12a^2 10 x (1/2)a^2 whats will be the solution

Mathematics

1 answer:

AVprozaik [17]4 years ago

4 0

Get Photomath it will give you the answer

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Use fundamental identities to fully simply the expression.

gtnhenbr [62]

$\bf \stackrel{\textit{pythagorean identities}}{sin^2(\theta)+cos^2(\theta)=1}\qquad \qquad \stackrel{\textit{symmetry identities}}{cos(-\theta)=cos(\theta ) } \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{array}{llrll} 3sin^3(t)csc(t)\qquad &+cos^2(t)&\qquad +2cos(-t)cos(t)\\\\ 3sin^3(t)\cdot \cfrac{1}{sin(t)}&&+2[cos(t)]cos(t)\\\\ 3sin^2(t)&&+2cos^2(t) \end{array} \\\\[-0.35em] ~\dotfill\\\\ 3sin^2(t)+cos^2(t)+2cos^2(t)\implies 3sin^2(t)+3cos^2(t) \\\\\\ 3[sin^2(t)+cos^2(t)]\implies 3[1]\implies 3$

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4 years ago

Demons answer this for me.

denis-greek [22]

Answer:

Step-by-step explanation:

if you have negative numbers of x (this means negative meters) then the equation transform them into negative yards.

then replace y by 40yd in the equation like

40 = 1.09x

x = 40/1.09 = 36.7 meters

so in a 40yd race, there are 36.7 meters.

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3 years ago

Find the value of the power. 2400 The value of the power is

Aloiza [94]

The value of power is
0

7 0

3 years ago

According to the scatter plot below, which statement is correct The input, or dependent variable is enrollment, the output , or

stepan [7]

It’s B I just took the test

4 0

3 years ago

In the figure to the right, if AC=19 and BC=16, what is the radius? *For a circle*

ch4aika [34]

Answer:

The radius of the circle is 10.2 units

Step-by-step explanation:

The complete question is

In the figure to the right, if AC=19 and BC=16, what is the radius?

A circle has center A. Points B and D are on the circle, with B on the left and D near the bottom. Point C lies outside the circle such that the line segment A C passes through point D and the line segments A B and B C form a right angle.

The radius is approximately (Round to the nearest tenth as needed.)

The picture of the question in the attached figure

we know that

In the right triangle ABC

Applying the Pythagorean Theorem

$AC^2=AB^2+BC^2$