All categories

3 years ago

1/5(2x-10)+4x=-3(1/5x+4)

2 answers:

4 0

The question gives us the equation $\frac{1}{5}(2x-10)+4x=-3(\frac{1}{5}x+4)$ and are told to solve for the variable $x$ . To do this, we need to simplify the equation by using the distributive property. This gives us $\frac{2}{5}x-2+4x=-\frac{3}{5}-12$ . Now, we combine like terms. When we do this, we get $\frac{22}{5}x-2=-\frac{3}{5}x-12$ . Next, we add 12 to both sides. This gives us $\frac{22}{5}x+10=-\frac{3}{5}x$ . Then, we subtract \frac{22}{5} from both sides and simplify. This gives us $10=-5x$ . Now, we divide both sides by $-5$ to get x = -2. Therefore, the answer to this problem is $\boxed{x=-2}$ . Hope this helped!

borishaifa [10]3 years ago

3 0

First solve to your equation by step by step.

1/5(2x-10)+4x=-3(1/5x+4)
= x=-2

Answer is x=-2

It will help you.

You might be interested in

Can Somebody Help With Geometry?

Cerrena [4.2K]

Answer:

SAS

Step-by-step explanation:

II side (given)

Angles between the congruent sides (vertically opposite angle)

III side (given)

Hence by, SAS the Triangles are congruent.

7 0

3 years ago

Mark is playing baseball after school. He wants to find the total distance he will need to ride his bike from school to the park

Ymorist [56]

Use the distance formula (derived from Pythagorean Theorem:
$d = \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} } \\ = \sqrt{ {(10 - 8)}^{2} + {(11 - 7)}^{2} } \\ = \sqrt{ {(2)}^{2} + {(4)}^{2} } = \sqrt{4 + 16} \\ d = \sqrt{20}$
$d = \sqrt{20} = \sqrt{4} \times \sqrt{5} = 2 \sqrt{5}$

4 0

3 years ago

Read 2 more answers

a term in an arithmetic sequence is 488. the first term of the sequence is 7 and the common difference is 13. find the term numb

lisabon 2012 [21]

Answer: the term number is 38

Step-by-step explanation:

Let the number of the term be x

The value of the xth term = 488

In an arithmetic sequence, the terms differ by a common difference, d. This means that the difference between two consecutive terms, d is constant.

The formula for the nth term is

Tn = a + (n-1)d

Where

Tn = the nth term of the arithmetic sequence

a = the first term of the arithmetic sequence.

d = common difference.

From the information given,

a = 7

d = 13

We are looking for the xth term.

Tx = 488 = 7 + (x-1)13

488 = 7 + 13x - 13

Collecting like terms on the left hand side and right hand side of the equation,

13x = 488 -7 + 13

13x = 494

x = 38

The value of the 38th term is 488.

5 0

4 years ago

Is it possible to solve a quadratic equation that is not factorable over the set of integers? Explain.

zvonat [6]

Answer:

Yes

Step-by-step explanation:

Yes it is possible to solve a quadratic equation that is not factorable over the set of integers.

The solution may vary like Integers, rationals, irrationals or complex solutions.

To find two roots of the equation we can always use the formula given below to solve a quadratic equation,

For the quadratic equation,

$ax^2+bx+c=0$ , we have,

$x= \frac{-b\pm \sqrt{b^2-4 ac}}{2a}$

If the discriminant is greater than $b^2$ , we get complex roots.

8 0

4 years ago

If f(x)=x/3-2 and g(x)=3x^2+2x-6, find (f+g)(x)

iren2701 [21]

Use cymath it might gives you the answer and shows you how to do it step by step.

6 0

3 years ago