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3 years ago

Nuu.

Mathematics

1 answer:

Zolol [24]3 years ago

3 0

Answer:27/5

Step-by-step explanation:It just is that

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SOMEONE PLEASE HELP.>>>>>>

alexira [117]

Answer:

The answer would be B

6 0

3 years ago

I need help :/ i have five questions

IrinaK [193]

Answer:

Step-by-step explanation:

1. The given equation is:

$\frac{3}{5}{\times}\frac{4}{5}$

Multiplying both the terms, we get

$\frac{12}{25}$

Option A is correct.

2. The given equation is:

$2\frac{1}{3}{\times}\frac{3}{4}$

Converting the improper fraction into the mixed fraction,

$\frac{7}{3}{\times}\frac{3}{4}$

= $\frac{7}{4}$

Option C is correct.

3. The given equation is:

$\frac{3}{16}{\times}\frac{4}{21}$

Multiplying both the terms, we get

= $\frac{1}{28}$

4. The given equation is:

$3\frac{8}{9}{\times}4\frac{4}{5}$

Converting the improper fraction into the mixed fraction,

$\frac{35}{9}{\times}\frac{24}{5}$

= $\frac{56}{3}$

5. One package contains= $5\frac{1}{4}=\frac{21}{4}$ pounds of raisins.

$4\frac{2}{3}$ packages contains= $4\frac{2}{3}{\times}5\frac{1}{4}$ pounds of raisins.

= $\frac{14}{3}{\times}\frac{21}{4}$

= $\frac{49}{2}$

Therefore, the baker used $\frac{49}{2}$ pounds of raisins.

5 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

Easy question help fast please

zubka84 [21]

Answer:

<BOC = 100*

Step-by-step explanation:

5 0

2 years ago

Can anyone help me with part b? I will mark you as brainliest!

Kruka [31]

Answer:

There are 75 total cubes in the box.

Step-by-step explanation:

If 0.4 is 30, then divide 30 by 4 and then multiply that by 3, because that would be the answer for red or yellow, then just multiply 22.5 x 2 to find both red and yellow. Finally, add 30 and 45 for your answer.

8 0

3 years ago