All categories

3 years ago

Please help. Is algebra.

Mathematics

2 answers:

ExtremeBDS [4]3 years ago

8 0

Answer:

answer is

Step-by-step explanation:

-8b

have great day

lesantik [10]3 years ago

4 0

The answer is -8b because when you divide -b3 and b3 it cancels out, so you left with 2b • -4 and that’s -8b.

You might be interested in

Solve the equation by factoring 0=x2+8x+16

stepan [7]

Answer:

Alright all you nee to do is move all terms

which gives you a Answer of x = - 4 Hope this helps :)

Step-by-step explanation:

If u want i can explained with more detail

6 0

4 years ago

Read 2 more answers

Which of these is straight of a line and has two endpoint

Alla [95]

This question is impossible to answer without knowing or being able to see the question.

8 0

3 years ago

Read 2 more answers

A CD case measures four and four-fifths inches by five and three-fifths inches.What is the decimal form of the measurements

Sedaia [141]

4 4/5 is 4.8 and 5 3/5 is 5.6

6 0

3 years ago

Consider the circle with the radius of 30 cm

pentagon [3]

Answer:

60π ≈ 188.4955592 cm

Step-by-step explanation:

Circumference of a circle is given by 2 ∙ π ∙ r

Therefore, you have to plug in the given value of 30 into radius (r):

2 ∙ π ∙ 30

And multiply:

60π ≈ 188.4955592 cm

Hope this helps!

5 0

3 years ago

If sine of x equals square root of 2 over 2, what is cos(x) and tan(x)? Explain your steps in complete sentences. (10 points)

tensa zangetsu [6.8K]

$\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\quad 
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\\\\
-------------------------------\\\\$

$\bf sin(x)=\cfrac{\sqrt{2}}{2}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\impliedby \textit{now, let's find the adjacent side}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\pm\sqrt{(2)^2-(\sqrt{2})^2}=a\implies \pm\sqrt{4-2}=a\implies \pm\sqrt{2}=a$

now, the adjacent side, or the x-coordinate for the angle, can be negative and positive, is positive in the I quadrant and negative in the II quadrant, with a positive y-coordinate... so it can be either, the +/- one.

$\bf cos(x)=\cfrac{\pm\sqrt{2}}{2}\qquad tan(x)=\cfrac{\frac{\sqrt{2}}{2}}{\pm\frac{\sqrt{2}}{2}}\implies tan(x)=\pm 1$

6 0

3 years ago