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3 years ago

5

SOLVE EACH SYSTEM USING SUBSTITUTION

1 answer:

NNADVOKAT [17]3 years ago

3 0

1. since the equations equal each other
2x+19=x+7
-x -x
x+19=7
-19 -19
x= -12
sub into y= x+7
y= -12+7
y= -5

2. x+2y=-7
<span>x+y+23=0
i don't know what the second equation is equal to, i'll assume 0

2y=x-7
y=1/2x-7/2
sub this into x+y+23=0
x+1/2x-7/2+23=0
3/2x+39/2=0
-39/2 -39/2
3/2x= -39/2
x=-39/2 / 3/2
x= -13
sub into y=1/2x-7/2
y=1/2(-13)-7/2
y= -10

3. </span>x+5y=2
<span>x=-4y+5
</span>sub x=-4y+5 into x+5y=2
-4y+5+5y=2
y+5=2
-5 -5
y= -3
sub y =-3 into x =-4y+5
x=-4(-3)+5
x= 17

4. 3x+y=9
<span>y=-5x+9
</span>sub -5x+9 into 3x+y=9
3x+(-5x+9)=9
3x-5x+9=9
-2x+9=9
-9 -9
-2x=0
x=0
sub into y=-5x+9
y=-5(0)+9
y=9

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Mr Bryce planted a garden in his backyard the length and width of the garden are shown below. what is the area, in square meters

VashaNatasha [74]

Answer:

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Step-by-step explanation:

Area = (lenght × widht)

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simplified

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Hope this helps

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2 years ago

Find values of  that satisfy the equation for 0º    360º and 0  θ  2 . Give answers in degrees and radians.

suter [353]

Answer:

$\theta = 30^\circ, 330^\circ$

$\theta = \frac{\pi}{6}, \frac{11\pi}{6}$

Step-by-step explanation:

Given [Missing from the question]

Equation:

$cos\theta = \frac{\sqrt 3}{2}$

Interval:

$0 \le \theta \le 360$

$0 \le \theta \le 2\pi$

Required

Determine the values of $\theta$

The given expression:

$cos\theta = \frac{\sqrt 3}{2}$

... shows that the value of $\theta$ is positive

The cosine of an angle has positive values in the first and the fourth quadrants.

So, we have:

$cos\theta = \frac{\sqrt 3}{2}$

Take arccos of both sides

$\theta = cos^{-1}(\frac{\sqrt 3}{2})$

$\theta = 30$ --- In the first quadrant

In the fourth quadrant, the value is:

$\theta = 360 -30$

$\theta = 330$

So, the values of $\theta$ in degrees are:

$\theta = 30^\circ, 330^\circ$

Convert to radians (Multiply both angles by $\pi/180$ )

So, we have:

$\theta = \frac{30 * \pi}{180}, \frac{330 * \pi}{180}$

$\theta = \frac{\pi}{6}, \frac{33 * \pi}{18}$

$\theta = \frac{\pi}{6}, \frac{11 * \pi}{6}$

$\theta = \frac{\pi}{6}, \frac{11\pi}{6}$

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2 years ago