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3 years ago

DOES SOMEONE KNOW THISSS!!!

Mathematics

2 answers:

Inessa [10]3 years ago

5 0

4 and 5
hope this helped in some way..

Novay_Z [31]3 years ago

4 0

Answer:

Step-by-step explanation:

$a$

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6x+6=-5+3x+17<br><br> I need help

yanalaym [24]

Answer:

Get all the terms with x on the left. Get all the terms without x on the right. Simplify by adding and subtracting the terms without x and adding and subtracting the terms with x. Finally divide to get the value of x.

Step-by-step explanation:

6x+6 =- 5+3x+17

6x+6 -6 =- 5+3x+17 -6

6x = 5+3x+11

6x = 3x + 16

6x - 3x = 3x + 16 -3x

3x = 16

3x / 3 = 16 / 3

x = 15/3 + 1/3

x = 5 1/3

x = 5.33333...

4 0

2 years ago

Evaluate the expression. 5^2−4⋅6+11 Enter your answer in the box.

mart [117]

Explanation:

This question's it's about the order of operations. P-parenthesis, E-exponents, M-multiply, D-divide, A-add, and S-subtract. It can also go left to right.

Do exponents first.

$5^2=5*5=25$

$25-4*6+11$

Then, multiply left to right.

$4*6=24$

$25-24+11$

And add/subtract left to right.

$25-24=1$

$1+11=12$

$\boxed{=12}$

Hope this helps!

Thanks!

Have a great day!

6 0

2 years ago

Whoever answers first can get brainliest!

Andrew [12]

Answer:

Step-by-step explanation:

8 0

2 years ago

50+51+52+...+101=...<br>Can you help me? I do not understand

Dmitriy789 [7]

Suppose

$S=50+51+52+\cdots+100+101$

At the same time, we can write

$S^*=101+100+99+\cdots+51+50$

Note that $S=S^*$ (just reverse the sum). Let's pair the first terms of $S$ and $S^*$ , and the second, and the third, and so on:

$S+S^*=(50+101)+(51+100)+(52+99)+\cdots+(100+51)+(101+50)$

Now, each grouped term in the sum on the right side adds to 151. There are 52 grouped terms on that same side (because there are 50 numbers in the range of integers 51-100, plus 50 and 101), which menas

$S+S^*=52\cdot151$