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3 years ago

12

Evaluate the expression 6(x) when x equals 3. Pls... I need help.

1 answer:

Leto [7]3 years ago

7 0

Answer:

18

Step-by-step explanation:

substitute x = 3 into the expression

6x = 6(3) = 6 × 3 = 18

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in 1900, greenville had a population of 4000 people. Greenville's population increased by 100 people each year. Springfield had

jasenka [17]

Answer: year 1948

Step-by-step explanation:

1900 is our t = 0

The population of Greenville can be written as:

G = 4000 + 100*t

the equation for Springfield can be written as:

S = 1000*(1.04)^t

we want to find the value of t that makes S(t) = G(t)

4000 + 100*t = 1000*(1.04)^t

(4000 + 100*t) = 1.04^t

4 + 0.1*t = 1.04^t

now, you can graph both of this equations (left side and right side) and see in wich value of t the graphs intersect eachother, or you also may use different values of t until the values are about the same for both sides, which is the thing i did with the equation:

4 = 1.04^t - 0.1*t

You will find that the correct value is t = 48

So we can assume that in year 1948 the populations will be about the same.

5 0

3 years ago

Which statements about functions g(x) = x2 - 4x + 3 and f(x) = x2 - 4x are true? Select all that apply.

Tanya [424]

Answer:

A and B

Step-by-step explanation:

We are given that

$g(x)=x^2-4x+3$

$g(x)=(x^2-2\times x\times 2+4)-4+3=(x-2)^2-1$

Compare with it

$y=(x-h)^2+k$

Where vertex=(h,k)

We get

Vertex of g=(2,-1)

$f(x)=x^2-4x=(x^2-2\times x\times 2+4)-4=(x-2)^2-4$

Vertex of f=(2,-4)

Equation of axis of symmetry=x-coordinate of vertex

Axis of symmetry of g

x=2

Axis of symmetry of f

x=2

Differentiate w.r.t x

$g'(x)=2x-4=0$

$2x-4=0\implies 2x=4$

$x=\frac{4}{2}=2$

$f'(x)=2x-4$

$2x-4=0\implies 2x=4$

$x=\frac{4}{2}=2$

$g''(x)=2>0$

$f''(x)=2>0$

f and g have both minima at x=2

Hence, option A and B are true.

8 0

4 years ago

find the surface area of a right regular hexagonal pyramid with sides 3 cm and slant heights 6cm. show all of your work.

DedPeter [7]

Answer:

The surface area of right regular hexagonal pyramid = 82.222 cm³

Step-by-step explanation:

Given as , for regular hexagonal pyramid :

The of base side = 3 cm

The slant heights = 6 cm

Now ,

The surface area of right regular hexagonal pyramid = $\frac{3\sqrt{3}}{2}\times a^{2} + 3 a \sqrt{h^{2}+ 3\times \frac{a^{2}}{4}}$

Where a is the base side

And h is the slant height

So, The surface area of right regular hexagonal pyramid = $\frac{3\sqrt{3}}{2}\times 3^{2} + 3 \times 3 \sqrt{6^{2}+ 3\times \frac{3^{2}}{4}}$

Or, The surface area of right regular hexagonal pyramid = $\frac{3\sqrt{3}}{2}\times 9 + 9 \sqrt{36+ 3\times \frac{9}{4}}$

Or, The surface area of right regular hexagonal pyramid = 23.38 + 9 × $\frac{171}{4}$

∴ The surface area of right regular hexagonal pyramid = 23.38 + 9 × 6.538

I.e The surface area of right regular hexagonal pyramid = 23.38 + 58.842

So, The surface area of right regular hexagonal pyramid = 82.222 cm³ Answer

6 0

4 years ago

The Width of a rectangle is 2w+3. The length is equal to 4 times its width. Write an expression to represent the perimeter of th

kupik [55]

Since the perimeter is the sum of the length of all sides we multiply width and length by 2 and then add them together

4 0

3 years ago

SOMEONE PLEASE JUST ANSWER THIS FOR BRAINLIEST!!!

ohaa [14]

You’re answer would be -2a^2 + 48 - 6a^4

4 0

3 years ago