All categories

3 years ago

What does x equal 10x-5=65

Mathematics

2 answers:

Lynna [10]3 years ago

7 0

Step-by-step explanation:

Add 5 to both sides to get 10x alone: 10x = 70
Divide both sides by ten to get x alone: x = 7
The final answer is x = 7.

alukav5142 [94]3 years ago

7 0

Step-by-step explanation:

$10x - 5 = 65 \\ \\ \therefore \: 10x = 65 + 5 \\ \\ \therefore \: 10x = 70 \\ \\ \therefore \:x = \frac{70}{10} \\ \\ \huge \red{ \boxed{ \therefore \:x = 7}}$

You might be interested in

You roll a 20-sided die and a 6-sided die. What is the probability that the sum of the dice is less than 5?

postnew [5]

Answer:

8 in 26 i think.

Step-by-step explanation:

im pretty sure if it is supposed to be less than five then you need to add the number of digits and those are the odds. pretty sure.

6 0

3 years ago

Write the equation for the line that passes through (-6, 1) and is perpendicular to

Dahasolnce [82]

Answer

(3,-2)

Explanation step by step

y= -2x-9

= -2(-6,1)-9

= (12,-2)-9

=(3,-2)

8 0

3 years ago

Simplify w1/2 and w1/3

mars1129 [50]

Answer:

They are on their simplified form.

Step-by-step explanation:

Since these numbers are prime numbers we can't simplify them. But we can change them to decimal.

1/2=0.5

1/3=0.3 (in which 3 repeats itself)

Hope this helps ;) ❤❤❤

4 0

3 years ago

X to the power of 2+ 2X plus 3X to the power of 2+ 4X subtract 2X to the power of 2+ X

eimsori [14]

Answer:

4x+12

Step-by-step explanation:

$2x + 2 \times { }^{2} + 6x + 12 x ^{2} - 4x + - 2x ^{2}$

8 0

4 years ago

three bags of sweets weigh 27/4 kg. two of them have the same weight and the third bag is heavier than each of the bags of equal

Nana76 [90]

I'm here buddy,

so, let's take the value of the two bags with equal weight as x.

= x + x + (x + $\frac{6}{5}$ ) = $\frac{27}{4}$

= 3x + $\frac{6}{5}$ = $\frac{27}{4}$

= 3x = $\frac{27}{4}$ - $\frac{6}{5}$

( let's take the LCM of 4 and 5 = 20

= 3x = $\frac{135}{20}$ - $\frac{24}{20}$

= 3x = $\frac{111}{20}$

= x = $\frac{111}{20}$ ÷ $\frac{3}{1}$ = $\frac{111}{20}$ × $\frac{1}{3}$ = $\frac{37}{20}$

So, the weight of the equal bags are $\frac{37}{20}$ and the weight of the third bag ( heavy one ) is $\frac{37}{20}$ + $\frac{6}{5}$ = $\frac{37}{20}$ + $\frac{24}{20}$ = $\frac{61}{20}$

1st bag = $\frac{37}{20}$ kg

2nd bag = $\frac{37}{20}$ kg

3rd bag = $\frac{61}{20}$ kg

Hope it helps...

7 0

3 years ago