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2 years ago

Which best describes how to solve the equation for x.

Mathematics

1 answer:

prohojiy [21]2 years ago

7 0

Answer:

B. multiply both sides by c.

Step-by-step explanation:

To solve the equation for x, you have to isolate the x variable on one side of the equation. To do this, do the opposite of any operation applied to x to remove other variables from the x's side of the equation:

x/c = d

multiply both sides of the equation by c to isolate x:

(x/c) * c = d * c

x = dc

Now x can be solved for.

Hope this helps :)

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Simplify the following ratios:

jenyasd209 [6]

A) 3:4
b) 2:7
c) 3.5
d) 2:3

6 0

2 years ago

Put the following equation of a line into slope-intercept form, simplifying all fractions. 3y - 40 = 9

Nina [5.8K]

Answer:

3y=49

Step-by-step explanation:

yeppp I'm pretty sure

8 0

3 years ago

Round 10.09 To The Nearest Whole Number

ziro4ka [17]

Answer=10

When rounding to the nearest whole number, we need to look to the number in the tenths place

10.09

If the number is 5 or greater, we round up. If the number is 4 or less, we round down.

The number is 0, so we would round down.

10.09 becomes 10

3 0

3 years ago

Read 2 more answers

Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$

Observe that, because $n \geq 2$ and is an integer,

$n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5$

In concusion, the statement is true if and only if n is a non negative integer such that $n\leq 77$

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute $(4n)^4- \sum^{n}_{i=0} (2i)^4$ for 77 and 78 you will obtain:

$(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064$

$(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992$

7 0

3 years ago

Erik and Caleb were trying to solve the equation: 0=(3x+2)(x-4) Erik said, "The right-hand side is factored, so I'll use the zer

Mumz [18]

Answer:

C) Both

Step-by-step explanation:

The given equation is:

$0=(3x+2)(x-4)$

To solve the given equation, we can use the Zero Product Property according to which if the product <em>A.B = 0</em>, then either A = 0 OR B = 0.

Using this property:

$(3x+2) = 0 \Rightarrow \bold{x = -\frac{2}{3}}\\(x-4) = 0 \Rightarrow \bold{x = 4}$

So, Erik's solution strategy would work.

Now, let us discuss about Caleb's solution strategy:

Multiply $(3x+2)(x-4)$ i.e. $3x^2-12x+2x-8$ = $3x^2-10x-8$

So, the equation becomes:

$0=3x^2-10x-8$

Comparing this equation to standard quadratic equation:

$ax^2+bx+c=0$

a = 3, b = -10, c = -8

So, this can be solved using the quadratic formula.

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

$x=\dfrac{-(-10)\pm\sqrt{(-10)^2-4\times3 \times (-8)}}{2\times 3}\\x=\dfrac{-(-10)\pm\sqrt{196}}{6}\\x=\dfrac{10\pm14}{6} \\\Rightarrow x= 4, -\dfrac{2}{3}$

The answer is same from both the approaches.

So, the correct answer is:

C) Both

3 0

2 years ago