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Valentin [98]
3 years ago
12

Simplify : -26b^2 + (-27b^2)

Mathematics
2 answers:
____ [38]3 years ago
5 0

<u><em>-53b² is the correct answer.</em></u><em>  </em>First you had to remove parenthesis first, and it gave us, -26b^2-27b^2. And then add similar elements, and it gave us the answer is -26b^2-27b^2=-53b^2, or -53b² is the right answer. Hope this helps! And thank you for posting your question at here on brainly, and have a great day. -Charlie

IgorLugansk [536]3 years ago
3 0
Step by step solution :

Step 1 :

Equation at the end of step 1 :

(0 - (26 • (b2))) + (0 - 33b2)
Step 2 :

Equation at the end of step 2 :

(0 - (2•13b2)) + -33b2
Step 3 :

Final result :

-53b2
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I need help with #6 please.
Juliette [100K]

9514 1404 393

Answer:

  6. (A, B, C) ≈ (112.4°, 29.5°, 38.0°)

  7. (a, b, C) ≈ (180.5, 238.5, 145°)

Step-by-step explanation:

My "work" is to make use of a triangle solver calculator. The results are attached. Triangle solvers are available for phone or tablet and on web sites. Many graphing calculators have triangle solvers built in.

__

We suppose you're to make use of the Law of Sines and the Law of Cosines, as applicable.

6. When 3 sides are given, the Law of Cosines can be used to find the angles. For example, angle A can be found from ...

  A = arccos((b² +c² -a²)/(2bc))

  A = arccos((8² +10² -15²)/(2·8·10)) = arccos(-61/160) = 112.4°

The other angles can be found by permuting the variables appropriately.

  B = arccos((225 +100 -64)/(2·15·10) = arccos(261/300) ≈ 29.5°

The third angle can be found as the supplement to the other two.

  C = 180° -112.411° -29.541° = 38.048° ≈ 38.0°

The angles (A, B, C) are about (112.4°, 29.5°, 38.0°).

__

7. When insufficient information is given for the Law of Cosines, the Law of Sines can be useful. It tells us side lengths are proportional to the sine of the opposite angle. With two angles, we can find the third, and with any side length, we can then find the other side lengths.

  C = 180° -A -B = 145°

  a = c(sin(A)/sin(C)) = 400·sin(15°)/sin(145°) ≈ 180.49

  b = c(sin(B)/sin(C)) = 400·sin(20°)/sin(145°) ≈ 238.52

The measures (a, b, C) are about (180.5, 238.5, 145°).

7 0
2 years ago
Solve the following and explain your steps. Leave your answer in base-exponent form. (3^-2*4^-5*5^0)^-3*(4^-4/3^3)*3^3 please st
Naily [24]

Answer:

\boxed{2^{\frac{802}{27}} \cdot 3^9}

Step-by-step explanation:

<u>I will try to give as many details as possible. </u>

First of all, I just would like to say:

\text{Use } \LaTeX !

Texting in Latex is much more clear and depending on the question, just writing down without it may be confusing or ambiguous. Be together with Latex! (*^U^)人(≧V≦*)/

$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

\left(\dfrac{1}{3^2}  \cdot \dfrac{1}{4^5}  \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3

Also,

\boxed{a^{0} = 1, a\neq 0 }

\left(\dfrac{1}{9}  \cdot \dfrac{1}{1024}  \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27

Note

\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq  0}

\left(\dfrac{1}{9216}   \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27

\left(\dfrac{1}{9216}   \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)

\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)

\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)

Note

\boxed{\dfrac{1}{\dfrac{1}{a} }  = a}

9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)

\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)

Once

9216=2^{10}\cdot 3^2 \implies  9216^3=2^{30}\cdot 3^6

\boxed{(a \cdot b)^n=a^n \cdot b^n}

And

$4^{\frac{4}{27}} = 2^{\frac{8}{27} $

We have

\left(\dfrac{ 2^{30} \cdot 3^6\cdot 27}{2^{\frac{8}{27} } }\right)

Also, once

\boxed{\dfrac{c^a}{c^b}=c^{a-b}}

2^{30-\frac{8}{27}} \cdot 3^6\cdot 27

As

30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27}  =\dfrac{802}{27}

2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3

2^{\frac{802}{27}} \cdot 3^9

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