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kupik [55]
2 years ago
7

What's the answer to this question?

Mathematics
1 answer:
o-na [289]2 years ago
4 0
The average number of shots is 4 shots

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Simplify #53: 3(x + 2) - 4x<br><br><br><br> Help please
labwork [276]

Answer:

-x+6

Step-by-step explanation:

3(x+2)-4x

3x+6-4x

-x+6

8 0
2 years ago
Can someone please help me find the simple interest
Stells [14]

\bf ~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$300\\ r=rate\to 4\%\to \frac{4}{100}\dotfill &0.04\\ t=years\dotfill &3 \end{cases} \\\\\\ I=(300)(0.04)(3)\implies I=36

5 0
3 years ago
Simplify 3x + 5x + 14x
navik [9.2K]
Simply collect all the like terms together
So..3+5+14 is 22x
3 0
3 years ago
Read 2 more answers
I need help on this loll
Serhud [2]

Answer: The translation rule that maps point D ( 7 , − 3 ) onto point D ' ( 2 , 5 )

is (x , y) → (x - 5 , y + 8)

Step-by-step explanation:

Let us revise the translation

If the point (x , y) translated horizontally to the right by h units  then its image is (x + h , y)

If the point (x , y) translated horizontally to the left by h units  then its image is (x - h , y)

If the point (x , y) translated vertically up by k units  then its image is (x , y + k)

If the point (x , y) translated vertically down by k units  then its image is (x , y - k)

(x , y) → (x ± h , y ± k) the right arrow symbol used to show the

translation from a point to its image

Example:

∵ P (0 , 0) → P' (1 , 2)

∴ The rule is (x , y) → (x + 1 , y + 2)

Let us find the translation rule that maps point D ( 7 , − 3 ) onto

point D' (2 , 5)

∵ Point (x , y) = (7 , -3)

∵ Its image after translation (x + h , y + k) = (2 , 5)

∴ x + h = 2

∵ x = 7

∴ 7 + h = 2

- Subtract 7 from both sides

∴ h = -5

∵ y + k = 5

∵ y = -3

∴ -3 + k = 5

- Add 3 to both sides

∴ k = 8

∴ The rule of translation is (x , y) → (x - 5 , y + 8)

The translation rule that maps point D ( 7 , − 3 ) onto point D ' ( 2 , 5 )

is (x , y) → (x - 5 , y + 8)

3 0
2 years ago
Construct a polynomial function with the stated properties. Reduce all fractions to lowest terms. Third-degree, with zeros of −3
RSB [31]

Answer:

\Large \boxed{-\dfrac{5}{4}(x+3)(x+1)(x-2)}

Step-by-step explanation:

Hello,

Based on the indication, we can write this polynomial as below, k being a real number that we will have to identify (degree = 3 and we have three zeroes -3, -1, and 2).

   \Large \boxed{k(x+3)(x+1)(x-2)}

We know that the point (1,10) is on the graph of this function, so we can say.

k(1+3)(1+1)(1-2)=10}\\\\4*2*(-1)*k=10\\\\-8k=10\\\\k=\dfrac{10}{-8}=-\dfrac{5}{4}

Then the solution is:

\large \boxed{-\dfrac{5}{4}(x+3)(x+1)(x-2)}

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

8 0
3 years ago
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