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yanalaym [24]
3 years ago
7

(4x+5)(4x−5) I'm stuck can someone please help me ​

Mathematics
2 answers:
Marizza181 [45]3 years ago
5 0

Answer:6small2−25

Step-by-step explanation:

Vesnalui [34]3 years ago
3 0

Answer:

16x² - 25

Step-by-step explanation:

(4x+5)(4x−5)

16x² - 20x + 20x - 25

16x² - 25

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Find the slope of the line that passes through (3, 6) and (8, 7)
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Answer:

The slope of this line is <u>1/5</u>

Step-by-step explanation:

You can find the answer easily by graphing the problem. In doing so, you will find the slope (rise over run) to be one over five. This is because you rise one unit, and run five.

8 0
2 years ago
Sue needs 1 and 1/4 flour for a batch of cookies How many batches can she make with 9 cup of flour
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3 years ago
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
If the diameter of the sphere is 12cm what is the volume of the cylinder
german

r^{2}  \times \pi \times (h = height \: of \: the \: cilinder)
6 0
3 years ago
Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul
natali 33 [55]

Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

\bf B=\displaystyle\frac{L}{4\pi d^2}

where

<em>L = luminosity of the star (related to the Sun) </em>

<em>d = distance in ly (light-years) </em>

The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.

Hence the apparent brightness of  Alpha Centauri A is

\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728

According to the inverse square law for light intensity

\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}

where

\bf I_1= light intensity at distance \bf d_1  

\bf I_2= light intensity at distance \bf d_2  

Let \bf d_2  be the distance we would have to place the 50-watt bulb, then replacing in the formula

\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

7 0
3 years ago
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