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3 years ago

5

HELP NEEDED!!!!

1 answer:

strojnjashka [21]3 years ago

8 0

Answer:

D

Step-by-step explanation:

Wkx2di2jd1nd1do jf2kd2ifj2dkdu2dj2d

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The coordinates below represent two linear equations.

vovikov84 [41]

First write the equations in the slope-intercept form:
$y=mx+b \\ \\
m=\frac{y_2-y_1}{x_2-x_1}$

Line 1 passes through the points (-6,-6) and (-4,-3).
$m=\frac{-3-(-6)}{-4-(-6)}=\frac{-3+6}{-4+6}=\frac{3}{2} \\ \\
(-6,-6) \to x=-6, \ y=-6 \\
-6=\frac{3}{2} \times (-6) + b \\
-6=-9+b \\
b=3 \\ \\ y=\frac{3}{2}x+3$

Line 2 passes through the points (0,3) and (2,6).
$m=\frac{6-3}{2-0}=\frac{3}{2} \\ \\
(0,3) \to x=0, \ y=3 \\
3=\frac{3}{2} \times 0 + b \\
b=3 \\ \\
y=\frac{3}{2}x+3$

These are two identical lines so the system of equations has infinitely many solutions. The answer is D.

7 0

2 years ago

Read 2 more answers

I know he is incorrect but I just can’t say why please help

egoroff_w [7]

Answer:

Step-by-step explanation:

Umm, Wheres Tanner's graph? Please post it so that I could answer your question!! :)

3 0

2 years ago

Can somebody pls help me with this question I dont get it

alekssr [168]

Answer:

1. 195=6.50·c

2. 30

Step-by-step explanation:

195=6.5·c

Divide both sides by 6.5

30=c

6 0

2 years ago

62. Height Kirsten places her surveyor's telescope on the top of

Solnce55 [7]

Answer:

21 feet

Step-by-step explanation:

Given :

Height of the tripod above the ground = 5 ft

Elevation = 8°

Distance of the tree from the tripod = 120 ft

Therefore from the figure, we can find the height of the tress as:

$\tan 8 ^\circ = \frac{BC}{AB}$

$0.1405 = \frac{BC}{120}$

∴ BC ≈ 16

Therefore, CE = BC+BE

= 16 + 5

= 21

Therefore the height of the tree is 21 feet.

4 0

2 years ago

When a person is breathing normally the amount of air in their lawns varies sinusoidally. When full Karen’s lungs hold 2.8 L of

makkiz [27]

Answer:

$A(t) = 2.2\sin \frac{(t - 2)\pi }{6} + 0.6$

Step-by-step explanation:

Let the function of quantity in the lung of air be A(t)

So $A(t) \alpha \sin (\frac{t - \alpha }{k} )$

so, A(t) = Amax sin t + b

A(t) = 2.8t⇒ max

A(t) = 0.6t ⇒ min

max value of A(t) occur when sin(t) = 1

and min value of A(t) = 0

So b = 0.6

and A(max) = 2.2

$A(t) = 2.2\sin \frac{(t)}{k} + 0.6$

at t = 2 sec volume of a is 0.6

So function reduce to

$A(t) = 2.2\sin \frac{(t - 2)}{k} + 0.6$

and t = 5 max value of volume is represent

so,

$\sin \frac{t - \alpha }{k} = 1$

$\frac{t - 2}{k} = \frac{\pi }{2}$ when t = 5

$\frac{6}{\pi } = k$

so the equation becomes

$A(t) = 2.2\sin \frac{(t - 2)\pi }{6} + 0.6$

7 0

2 years ago