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VikaD [51]
3 years ago
12

4x+8=22-6 solve for x

Mathematics
1 answer:
Digiron [165]3 years ago
6 0
=> 4x = 16 - 8
=>4x= 8
=> x= 2

in short the answer is 2
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At the start of 2012, the US federal budget had a deficit of more than $1.5×1013. Convert this number to decimal form.
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Answer:

Just multiple it

Step-by-step explanation:

1.5 × 1013

1519.5 has your result

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3 years ago
There are 1,500 students in a school. 65% are girls. How many girls are there in the school?
Reil [10]
1500 x 0.65= Number of girls
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Which type of transformation preserves the symmetry of an even function f(x) but does not preserve the symmetry of an odd functi
Alexus [3.1K]

The type of transformation that preserves the symmetry of an even function f(x) but does not preserve the symmetry of an odd function g(x) is vertical translation.

<h3>What is vertical translation?</h3>

Vertical translation of a graph is done by moving the base graph up or down in the y-axis direction. Each point on a graph is moved k units vertically to translate the graph by that many units.

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brainly.com/question/4397823

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7 0
1 year ago
Which relationship describes angles 1 and 2?
pshichka [43]
Complementary angles
6 0
3 years ago
Save me the headache
maxonik [38]

(9\sin2x+9\cos2x)^2=81

Taking the square root of both sides gives two possible cases,

9\sin2x+9\cos2x=9\implies\sin2x+\cos2x=1

or

9\sin2x+9\cos2x=-9\implies\sin2x+\cos2x=-1

Recall that

\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta

If \alpha=2x and \beta=\dfrac\pi4, we have

\sin\left(2x+\dfrac\pi4\right)=\dfrac{\sin2x+\cos2x}{\sqrt2}

so in the equations above, we can write

\sin2x+\cos2x=\sqrt2\sin\left(2x+\dfrac\pi4\right)=\pm1

Then in the first case,

\sqrt2\sin\left(2x+\dfrac\pi4\right)=1\implies\sin\left(2x+\dfrac\pi4\right)=\dfrac1{\sqrt2}

\implies2x+\dfrac\pi4=\dfrac\pi4+2n\pi\text{ or }\dfrac{3\pi}4+2n\pi

(where n is any integer)

\implies2x=2n\pi\text{ or }\dfrac\pi2+2n\pi

\implies x=n\pi\text{ or }\dfrac\pi4+n\pi

and in the second,

\sqrt2\sin\left(2x+\dfrac\pi4\right)=-1\implies\sin\left(2x+\dfrac\pi4\right)=-\dfrac1{\sqrt2}

\implies2x+\dfrac\pi4=-\dfrac\pi4+2n\pi\text{ or }-\dfrac{3\pi}4+2n\pi

\implies2x=-\dfrac\pi2+2n\pi\text{ or }-\pi+2n\pi

\implies x=-\dfrac\pi4+n\pi\text{ or }-\dfrac\pi2+n\pi

Then the solutions that fall in the interval [0,2\pi) are

x=0,\dfrac\pi4,\dfrac\pi2,\dfrac{3\pi}4,\pi,\dfrac{5\pi}4,\dfrac{3\pi}2,\dfrac{7\pi}4

5 0
3 years ago
Read 2 more answers
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