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3 years ago

Please help and thank you very much,

Mathematics

1 answer:

finlep [7]3 years ago

7 0

I’m going with 92 sorry if it’s wrong

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Which graph shows the solution to the system of linear inequalities?

tangare [24]

Answer:

See below

Step-by-step explanation:

Given system:

y ≥ 2x + 1
y ≤ 2x – 2

The solution is the common area shaded by the inequalities.

The lines are parallel because of same slope.

Both lines are solid because of equal sign in both inequalities.

The first inequality has a y-intercept of 1 and shaded area is to the left of the line since the value of y is greater as x increases.

The second inequality has a y-intercept of -2 and shaded area is to the right of the line since the value of y is greater as x increases.

The matching graph is the second picture (or attached below) and there is no solution.

8 0

3 years ago

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7 (3) (-2) PLEASE HELP ASAP!!!!

Alex777 [14]

Answer: the pic is the answer and

Step-by-step explanation:I explaned on the pic HOPE THIS HELPS WITH UR QUESTION

7 0

3 years ago

The circumference of a circle is 3.14 inches. What is the area of the circle?

SVETLANKA909090 [29]

Answer:

A = .785 inches ^2

Step-by-step explanation:

C = pi *d

We know the diameter = 2*r

C = pi*2*r

Substituting pi = 3.14 we can solve for r

3.14 = 2*pi*r

3.14 = 2*3.14 *r

Divide by 3.14 on each side

3.14/3.14 = 2*3.14 *r/3.14

1 = 2r

Divide by 2

1/2 =2r/2

1/2 =r

Now we can find the area

A = pi *r^2

A = 3.14 *(1/2)^2

A= 3.14 *1/4

A = .785 inches ^2

6 0

3 years ago

What is the quotient of 1 1/3 divided 3/5 Don’t know

yawa3891 [41]

Answer:

20/9

Step-by-step explanation:

I can't really explain this, but you should you the desmos scientific calculator, just look it up. For other fractions too, you can turn into decimals, and then divide from there which is easier for me.

Hope this helps!

3 0

3 years ago

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Question 9

Alenkasestr [34]

Answer:

$PX=4\\PY=4\sqrt{3}$

Step-by-step explanation:

Let $\angle PYZ=Q$

$\Rightarrow \angle XYP=90-\angle PYZ\\\angle XYP=90-\theta$

In the $\Delta PYZ$

$\tan\theta =\frac{opposite}{adjacent}=\frac{PZ}{PY}......(1)$

In the $\Delta YPX$

$\tan(90-\theta)=\frac{opposite}{adjacent}=\frac{PX}{PY}$

$\cot=\frac{PX}{PY}....(2)$

eqn(1) $\times$ eqn(2)

$\tan\times\cot\theta=\frac{PZ}{PY}\times\frac{PX}{PY}\\\\1=\frac{PZ\times PX}{PY^2}\ \ (as\ tan\theta\ =\frac{1}{\cot\theta})\\\\PY^2=PZ\times PX\\PY^2=12PX.......(3)$

Now in $\Delta XYP$

use Pythagorean theorem

$XY^2=PY^2+PX^2\\8^2=PY^2+PX^2\ \ (as\ XY\ =8)\\PX^2+PY^2=64\\\\PY^2=12PX\ \ (eqn(3)\\\\\Rightarrow PX^2+12PX=64\\\\PX^2+12PX-64=0\\\\PX^2+16-4PX-64=0\\\\(PX+16)(PX-4)=0\\\\PX=4,\ -16$

Length can not be negative

Hence $PX=4$

$PY^2=12PX=12\times4=48\\\\PY=\sqrt{48}=4\sqrt{3}$

Hence $PX=4,\ PY=4\sqrt{3}$

8 0

3 years ago