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Ray Of Light [21]
3 years ago
7

3(5+6) ??? its distributive property

Mathematics
2 answers:
Luden [163]3 years ago
6 0
To answer this we will use the distributive poverty. Lets do it:-

3(5+6)
3 · 5 = 15
3 · 6 = 18
15 + 18 = 33

So, 3(5+6) = 33.

Hope I helped ya!! 
kaheart [24]3 years ago
4 0
Yes, we apply the distributive property to simplify this expression.

3*5 = 15
6*3 = 18

15 + 18 is the final answer.
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7 0
2 years ago
Identify the x-intercepts of the function below f(x)=x^2+12x+24
damaskus [11]

<u>ANSWER:  </u>

x-intercepts of  \mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})

<u>SOLUTION:</u>

Given, f(x)=x^{2}+12 x+24 -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Here, for (1) a = 1, b= 12 and c = 24

X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}

\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}

Hence the x-intercepts of  \mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})

8 0
2 years ago
Chapter 11 assesment guide 5th grade
Alex73 [517]

Answer:

i need more info

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
a 240 centimeter board is to be cut into two pieces so that the longer piece is twice as long as the shorter piece. find the len
amid [387]
You need two pieces and 1 has to be 2x longer than the other 
set x as the smaller piece (easier to always go with the smaller value as x)
the other piece is 2x (double in size)
your equations would be 3x=240>>> x= 60

so smaller piece/ x= 60
bigger piece/ 2x= 120
7 0
3 years ago
Read 2 more answers
To two decimal places, find the value of k that will make the function f(x) continuous everywhere. f of x equals the quantity 3x
Fiesta28 [93]

Given function is

f(x)=\left\{\begin{matrix}3x+k & x\leq -4 \\ kx^2-5 & x> -4\end{matrix}\right.

now we need to find the value of k such that function f(x) continuous everywhere.

We know that any function f(x) is continuous at point x=a if left hand limit and right hand limits at the point x=a are equal.

So we just need to find both left and right hand limits then set equal to each other to find the value of k

To find the left hand limit (LHD) we plug x=-4 into 3x+k

so LHD= 3(-4)+k

To find the Right hand limit (RHD) we plug x=-4 into

kx^2-5

so RHD= k(-4)^2-5

Now set both equal

k(-4)^2-5=3(-4)+k

16k-5=-12+k

16k-k=-12+5

15k=-7

k=-\frac{7}{15}

k=-0.47

<u>Hence final answer is -0.47.</u>




7 0
3 years ago
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