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3 years ago

What is the solution to the equation 5(x + 4) = 5x – 3?

Mathematics

2 answers:

vodka [1.7K]3 years ago

8 0

Answer: C, there is no solution

Step-by-step explanation:

First you want to distribute on the left side of the equation. 5(x+4) = 5x - 3 when distributed becomes 5x + 20 = 5x - 3. Next you want to get rid of the variable on one side, but since its the same, when you subtract 5x from both sides you get 20 = -3, which is not equal, meaning there is no solution.

Alex73 [517]3 years ago

6 0

$5(x+4)=5x-3\\5x+20=5x-3\\5x+20-5x+3=0\\0+23=0\\23=0$

So the answer is C. There is no solution, because the equation is false

A and B choices both have the answer, the value of x.

D choice only when the equation is true like 2=2.

You might be interested in

A wooden block measures 2 inches by 2 inches by 3 inches. A wedge is cut off from each corner of the block by slicing at points

seraphim [82]

Answer:

Number of edges in the resulting solid = 12

Step-by-step explanation:

A wooden block measures 2 inches by 2 inches by 3 inches. A wedge is cut off from each corner of the block by slicing at points that are 1 inch from each corner.

We are supposed to find the number of edges that are in the resulting solid.

In some corners , if we take out inch from both, then there would be nothing remaining.

The two faces along the height would be removed and it will become sharpened.

So the number of edges = 4 + 4 + 4 = 12

7 0

3 years ago

What term is 1/1024 in the geometric sequence,-1,1/4,-1/6..?

Trava [24]

Answer:

$\large\boxed{\text{sixth term is equal to}\ \dfrac{1}{1024}}$

Step-by-step explanation:

The explicit formula for a geometric sequence:

$a_n=a_1r^{n-1}$

$a_n$ - n-th term

$a_1$ - first term

$r$ - common ratio

$r=\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=...=\dfrac{a_n}{a_{n-1}}$

We have

$a_1=-1,\ a_2=\dfrac{1}{4},\ a_3=-\dfrac{1}{6},\ ...$

The common ratio:

$r=\dfrac{\frac{1}{4}}{-1}=-\dfrac{1}{4}\\\\r=\dfrac{-\frac{1}{6}}{\frac{1}{4}}=-\dfrac{1}{6}\cdot\dfrac{4}{1}=-\dfrac{2}{3}\neq-\dfrac{1}{4}$

<h2>It's not a geometric sequence.</h2>

If $a_3=-\dfrac{1}{16}$ then the common ratio is $r=\dfrac{-\frac{1}{16}}{\frac{1}{4}}=-\dfrac{1}{16}\cdot\dfrac{4}{1}=-\dfrac{1}{4}$

Put to the explicit formula:

$a_n=-1\left(-\dfrac{1}{4}\right)^{n-1}$

Put $a_n=\dfrac{1}{1024}$ and solve for <em>n </em>:

$-1\left(-\dfrac{1}{4}\right)^{n-1}=\dfrac{1}{1024}\qquad\text{use}\ a^n:a^m=a^{n-m}\\\\-\left(-\dfrac{1}{4}\right)^n:\left(-\dfrac{1}{4}\right)^1=\dfrac{1}{1024}\\\\-\left(-\dfrac{1}{4}\right)^n\cdot(-4)=\dfrac{1}{1024}\\\\(4)\left(-\dfrac{1}{4}\right)^n=\dfrac{1}{1024}\qquad\text{divide both sides by 4}\ \text{/multiply both sides by}\ \dfrac{1}{4}/\\\\\left(-\dfrac{1}{4}\right)^n=\dfrac{1}{4096}\\\\\dfrac{(-1)^n}{4^n}=\dfrac{1}{4^6}\qquad n\ \text{must be even number. Therefore}\ (-1)^n=1$