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3 years ago

I GIVR BRAINLIST AND POINTS

Mathematics

1 answer:

PIT_PIT [208]3 years ago

4 0

Answer:

Step-by-step explanation:

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7(2x-5)=21 What is x?

Over [174]

Answer:

x = 4

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Step-by-step explanation:

Step 1: Define Equation

7(2x - 5) = 21

Step 2: Solve for x

Distribute 7: 2x - 5 = 3
Isolate x term: 2x = 8
Isolate x: x = 4

Step 3: Check

Plug in x into the original equation to verify it's a solution.

Substitute in x: 7(2(4) - 5) = 21
Multiply: 7(8 - 5) = 21
Subtract: 7(3) = 21
Multiply: 21 = 21

Here we see that 21 does indeed equal 21.

∴ x = 4 is the solution to the equation.

6 0

3 years ago

Jordan’s credit card has an APR of 10.59%, compounded monthly. He is required to make a minimum payment of 3.96% of his current

lys-0071 [83]

Answer:

b. 988.97

Step-by-step explanation:

got a 100 on the test

6 0

3 years ago

You may use a spreadsheet for this one. If $60 is put in an account that gets 7% per year, and I add $15 at the end of each year

Nutka1998 [239]

You will have approximately $257 at the end of 8 years.

3 0

4 years ago

What is the true solution to 2 l n e Superscript l n 5 x Baseline = 2 l n 15 x = 0 x = 3 x = 9 x = 15

weeeeeb [17]

Answer:

x = 3

Step-by-step explanation:

I got it right on edge

4 0

4 years ago

Find the equation for the plane that contains the line x=−1+3t , y=1+2t, z=2+4t and is perpendicular to the plane containing the

Ivan

Let $L$ be the line given by the vector equation

$(-1,1,2) + \lambda(3,2,4) \ , \lambda \in \mathbb{R}$ .

First, we use the director vectors of the lines L1 and L2 to get the

vector equation of the plane containing them, which we denote by $\Pi_1$ . This is,

$\\\\\Pi_1 : (1,-1,5) + \alpha (2,-1,6) + \beta (1,1,-3) \ , \alpha, \beta \in \mathbb{R}\\\\\\$

We observe that $\vec{N} = (2,-1,6)\times(1,1,-3) = (-3,12,3) \ne (0,0,0)$ . Therefore, the vector equation of $\Pi_1$ defines a plane and $\vec{N}$ is a normal vector to $\Pi_1.$

Finally, the vector equation for the wanted plane, which we denote by $\Pi$ , is

$\Pi : (-1,1,2) + r(3,2,4) + s(-3,12,3), r,s \in \mathbb{R} \ .$

Thus, if $s = 0$ , then $L \subset \Pi$ and since $\vec{N}$ is parallel to $\Pi$ , then it is perpendicular to $\Pi_1$ .

8 0

3 years ago