All categories

1 year ago

Solve for x in the following problem: 9 = 3/5x

Mathematics

2 answers:

mel-nik [20]1 year ago

7 0

Answer:

x = 15

Step-by-step explanation:

$\sf 9 = \dfrac{3}{5}x\\\\ \text{Multiply both sides by 5},\\\\9*5 = \dfrac{3}{5}*5*x\\\\ 9 *5 = 3x\\\\\text{Divide both sides by 3}\\\\\dfrac{9*5}{3}=x\\\\3*5=x\\\\\boxed{\bf x = 15}$

schepotkina [342]1 year ago

6 0

Answer:

dont know exactly what way ur studying to solve for it rn but heres a simple answer and if theres a diff process yall r using then ermmm sorry bud

x= 1/15

You might be interested in

Which statement about these equations is true?

Juliette [100K]

Answer:

Step-by-step explanation:

A:y=3x+4

7 0

2 years ago

Read 2 more answers

Pythagorean Theorem help please!

Sergio [31]

$14^2+48^2=c^2$
196+2304= $c^2$
2500= $c^2$
50 = c

8 0

3 years ago

Simplity:<br> 17 (-12 + 3)

Schach [20]

Answer:

$\Huge\boxed{\mathsf{-153}}}$

Step-by-step explanation:

$\Large\boxed{\mathsf{LESSON: ORDER \quad OF \quad OPERATIONS}}}$

$\Large\boxed{\mathsf{SUBJECT: MATH}}}$

To use the order of operations stands for (Parenthesis, Exponents, Multiply, Divide, Add, and Subtract) from left to right.

First, do parenthesis.

17(-12+3)

(-12+3)=-9

17(-9)

Then, multiply.

17*9=153

17(-9)=-153

-153

$\Large\boxed{\mathsf{\Rightarrow -153}}$

The correct answer is -153.

3 0

3 years ago

Three roots of the polynomial equation x4 - 4x3 - 2x2 + 12x + 9 = 0 are 3, -1 and -1. Explain why the fourth root must be a real

LiRa [457]

Descares Theorem says that the number of roots of a polynomial is equal to its highest exponent, and the roots could be real and/or imaginary (complex).

However is their is one complex root (a + i.b), it's conjugate is also
a root (a-i.b).

In the equation x⁴ - 4x³ - 2x² + 12x + 9 = 0, we have already 3 roots, then the 4rth cannot be a complex one since we don't have a fifth for its conjugate.

3 0

2 years ago

Write the inequality |x - 2| < 3 without absolute value symbols.

kotegsom [21]

$|x-2| < 3\Rightarrow x-2 < 3\ \wedge\ x-2 > -3\ \ \ \ |+2\\\\x < 5\ \wedge\ x > -1$

$x\in(-1,\ 5)$

4 0

2 years ago