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weqwewe [10]
3 years ago
12

How do I figure out 1b?

Mathematics
1 answer:
taurus [48]3 years ago
4 0
300/5= 60
60x4(children)=240
240 seats will be for children and 60 seats for adults.
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Which does not describe the same situation y=3x+5
Naddika [18.5K]
Given the function. y = 3x + 5

For, x = -3; y = 3(-3) + 5 = -9 + 5 = -4
For x = 1; y = 3(1) + 5 = 3 + 5 = 8
For x = 4; y = 3(4) + 5 = 12 + 5 = 17

Thus the table representing the function is the table with: -3, 1 and 4 as x-values and -4, 8, 17 as y-values.

For x = 0; y = 3(0) + 5 = 0 + 5 = 5
For y = 0; 0 = 3x + 5; 3x = -5 and x = -5/3

Thus the graph of the function is a straight line passing through points (0, 5) and (-5/3, 0).

To illustrate the fuction as a word statement we say that y is five more than three times x


From the given descriptions, the graph does not represent the graph of y = 3x + 5.

Therefore, the one that does not describe the same situation is the graph.
3 0
3 years ago
Solve 4 cos2x-3 = 0 for all real values of x.
vichka [17]

4 cos² x - 3 = 0

4 cos² x = 3

cos² x = 3/4

cos x = ±(√3)/2

Fixing the squared cosine doesn't discriminate among quadrants.  There's one in every quadrant

cos x = ± cos(π/6)

Let's do plus first.  In general, cos x = cos a has solutions x = ±a + 2πk integer k

cos x = cos(π/6)

x = ±π/6 + 2πk

Minus next.

cos x = -cos(π/6)

cos x = cos(π - π/6)

cos x = cos(5π/6)

x = ±5π/6 + 2πk

We'll write all our solutions as

x = { -5π/6, -π/6, π/6, 5π/6 } + 2πk integer k

3 0
3 years ago
Find the center and radius of the circle 3 x squared plus 3 y squared plus 6 y equals 03x2+3y2+6y=0.
denis23 [38]

your answer would be 16

6 0
3 years ago
Read 2 more answers
3 + x over 2 = 5<br> please help? two step equations
elena-14-01-66 [18.8K]
I think But I'm not 100% sure yet it's a guess

Y=3+x
Root: (-3, 0)
Intercept:(0, 3)
5 0
3 years ago
The gravitational force exerted by the planet Earth on a unit mass at a distance r from the center of the planet is:
jolli1 [7]

Answer:

Yes, F is a continuous function of r

Step-by-step explanation:

We are given that

When r<R

F(r)=\frac{GMr}{R^3}

r\geq R

F(r)=\frac{GM}{r^2}

Where M=Mass of the earth

R=Radius of earth

G=Gravitational constant

We have to find the function is continuous of r or not.

LHL

\lim_{r\rightarrow R-}\frac{GMr}{R^3}=\frac{GMR}{R^3}=\frac{GM}{R^2}

RHL

\lim_{r\rightarrow R+}\frac{GM}{R^2}=\frac{GM}{R^2}

F(R)=\frac{GM}{R^2}

When a function is continuous at x=a

Then, LHL=RHL=f(a)

RHL==LHL=F(R)

Hence, the function is continuous of r.

8 0
3 years ago
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