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cosine of .97444

\(\begin{array}{rcll} \cos{(\varphi)} &=& 0.97444 \quad & | \quad \pm\arccos{} \\ \varphi &=& \pm\arccos{(0.97444)} \pm k\cdot 360^{\circ} \quad & | \quad k=0,1,2,\dots \\ \varphi &=& \pm 12.9821752590^{\circ} \pm k\cdot 360^{\circ} \\\\ \mathbf{\varphi_1 }&\mathbf{=}& \mathbf{ 12.9821752590^{\circ} \pm k\cdot 360^{\circ} }\\\\ \varphi_2 &=& -12.9821752590^{\circ} \pm k\cdot 360^{\circ} \\ \varphi_2 &=& 360^{\circ}-12.9821752590^{\circ} \pm k\cdot 360^{\circ} \\ \mathbf{\varphi_2 }&\mathbf{=}& \mathbf{347.017824741\pm k\cdot 360^{\circ}} \end{array}\)

9980

Zora is budgeting her available time for the next couple days. She spent 2 over 5 of her time on homework and then 3 over 10 of her time on the internet. Homework and time on the internet consumed 14 hours. How much available time did Zora have to budget

x = budget

\(\begin{array}{rcll} \dfrac{2}{5}\cdot x + \dfrac{3}{10} \cdot x &=& 14\ h \\ x\cdot \left( \dfrac{2}{5} + \dfrac{3}{10} \right) &=& 14\ h \\ x\cdot \left( \dfrac{2\cdot 10+5\cdot 3}{5\cdot 10} \right) &=& 14\ h \\ x\cdot \left( \dfrac{20+15}{50} \right) &=& 14\ h \\ x\cdot \dfrac{35}{50} &=& 14\ h \\ x\cdot \dfrac{5\cdot 7}{5\cdot 10} &=& 14\ h \\ x\cdot \dfrac{ 7 }{ 10 } &=& 14\ h \quad & | \quad \cdot \dfrac{ 10 }{ 7 } \\ x &=& \dfrac{ 10 }{ 7 }\cdot 14\ h \\ x &=& 10\cdot 2\ h \\ \mathbf{ x }& \mathbf{=} & \mathbf{ 20\ h } \\ \end{array}\)

\(\begin{array}{rcll} 20\ h \cdot \dfrac{2}{5} &=& 8\ h \\\\ 20\ h \cdot \dfrac{3}{10} &=& 6\ h \\\\ 8\ h +6\ h &=&14\ h \end{array}\)

Zora has 20 hours

(10√1.3333-1)100=2.919 can someone explain to me how this happened please

\(\begin{array}{rcll} (\sqrt[10]{1.3333}-1)\cdot 100 &=& 100\cdot (\sqrt[10]{1.3333}-1) \\ &=& 100\cdot (1.02918343597-1) \\ &=& 100\cdot 0.02918343597 \\ &=& 2.918343597 \\ &\approx& 2.918 \end{array}\)

\(\begin{array}{rcll} (\sqrt[10]{1.3333\dots }-1)\cdot 100 &=& 100\cdot (\sqrt[10]{1.3333\dots}-1) \\ &=& 100\cdot (\sqrt[10]{1\frac{1}{3}}-1) \\ &=& 100\cdot (\sqrt[10]{\frac{4}{3}}-1) \\ &=& 100\cdot (1.02918600896-1) \\ &=& 100\cdot 0.02918600896 \\ &=& 2.91860089648 \\ &\approx& 2.919 \end{array}\)

(10√1.3333-1)100=2.919 can someone explain to me how this happened please

You have it written down wrong!! It should be like this:

[1.3333]^(1/10] -1 x 100 =1.02919 - 1=.02919 x 100=2.919.

You were reading this bit wrong:(10√1.3333-1). You are supposed to take the 10th root of 1.3333 - 1, NOT the sqrt(1.3333 - 1) x 10.

And that is what you were doing wrong.

\(1+1=2\)

 where is the table ? :/

Make the number 48 with the numbers 1, 6, 7, 8.

Only use addition, subtraction, multiplication, and division. You can use each number once! You have to use all the numbers

\(\begin{array}{rcll} 48 = \dfrac{8} { \left( \dfrac{7}{6} \right) - 1 } \\ \end{array}\)

\(\begin{array}{rcll} 48 = \dfrac{6} { 1- \left( \dfrac{7}{8} \right) } \\ \end{array}\)

What do you mean by "square rooted by 2"?

Net force, F = (7.87 - 3.08)*10⁴ N = 4.79*10⁴ N

Acceleration, a = F/mass  so  a = 4.79*10⁴/1.63*10⁴ m/s²  ≈  2.93865 m/s²

Use v² = u² + 2*a*s   where v = final velocity, u = initial velocity and s = distance travelled to get the final velocity.

i.e.  v = sqrt(64.5² + 2*2.93865*590) m/s

1 mole corresponds to 12 + 2*16 grams (i.e. 44 grams)    (12 for carbon,  16 for oxygen)

Hence there are 28/44 moles in 28 grams of CO₂  (28/44 ≈ 0.636)  

Depends on the context!  Do you have a specific example in mind?

Amen 2 that Emily! Hope u 2 have a goodnight. 

Make the number 48 with the numbers 1, 6, 7, 8.
Only use addition, subtraction, multiplication, and division. You can use each number once!

How do you determine the correct way to simplify two radical numbers while dividing them and how do I know if the number remains squared or becomes a whole number? Perhaps this should be used for an example? 4√63 / 6√36

\(\begin{array}{rcll} \dfrac{ 4 \cdot \sqrt{63} } { 6 \cdot \sqrt{36} } &=& \dfrac{ 4 }{ 6 }\cdot \dfrac{ \sqrt{63} } { \sqrt{36} } \\\\ &=& \dfrac{ 2 }{ 3 }\cdot \dfrac{ \sqrt{63} } { \sqrt{36} } \\\\ &=& \dfrac{ 2 }{ 3 }\cdot \sqrt { \dfrac{ 63 } { 36 } } \\\\ &=& \dfrac{ 2 }{ 3 }\cdot \sqrt { \dfrac{ 9\cdot 7 } { 9\cdot 4 } } \\\\ &=& \dfrac{ 2 }{ 3 }\cdot \sqrt { \dfrac{ 7 } { 4 } } \\\\ &=& \dfrac{ 2 }{ 3 }\cdot \dfrac{ \sqrt{7} } { \sqrt{4} } \\\\ &=& \dfrac{ 2 }{ 3 }\cdot \dfrac{ \sqrt{7} } { 2 } \\\\ &=& \dfrac{ \sqrt{7} } { 3 } \\\\ \end{array}\)

Make the number 48 with the numbers 1, 6, 7, 8.

Only use addition, subtraction, multiplication, and division.

You can use each number once!

Make the number 48 with the numbers 1, 6, 7, 8.

Only use addition, subtraction, multiplication, and division.

You can use each number once!

You have to use all the numbers

\(\begin{array}{rcll} 48 = \dfrac{6} { 1- \left( \dfrac{7}{8} \right) } \\ \end{array}\)

x+2y=7

2x+2y=13

\(\begin{array}{lrcll} (1)& x+2y &=& 7 \quad & | \quad -x\\ & 2y &=& 7 -x \\\\ (2)& 2x+2y &=& 13 \\ & 2x +(7-x) &=& 13 \\ & 2x + 7-x &=& 13 \\ & x + 7 &=& 13 \quad & | \quad -7\\ & x &=& 13 -7\\ & \mathbf{x} & \mathbf{=} & \mathbf{6} \\\\ & 2y &=& 7 -x \\ & 2y &=& 7 -6 \\ & 2y &=& 1 \quad & | \quad :2\\ & y &=& \frac12\\ & \mathbf{y} & \mathbf{=} & \mathbf{0.5} \\\\ \end{array}\)

Solve the following system:
{4 x+3 y = -8
y-8 x = -12

In the second equation, look to solve for y:
{4 x+3 y = -8
y-8 x = -12

Add 8 x to both sides:
{4 x+3 y = -8
y = 8 x-12

Substitute y = 8 x-12 into the first equation:
{4 x+3 (8 x-12) = -8
y = 8 x-12

4 x+3 (8 x-12) = (24 x-36)+4 x = 28 x-36:
{28 x-36 = -8
y = 8 x-12

In the first equation, look to solve for x:
{28 x-36 = -8
y = 8 x-12

Add 36 to both sides:
{28 x = 28
y = 8 x-12

Divide both sides by 28:
{x = 1
y = 8 x-12

Substitute x = 1 into the second equation:
Answer: |x = 1          and             y=-4

[1 x 10^9] x [1 - 15/100]=8.5 x 10^8

If I spent 25 € and it cost me 236,50 sek, would the conversion rate be 9.46?

If so, would 11,54 sek equal 1,4894291754756871 € and 14,09 sek equal 1,219873150105708 € and when rounded the two final values would be 1,49 € and 1,22 €?

please help a dummy out

236.50 / 25 =9.46 The exchange rate.

11.54 / 9.46 =1.21987315 Euros

14.09 / 9.46=1.4894291 Euros

Above, you just got them reversed. And rounding them off, would be:

1.22 Euros and 1.49 Euros, which you got right.

Thank-you

A school counselor surveyed 90 randomly selected students about the languages they speak. Of the students surveyed, 16 speak more than one language fluently. Based on these results, how many of the 1,800 students at the school can be expected to speak more than one language fluently?