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xenn [34]
3 years ago
6

CATHERINE'S MOTHER IS WRAPPING A BOX THAT IS 12 INCHES LONG X 8 INCHES WIDE X 10 INCHES IN HIGH. HOW MANY SQUARE INCHES OF WRAPP

ING PAPER WILL CATHERINE NEED TO COVER THE BOX ?
Mathematics
1 answer:
taurus [48]3 years ago
8 0
12 x 8= 96
10 x 8 =80
12 × 10= 120
120+80+96= 296
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Given the parent function of f(x) = x³, what is the value of k in the translated graph of f(x - h) + k?
jekas [21]

The value of k in the translated graph of f(x - h) + k is; k = 2

<h3>How to carry out Translations on Graphs?</h3>

We are given the parent function of f(x) = x³.

Now, the point (0, 0) in the graph of f(x) is the point (3, 2) in the translated graph. Thus;

The rule of translation is;

(x, y) → (x + 3, y + 2)

That means;

The translation is 3 units to the right and 2 units up

The equation of the translated graph is equal to;

f(x) = (x - 3)³ + 2

Thus;

The value of k is equal to 2.

Read more about Graph Translations at; brainly.com/question/4025726

#SPJ1

7 0
2 years ago
What is the coefficient of x2y3 in the expansion of (2x + y)5?
Zigmanuir [339]

Option C:

The coefficient of x^{2} y^{3} is 40.

Solution:

Given expression:

(2 x+y)^{5}

Using binomial theorem:

(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}

Here a=2 x, b=y

Substitute in the binomial formula, we get

(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}

                                                            $+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}

Let us solve the term one by one.

$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}

$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y

$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}

$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}

$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}

$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}

Substitute these into the above expansion.

(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}

The coefficient of x^{2} y^{3} is 40.

Option C is the correct answer.

5 0
3 years ago
Help me fasssssst huryyyy up
GenaCL600 [577]

Answer: 20%

1/(1+4) = the answer

its just the amount of lemon juice over total volume

4 0
3 years ago
Which graph shows the solution set of the compound inequality or 1.5x-1 &gt; 6.5 or 7x+3 &lt; -25 ?
Dmitry [639]
1.
<span>1.5x-1 > 6.5 
1.5x>1+6.5
</span>1.5x>7.5, divide by 1.5

x>5, is represented by the region to the right of the vertical line x=5 

2. 

<span>7x+3 < -25
7x<-25-3
7x<-28, divide by 4:

x<-4

</span>x<-4, is represented by the region to the left of the vertical line x=-4

Answer: check the picture

6 0
3 years ago
Read 2 more answers
Plz help me
Snowcat [4.5K]

Answer:

9/4 as a fraction or 2.25 as a decimal

Step-by-step explanation:

6 0
4 years ago
Read 2 more answers
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