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Sholpan [36]
3 years ago
7

Which applies the power of a power rule properly to simplify the expression (710)5?

Mathematics
1 answer:
dmitriy555 [2]3 years ago
5 0

Answer:

D;(710)5= 710 · 5 = 750

Step-by-step explanation:

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Given a soda can with a volume of 15 and a diameter of 2, what is the volume of a cone that fits perfectly inside the soda can?
Naddika [18.5K]
Check the picture below.

so, a cone, that fits inside the cylinder, will have the same height "h" and radius "r".  Thus


\bf \textit{volume of a cylinder}\\\\
V=\pi r^2 h\qquad \boxed{\pi r^2 h}=15
\\\\\\
\textit{volume of a cone}\\\\
V=\cfrac{\pi r^2 h}{3}\implies V=\cfrac{\boxed{15}}{3}

8 0
3 years ago
Maria works as an electrician and earns $24.68/h. If she worked for 15 hours on one job, how much did she earn? *
rodikova [14]

Answer:

she would have $370.20

Step-by-step explanation:

24.68*15=370.2

8 0
3 years ago
Circle D is shown with the measure of the minor arcs which segment are congruent
mrs_skeptik [129]
<span> EF=GF
................................................................</span>
5 0
3 years ago
Please prove this........​
Crazy boy [7]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π    →     C = π - (A + B)

                                    → sin C = sin(π - (A + B))       cos C = sin(π - (A + B))

                                    → sin C = sin (A + B)              cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

<u>Proof LHS → RHS</u>

LHS:                        (sin 2A + sin 2B) + sin 2C

\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C

\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C

\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C

\text{Given:}\qquad \qquad \quad 2\sin C\cdot \cos (A - B)+2\sin C\cdot \cos (A+B)

\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]

\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B

\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C    \checkmark

7 0
3 years ago
Show that the equation x^4/2021 − 2021x^2 − x − 3 = 0 has at least two real roots.
andreev551 [17]

The roots of an equation are simply the x-intercepts of the equation.

See below for the proof that \mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0} has at least two real roots

The equation is given as: \mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0}

There are several ways to show that an equation has real roots, one of these ways is by using graphs.

See attachment for the graph of \mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0}

Next, we count the x-intercepts of the graph (i.e. the points where the equation crosses the x-axis)

From the attached graph, we can see that \mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0} crosses the x-axis at approximately <em>-2000 and 2000 </em>between the domain -2500 and 2500

This means that \mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0} has at least two real roots

Read more about roots of an equation at:

brainly.com/question/12912962

6 0
3 years ago
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