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3 years ago

Place parentheses in the expression 7^2+1-5x3 value is 135

Mathematics

1 answer:

SpyIntel [72]3 years ago

3 0

Answer:

Step-by-step explanation:

[{((7^2)+1)-5}*3]

=[{(49+1)-5}*3]

=[{50-5}*3]

=[45*3]

=135

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How can 1/5x − 2 = 1/3x + 8 be set up as a system of equations? a . 5y − 5x = −10. 3y − 3x = 24 b. 5y − 5x = −10. 3y + 3x = 24 c

VMariaS [17]

The answer is d. 5y − x = −10. 3y − x = 24

$\frac{1}{5}x-2= \frac{1}{3}x+8=y$
⇒    $y= \frac{1}{5}x-2$
   $y= \frac{1}{3}x+8$
________________________
$y= \frac{x}{5} -2= \frac{x}{5} - \frac{10}{5} = \frac{x-10}{5}$
$y= \frac{x}{3} +8= \frac{x}{3} + \frac{24}{3} = \frac{x+24}{3}$
________________________
$y= \frac{x-10}{5}$ ⇒    $5y=x-10$
$y= \frac{x+24}{3}$   ⇒    $3y=x+24$
________________________
$5y-x=-10$
$3y-x=24$

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2 years ago

2^n=8^20. Solve for n

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2 years ago

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Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

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

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

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

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3 years ago

I need help with this please asap

Alex

Step-by-step explanation:

We can prove that the triangles are congruent using the SAS method.

It's given that GH and JH are congruent (S)

Angle GHF and Angle IHJ are equal since they are vertically opposite (A)

It's given that FH and HI are congruent (S)

Thus, the triangles are congruent, which implies that the remaining sides are equal.

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2 years ago

What does the line 4x y = 17 look like?

olganol [36]

The equation must be 4x + y = 17

I will use the y-intercept slope version

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Then the slope is negative (-4) and the y-intercept is 17.

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