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3 years ago

rob needs 345 centimeters of tape to wrap packages. how many rolls of tape should rob buy if each roll holds 1 meter of tape

1 answer:

8 0

If there are 100 cm in a meter and he had 345 cm to cover 345 divided by 100 is 3.45, Rob should buy 4 rolls to have enough to wrap the packages.

HOPE THIS COULD HELP.

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The question for the graph has a leading coefficient that is___. What are the possible degrees for the equation of this graph

mrs_skeptik [129]

Answer: Part one (NEGATIVE) Part two is C (EVEN DEGREES OF FOUR OR GREATER)

Step-by-step explanation:

3 0

3 years ago

Solve for u.<br> -6=-2u +4(u-5)

sergij07 [2.7K]

Answer: u=7

Step-by-step explanation:

-6= -2u+4(u-5)

-6= -2u+4u-20

-6= 2u-20

14= 2u

7=u

5 0

3 years ago

Read 2 more answers

1.How many combinations are possible from the letters HOLIDAY if the letters are taken?

Lesechka [4]

Using the combination formula, it is found that:

1. A. 7 combinations are possible.

B. 21 combinations are possible.

C. 1 combination is possible.

2. There are 245 ways to group them.

<h3>What is the combination formula?</h3>

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by:

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Exercise 1, item a:

One letter from a set of 7, hence:

$C_{7,1} = \frac{7!}{1!6!} = 7$

7 combinations are possible.

Item b:

Two letters from a set of 7, hence:

$C_{7,2} = \frac{7!}{2!5!} = 21$

21 combinations are possible.

Item c:

7 letters from a set of 7, hence:

$C_{7,7} = \frac{7!}{0!7!} = 1$

1 combination is possible.

Question 2:

Three singers are taken from a set of 7, and four dances from a set of 10, hence:

$T = C_{7,3}C_{10,4} = \frac{7!}{3!4!} \times \frac{10!}{4!6!} = 245$

There are 245 ways to group them.

More can be learned about the combination formula at brainly.com/question/25821700

6 0

1 year ago

Kendra was given this system of equations.

iren2701 [21]

Answer:

Step-by-step explanation:

hope u enjoy

4 0

2 years ago

Describe the graph of y={1/(2x-10)}-3 compared to the graph of y=1/x

tester [92]

$\bf ~~~~~~~~~~~~\textit{function transformations}
\\\\\\
% templates
f(x)= A( Bx+ C)+ D
\\\\
~~~~y= A( Bx+ C)+ D
\\\\
f(x)= A\sqrt{ Bx+ C}+ D
\\\\
f(x)= A(\mathbb{R})^{ Bx+ C}+ D
\\\\
f(x)= A sin\left( B x+ C \right)+ D
\\\\
--------------------$

$\bf \bullet \textit{ stretches or shrinks horizontally by } A\cdot B\\\\
\bullet \textit{ flips it upside-down if } A\textit{ is negative}\\
~~~~~~\textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if } B\textit{ is negative}$

$\bf ~~~~~~\textit{reflection over the y-axis}
\\\\
\bullet \textit{ horizontal shift by }\frac{ C}{ B}\\
~~~~~~if\ \frac{ C}{ B}\textit{ is negative, to the right}\\\\
~~~~~~if\ \frac{ C}{ B}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by } D\\
~~~~~~if\ D\textit{ is negative, downwards}\\\\
~~~~~~if\ D\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{ B}$

with that template in mind, let's check these two

$\bf \stackrel{parent}{y=\cfrac{1}{x}}\qquad \qquad\qquad \qquad \stackrel{transformed}{y=\cfrac{1}{\stackrel{B}{2}x\stackrel{C}{-10}}\stackrel{D}{-3}}\\\\
-------------------------------\\\\
B=2\qquad \textit{shrinks horizontally by }\frac{1}{2}
\\\\\\
C=-10\qquad \cfrac{C}{B}=\cfrac{-10}{2}\implies -5\qquad \textit{horizontally right-shifted by }5
\\\\\\
D=-3\qquad \textit{vertically down-shifted by }3$

7 0

2 years ago