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Anon25 [30]
2 years ago
14

Which expression results from using the distributive property 5(4x)

Mathematics
2 answers:
mojhsa [17]2 years ago
7 0

Answer:

20*5x

Step-by-step explanation:

multiply outer number by each item in the parenthesis

so 5*4=20 and

5*x=5X

nignag [31]2 years ago
6 0

20x would be ur answer.

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The degree of the polynomial function f(x) is 4.The roots of the equation f(x)=0 are −6, −2, 1, and 3.Which graph could be the g
nalin [4]
It’s a s 7.47 because of the five and six
5 0
2 years ago
First you to find the worksheet and download it<br> plase I need help
Goshia [24]

Answer:

a) The horizontal asymptote is y = 0

The y-intercept is (0, 9)

b) The horizontal asymptote is y = 0

The y-intercept is (0, 5)

c) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

d) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

e) The horizontal asymptote is y = -1

The y-intercept is (0, 7)

The x-intercept is (-3, 0)

f) The asymptote is y = 2

The y-intercept is (0, 6)

Step-by-step explanation:

a) f(x) = 3^{x + 2}

The asymptote is given as x → -∞, f(x) = 3^{x + 2} → 0

∴ The horizontal asymptote is f(x) = y = 0

The y-intercept is given when x = 0, we get;

f(x) = 3^{0 + 2} = 9

The y-intercept is f(x) = (0, 9)

b) f(x) = 5^{1  - x}

The asymptote is fx) = 0 as x → ∞

The asymptote is y = 0

Similar to question (1) above, the y-intercept is f(x) = 5^{1  - 0} = 5

The y-intercept is (0, 5)

c) f(x) = 3ˣ + 3

The asymptote is 3ˣ → 0 and f(x) → 3 as x → ∞

The asymptote is y = 3

The y-intercept is f(x) = 3⁰ + 3= 4

The y-intercept is (0, 4)

d) f(x) = 6⁻ˣ + 3

The asymptote is 6⁻ˣ → 0 and f(x) → 3 as x → ∞

The horizontal asymptote is y = 3

The y-intercept is f(x) = 6⁻⁰ + 3 = 4

The y-intercept is (0, 4)

e) f(x) = 2^{x + 3} - 1

The asymptote is 2^{x + 3}  → 0 and f(x) → -1 as x → -∞

The horizontal asymptote is y = -1

The y-intercept is f(x) =  2^{0 + 3} - 1 = 7

The y-intercept is (0, 7)

When f(x) = 0, 2^{x + 3} - 1 = 0

2^{x + 3} = 1

x + 3 = 0, x = -3

The x-intercept is (-3, 0)

f) f(x) = \left (\dfrac{1}{2} \right)^{x - 2} + 2

The asymptote is \left (\dfrac{1}{2} \right)^{x - 2} → 0 and f(x) → 2 as x → ∞

The asymptote is y = 2

The y-intercept is f(x) = f(0) = \left (\dfrac{1}{2} \right)^{0 - 2} + 2 = 6

The y-intercept is (0, 6)

7 0
2 years ago
What is the radius of a circle with an area of 113.04 cubic inches? Use 3.14 for pi.
frosja888 [35]

Answer:

I am deeply sorry for the late answer

but the answer is The answer is

r ≈ 6

Step-by-step explanation:

The answer is r ≈ 6


A ≈ 113.04 cubic in.  

this means that  

d ≈ 12

and

C ≈ 37.69

Therefore

r ≈ 6

Hope this helped.

6 0
3 years ago
A rectangular plot of land is to be enclosed by a fence. One side is along a river, and does not need to be enclosed. If the tot
Bad White [126]

Answer:

Width 150 meters.

Length = 350 meters.

Step-by-step explanation:

Let us assume the  width of  the fence = k meters

So, both sides = k + k = 2k meters

Also, the TOTAL fencing length = 600 m

So, the one side length of the fence = (600 - 2 k) meters

AREA = LENGTH x WIDTH

⇒ A(k)  = (600 - 2k) (k)

or, A =  -2k² + 600 k

The above equation is of the form: ax² +bx +  C

Here: a  = - 2 , b = 600 and C = 0

As a< 0, the parabola opens DOWNWARDS.

Here, x value is given as:  x = \frac{-b}{2a}

Solving for the value of k similarly, we get:

k = \frac{-b}{2a}  = \frac{600}{2(-2)}  = 150

Thus the desired width = k = 150 meters

So, the desired dimensions of the plot  is width 150 meters.

And length = 650 - 2k = 650 - 300 = 350 meters.

5 0
3 years ago
Can somebody tell me how to solve these kind of problems?
Ugo [173]

Answer: 918km^2

Step-by-step explanation:

area(a): 1700km^2\\percentage(p): 5.75 = \frac{5.75}{100}=0.0575\\years(y): 8

Let x be the new dimension of the area;

x=a-(a*p*y)

x=(1700km^2)-(1700km^2*0.0575*8)\\x=1700km^2-782km^2\\x=918km^2

6 0
3 years ago
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