All categories

2 years ago

PLSSS HELP IF YOU TURLY KNOW

Mathematics

2 answers:

mylen [45]2 years ago

8 0

Answer:

x = 5

Step-by-step explanation:

(3-6x)/3 = -11

first multiple both sides by 3

3 - 6x = -33

then subtract 3 from both sides

-6x = -30

then divide by 6

x = 5

Marysya12 [62]2 years ago

4 0

Answer: x = 6

Step-by-step explanation:

(3-6x)/(3) = -11

3-6x = -33 [Multiply both sides by -3]

-6x = -36 [Subtract both sides by 3]

x=6 [Divide both sides by -6]

You might be interested in

You are busy baking cookies. You have baked 32 cookies and you want to make sure that each of your eight friend’s cookies look t

Alona [7]

Answer: 128 pieces of candy for all the cookies.

8 0

3 years ago

What is x*x=15? (Basically make an equation that equals up to 15.)

rodikova [14]

Answer:

$\displaystyle \sqrt{15} = x$

Step-by-step explanation:

Be VERY careful. When it write out the equation that way, it will give the speculation of this:

$\displaystyle 15 = x^2$

Now, in this case, since you were extra explanatory about this [what is written in parentheses], two factors I know of that multiply to 15 are these:

** $\displaystyle 15 = 2 \times 7\frac{1}{2}$

I am joyous to assist you at any time. ☺️

3 0

3 years ago

Intersection point of Y=logx and y=1/2log(x+1)

GalinKa [24]

Answer:

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

The Problem:

What is the intersection point of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ ?

Step-by-step explanation:

To find the intersection of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ , we will need to find when they have a common point; when their $x$ and $y$ are the same.

Let's start with setting the $y$ 's equal to find those $x$ 's for which the $y$ 's are the same.

$\log(x)=\frac{1}{2}\log(x+1)$

By power rule:

$\log(x)=\log((x+1)^\frac{1}{2})$

Since $\log(u)=\log(v)$ implies $u=v$ :

$x=(x+1)^\frac{1}{2}$

Squaring both sides to get rid of the fraction exponent:

$x^2=x+1$

This is a quadratic equation.

Subtract $(x+1)$ on both sides:

$x^2-(x+1)=0$

$x^2-x-1=0$

Comparing this to $ax^2+bx+c=0$ we see the following:

$a=1$

$b=-1$

$c=-1$

Let's plug them into the quadratic formula:

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

$x=\frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}$

$x=\frac{1 \pm \sqrt{1+4}}{2}$

$x=\frac{1 \pm \sqrt{5}}{2}$

So we have the solutions to the quadratic equation are:

$x=\frac{1+\sqrt{5}}{2}$ or $x=\frac{1-\sqrt{5}}{2}$ .

The second solution definitely gives at least one of the logarithm equation problems.

Example: $\log(x)$ has problems when $x \le 0$ and so the second solution is a problem.

So the $x$ where the equations intersect is at $x=\frac{1+\sqrt{5}}{2}$ .

Let's find the $y$ -coordinate.

You may use either equation.

I choose $y=\log(x)$ .

$y=\log(\frac{1+\sqrt{5}}{2})$

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

6 0

2 years ago

Going against the current, a boat takes 6 hours to make a 120-mile trip. When the boat travels with the current on the return tr

Andrew [12]

Answer:

120 miles/6 hours = 20 mph. You need to do the same for the boat going with the current. 120 miles/ 5 hours = 24 mph.

Step-by-step explanation:

5 0

3 years ago

the area of a triangle is 124 square units. what would it's new area be if its base was half as long and its height was three ti

Montano1993 [528]

To solve this problem you must apply the proccedure shown below:

1. You have that the formula for calculate the area of a triangle is:

A=bh/2

Where A is the area of the triangle, b is the base of the triangle and h is the height of the triangle.

bh/2=124
bh=124x2
bh=248

2. The problem asks for the new area of the triangle if its base was half as long and its height was three times as long. Then, you have:

Base=b/2
Height=3h

3. Therefore, when you substitute this into the formula for calculate the area of a triangle, you obtain:

A'=bh/2

(A' is the new area)

A'=(b/2)(3)/2
A'=3bh/4

4. When you substitute bh=248 into A'=3bh/4, you obtain:

A'=186 units²

The answer is: 186 units²

3 0

3 years ago