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3 years ago

Follow the order of operations to find the value of the expression 6+(9÷3)

Mathematics

2 answers:

dedylja [7]3 years ago

6 0

Answer:

The answer is 9 (hope this helps) (brainliest please thx)

Step-by-step explanation:

PEMDAS is order of operations

p=parentheses

e=exponents

m and d = multiplication of which comes first

a and s = addition and subtraction of which comes first

now we had division and addition

we need to do division first since we follow the order

9 divided by 3 is 3

then we add with 6 and get 9

Sliva [168]3 years ago

4 0

Answer:

Step-by-step explanation:

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What is the kinetic energy of a 1500 kg object moving at a velocity of 10 m/s?

DiKsa [7]

Answer: The Kinetic Energy Formula is: KE = mass x velocity^2 / 2

Plug in the correct numbers into the variables

KE = 1500 kg x 10 m/s ^2 / 2

Square the 10 m/s

KE = 1500 kg x 100 m/s / 2

Multiply

KE = 150,000 / 2

Divide

KE = 75,000 Newton-meters or joules

Step-by-step explanation:

7 0

2 years ago

4) Find the m<5, m<7, and the m<6 in the diagram below.

bazaltina [42]

Answer:

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Step-by-step explanation:

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8 0

2 years ago

Find the number of elements in A 1 ∪ A 2 ∪ A 3 if there are 200 elements in A 1 , 1000 in A 2 , and 5, 000 in A 3 if (a) A 1 ⊆ A

lina2011 [118]

Answer:

a. 4600

b. 6200

c. 6193

Step-by-step explanation:

Let $n(A)$ the number of elements in A.

Remember, the number of elements in $A_1 \cup A_2 \cup A_3$ satisfies

$n(A_1 \cup A_2 \cup A_3)=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)$

Then,

a) If $A_1\subseteq A_2, n(A_1 \cap A_2)=n(A_1)=200$ , and if $A_2\subseteq A_3, n(A_2\cap A_3)=n(A_2)=1000$

Since $A_1\subseteq A_2\; and \; A_2\subseteq A_3, \; then \; A_1\cap A_2 \cap A_3= A_1$

$n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-200-200-1000-200=4600$

b) Since the sets are pairwise disjoint

$n(A_1 \cup A_2 \cup A_3)=\\n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\200+1000+5000-0-0-0-0=6200$

c) Since there are two elements in common to each pair of sets and one element in all three sets, then

$n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-2-2-2-1=6193$

8 0

3 years ago

Kira looked through online census information to determine the average number of people living in the homes in her city. What is

Goshia [24]

First we need to identify if the data is qualitative or quantitative.

The data is average number of people living in the homes.

Qualitative data as its name indicates is an attribute or characteristic. It can not be measured e.g color. Quantitative data is such a data which can be counted or measured.

Since the average number of people can be counted and measured, the data is Quantitative.

In an observational study the individuals are observed. In the given case, Kira did not observed the individuals to gather the data, rather she used an Online resource to gather the data.

Therefore, the correct answer will be:
Kira used published data that is quantitative.

4 0

3 years ago

Read 2 more answers

20 feet of ribbon, cutting ribbon into 7 1/2 inches or 6 3/4 inch lengths to make bracelets. Write an algebraic expression that

IRINA_888 [86]

<u>ANSWER: </u>

The algebraic expression for number of ribbons is $\frac{20 \text { feet }-\left\{\frac{20 \text { feet }}{x}\right\}}{x}$ and the algebraic expression for length of ribbon is $\frac{25 x}{3} \text { feet }$

<u>SOLUTION: </u>

Given, 20 feet of ribbon, cutting ribbon into $7\frac{1}{2}$ inches or $6\frac{3}{4}$ inch lengths to make bracelets.

Let us convert the mixed fraction to improper fractions.

$7 \frac{1}{2}=\frac{7 \times 2+1}{2}=\frac{15}{2} \text { and } 6 \frac{3}{4}=\frac{6 \times 4+3}{4}=\frac{27}{4}$

Let, the length of bracelets can be made be “x”

First we have to remove the excess length of ribbon and then we have to divide the remaining with length of bracelet we want.

$\text { number of ribbons }=\frac{\text { length of ribbon-excess ribbon.}}{\text {length of bracelet}}$

$\left.=\frac{20 \text { feet }-\left\{\frac{20 \text { feep }}{x}\right.}{x} \text { [we know that, }\{x\} \text { is fractional part of } x\right$ ]

So, the algebraic expression for number of ribbons is $\frac{20 \text { feet }-\left\{\frac{20 \text { feet }}{x}\right\}}{x}$

Now, let us find the algebraic expression for feet of ribbon to make 100 ribbons.

We have 100 bracelets of length x, then total length = 100 × x

length of ribbon = 100x inches

$\begin{array}{l}{=100 \mathrm{x} \times \frac{1}{12} \text { feet }} \\\\ {=\frac{25 x}{3} \text { feet }}\end{array}$

So, the algebraic expression for length of ribbon is $\frac{25 x}{3} \text { feet }$

Hence, the algebraic expression for number of ribbons is $\frac{20 \text { feet }-\left\{\frac{20 \text { feet }}{x}\right\}}{x}$ and the algebraic expression for length of ribbon is $\frac{25 x}{3} \text { feet }$

5 0

3 years ago

Follow the order of operations to find the value of the expression 6+(9÷3)​

Follow the order of operations to find the value of the expression 6+(9÷3)