k makes the statement true? k + (–4.1) = –9.7

Ugo [173]

K + (-4.1) is the exact same as saying k- 4.1, so what you have to do now is put the -4.1 in the opposite side, but with opposite sign too (+)
so k= -9.7 + 4.1
so in order to answer it you have to deduct the number with the lowest value taking in count both as positive and the sign will be the one from the highest value taking both as positive
9.7
-4.1
____
5.6
so 9.7 had a negative sign so the answer to your question is
k = -5.6
i hope i helped you

8 0

3 years ago

To find the area of a ____ multiply half the sum of the bases by the height

Oksana_A [137]

To find the area of a triangle multiply half the sum of the bases by its height, i think?

4 0

4 years ago

Read 2 more answers

❊ Simplify :

DiKsa [7]

Answer:

See Below.

Step-by-step explanation:

Problem 1)

We want to simplify:

$\displaystyle \frac{a+2}{a^2+a-2}+\frac{3}{a^2-1}$

First, let's factor the denominators of each term. For the second term, we can use the difference of two squares. Hence:

$\displaystyle =\frac{a+2}{(a+2)(a-1)}+\frac{3}{(a+1)(a-1)}$

Now, create a common denominator. To do this, we can multiply the first term by (a + 1) and the second term by (a + 2). Hence:

$\displaystyle =\frac{(a+2)(a+1)}{(a+2)(a-1)(a+1)}+\frac{3(a+2)}{(a+2)(a-1)(a+1)}$

Add the fractions:

$\displaystyle =\frac{(a+2)(a+1)+3(a+2)}{(a+2)(a-1)(a+1)}$

Factor:

$\displaystyle =\frac{(a+2)((a+1)+3)}{(a+2)(a-1)(a+1)}$

Simplify:

$\displaystyle =\frac{a+4}{(a-1)(a+1)}$

We can expand. Therefore:

$\displaystyle =\frac{a+4}{a^2-1}$

Problem 2)

We want to simplify:

$\displaystyle \frac{1}{(a-b)(b-c)}+\frac{1}{(c-b)(a-c)}$

Again, let's create a common denominator. First, let's factor out a negative from the second term:

$\displaystyle \begin{aligned} \displaystyle &= \frac{1}{(a-b)(b-c)}+\frac{1}{(-(b-c))(a-c)}\\\\&=\displaystyle \frac{1}{(a-b)(b-c)}-\frac{1}{(b-c)(a-c)}\\\end{aligned}$