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4 years ago

Help me ASAP!! Solve the square root equation

Mathematics

2 answers:

OverLord2011 [107]4 years ago

6 0

Answer:

no solution

Step-by-step explanation:

there are no values of x that make the equation true

ivolga24 [154]4 years ago

3 0

Answer:

There is no solution to this, I used a calculator and it is correct.

There are no values of

that make the equation true.

By the way, I used math-way in case you are wondering

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Ali and Janet are selling gifts at a local craft show. For every bar of soap that Ali sells, she earns $5. For every mug that Ja

Alexus [3.1K]

Answer:

95$

Step-by-step explanation:

Alli sells 5 bars, for each bar she earnd 5$ so she earns 5*5=25$

Janet sells 7 mugs, and for every mug she earns 10$ (twice then Alli), so she earns 7*10=70$

Together: 25+70=95$

7 0

3 years ago

Read 2 more answers

For what values of p is the value of the binomial 1.5p−1 smaller than the value of the binomial 1+1.1p by 4?

77julia77 [94]

Answer:

p = -5 (only)

Step-by-step explanation:

It looks like we want to count solutions for ...

(1.5p -1) = (1 +1.1p) -4

0.4p = -2

p = -2/0.4 = -5

For p having a value of -5, the binomial (1.5p -1) = -8.5 will be smaller than the value of (1 +1.1p) = -4.5 by 4.

4 0

3 years ago

What is 2567 divided by 4

Nastasia [14]

2,567 / 4 = 641.75 Hope this helps.

5 0

3 years ago

Read 2 more answers

Will mark Brainiest!! Help asap

lys-0071 [83]

Answer:

Step-by-step explanation: because thats my answer

7 0

3 years ago

P(x)= 3x^3-5x^2-14x-4

nexus9112 [7]

$\displaystyle\\
P(x)=3x^3-5x^2-14x-4\\\\
D_{-4}=\{-4;~-2;~\underline{\bf -1};~1;~2;~4\}\\\\
\text{We observe that } \frac{-1}{3} \text{ is a solution of the equation:}\\
3x^3-5x^2-14x-4=0\\\\
$

$\displaystyle\\
\text{Verification}\\\\
3x^3-5x^2-14x-4=\\\\
=3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\
=-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\
 =-\frac{6}{9}+\frac{42}{9}- \frac{36}{9}= \frac{42-6-36}{9}=\frac{42-42}{9}=\frac{0}{9}=0\\\\
\Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\
\Longrightarrow~~~P(x)~\vdots~(3x+1)$

$\displaystyle\\
3x^3-5x^2-14x-4=0\\
~~~~~-5x^2 = x^2 - 6x^2\\
~~~~~-14x =-2x-12x \\
3x^3+x^2 - 6x^2-2x-12x-4=0\\
x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\
(3x+1)(x^2-2x -4)=0\\\\
\text{Solve: } x^2-2x -4=0\\\\
x_{12}= \frac{-b\pm \sqrt{b^2-4ac}}{2a}=\\\\=\frac{2\pm \sqrt{4+16}}{2}=\frac{2\pm \sqrt{20}}{2}=\frac{2\pm 2\sqrt{5}}{2}=1\pm\sqrt{5}\\\\
x_1 =1+\sqrt{5}\\
x_2 =1-\sqrt{5}\\
\Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-\sqrt{5})(x-1+\sqrt{5})}$

7 0

3 years ago