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san4es73 [151]
3 years ago
15

This look at pic I need number 22

Mathematics
1 answer:
kupik [55]3 years ago
3 0
I THINK THAT IT IS (A)
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Rearrange x=3g+2 to make g the subject.
jok3333 [9.3K]

Answer:

3g = x - 2

Step-by-step explanation:

x = 3g + 2

3g + 2 = x

3g + 2 - 2 = x - 2

3g = x - 2

6 0
3 years ago
Four times a number subtracted from ten is twice the number
sertanlavr [38]

Answer:

4 times x -10 times 2

i think

Step-by-step explanation:

7 0
4 years ago
Is the line increasing or decreasing?<br> y=-2x+5<br> (pls explain because im totally lost)
saveliy_v [14]

Answer:

Step-by-step explanation:

Assuming this is the slope equation

-2 is the slope

x is point x

5 is the y intercept

y is point y

the regular equation it y = mx + b

the slope is negative so it is decreasing aka gooing down

sorryy if this is the wrong answer to your question

6 0
3 years ago
Read 2 more answers
Need helps with geometry so hard and due soon smart people only and I mark brainly
Dennis_Churaev [7]

Answer:

Hello!!! Princess Sakura here ^^

Step-by-step explanation:

The 1st one is:

3x-17+ x+40+2x-5=180\\6x+18=180\\6x=162\\x=27

The 2nd one is:

103-x+2x=6x-7\\103+x=6x-7\\103=5x-7\\110=5x\\x=22

5 0
4 years ago
Read 2 more answers
In a plane, points P and Q are 20 inches apart. If point R is randomly chosen from all the points in the plane that are 20 inche
Karolina [17]

Answer:

E. \frac{2}{3}

Step-by-step explanation:

Consider a circle having centre P and radius 20 inches,

So, the area of the circle,

A=\pi (20)^2\text{ square inches}

Also, suppose points Q and R are on the circumference,

i.e. PQ = PR = 20 inches,

If QR = 20 inches,

So, triangle PQR is an equilateral triangle,

⇒ m∠RPQ = 60°,

Now, suppose R' is another point on the circle,

Such that, ΔPQR ≅ Δ PQR',

⇒ m∠QPR' = 60°,

Thus, minor angle, m∠RPR' = m∠RPQ + m∠QPR' = 60° + 60° = 120°,

⇒ major angle, m∠RPR' = 360° - 120° = 240°,

So, the area of the circle where a point on the circumference is closer to P than it is to Q

=\frac{240^{\circ}}{360^{\circ}}\pi (20)^2

=\frac{2}{3}\pi (20)^2

Hence, the probability that a point is closer to P than it is to Q = \frac{\frac{2}{3}\pi (20)^2}{\pi (20)^2}

=\frac{2}{3}

i.e. OPTION E is correct.

7 0
3 years ago
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