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3 years ago

(x+2)+10=44 please help

Mathematics

1 answer:

murzikaleks [220]3 years ago

6 0

Answer:

$x=32$

Step-by-step explanation:

Given the equation:

$(x+2)+10=44$

Begin by subtracting 10 from both sides of the equation:

$x+2=34$

Then, subtract 2 from both sides of the equation to find the value of $x$ :

$x=32$

You can check your answer by substituting the x-value into the initial equation:

$(32+2)+10=44$

$34+10=44$

$44=44$

Since both sides of the equation are equal, our solution is correct.

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Pwease help me anyone. thanks :-) :-) :-) :-) :-) :-) :-) :-)!!!!

adell [148]

A)
$x+3$

b)
$x(x+3)=340\\
x^2+3x-340=0$

c)
$x(x+3)=340\\ x^2+3x-340=0\\
\Delta=3^2-4\cdot1\cdot(-340)=9+1360=1369\\\sqrt{\Delta}=37\\
x_1=\dfrac{-3-37}{2}=-20\\
x_2=\dfrac{-3+37}{2}=17$

It's 17.

3 0

3 years ago

6.2 + 4/5 = ? .........

LUCKY_DIMON [66]

Answer:

7, .2=2/10=1/5+4/5=1+6=7.

P.s.

Please give me brainliest for my next rank.

4 0

3 years ago

Read 2 more answers

Which of the following is not one of the 8th roots of unity?

Anika [276]

Answer:

1+i

Step-by-step explanation:

To find the 8th roots of unity, you have to find the trigonometric form of unity.

1. Since $z=1=1+0\cdot i,$ then

$Rez=1,\\ \\Im z=0$

and

$|z|=\sqrt{1^2+0^2}=1,\\ \\\\\cos\varphi =\dfrac{Rez}{|z|}=\dfrac{1}{1}=1,\\ \\\sin\varphi =\dfrac{Imz}{|z|}=\dfrac{0}{1}=0.$

This gives you $\varphi=0.$

Thus,

$z=1\cdot(\cos 0+i\sin 0).$

2. The 8th roots can be calculated using following formula:

$\sqrt[8]{z}=\{\sqrt[8]{|z|} (\cos\dfrac{\varphi+2\pi k}{8}+i\sin \dfrac{\varphi+2\pi k}{8}), k=0,\ 1,\dots,7\}.$

Now

at k=0, $z_0=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 0}{8}+i\sin \dfrac{0+2\pi \cdot 0}{8})=1\cdot (1+0\cdot i)=1;$

at k=1, $z_1=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 1}{8}+i\sin \dfrac{0+2\pi \cdot 1}{8})=1\cdot (\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};$

at k=2, $z_2=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 2}{8}+i\sin \dfrac{0+2\pi \cdot 2}{8})=1\cdot (0+1\cdot i)=i;$

at k=3, $z_3=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 3}{8}+i\sin \dfrac{0+2\pi \cdot 3}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};$

at k=4, $z_4=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 4}{8}+i\sin \dfrac{0+2\pi \cdot 4}{8})=1\cdot (-1+0\cdot i)=-1;$

at k=5, $z_5=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 5}{8}+i\sin \dfrac{0+2\pi \cdot 5}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};$

at k=6, $z_6=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 6}{8}+i\sin \dfrac{0+2\pi \cdot 6}{8})=1\cdot (0-1\cdot i)=-i;$

at k=7, $z_7=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 7}{8}+i\sin \dfrac{0+2\pi \cdot 7}{8})=1\cdot (\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};$

The 8th roots are

$\{1,\ \dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ i, -\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ -1, -\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2},\ -i,\ \dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2}\}.$

Option C is icncorrect.

5 0

3 years ago

Pls look at this picture and answer

Pepsi [2]

Answer:

625

Step-by-step explanation:

7 0

3 years ago

In the pictures down below<br><br>Can anyone help? Fractions and percents<br><br>

wariber [46]

1st Picture:
$3)- \frac{6}{7} +\frac {10}{11}
\\\\LCD:77
\\\\\frac{6}{7}* \frac{11}{11} + \frac{10}{11}*\frac{7}{7}= \frac{66}{77} + \frac{70}{77} = \frac{136}{77} = 1\frac{59}{77} 
 \\ \\ 4) 3\frac{6}{7}* 2 \frac{1}{10}= \frac{27}{7} * \frac{21}{10} = \frac{567}{70} =\frac{81}{10} = 8\frac{1}{10}
\\\\5)2\frac{1}{2}+3\frac{1}{3}
\\\\LCD: 6
\\\\ 2\frac{3}{6} + 3\frac{2}{6} =5\frac{5}{6}
\\\\6)5\frac{1}{8}*-6\frac{6}{11}=\frac{41}{8}* \frac{72}{11}=\frac{2952}{88}=\frac{369}{11}=33\frac{6}{11}$

2nd Picture:

$1)80 percent of 80 = .8*80 =64
\\\boxed{64}
\\\\2)95percent of 100=.95*100=95
\\\boxed{95}$

7 0

3 years ago

(x+2)+10=44 please help​

(x+2)+10=44 please help