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2 years ago

Which is a possible leading term for the graph shown? 2x3 –2x3 2x4 –2x4

Mathematics

2 answers:

r-ruslan [8.4K]2 years ago

6 0

Answer:

C. 2x^4

Step-by-step explanation:

Just took it. Edg 2020. Hope this helps :)

Y_Kistochka [10]2 years ago

4 0

Answer:

2x^4

Step-by-step explanation:

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2. Ashley had a summer lemonade stand where she sold small cups of lemonade for $1.25 and large

Drupady [299]

Answer:

$\boxed {\boxed {\sf 98 \ Small \ Cups}}\\\\\boxed {\boxed {\sf 57 \ Large \ Cups}}$

Step-by-step explanation:

Let's make a system of equations.

First, define the variables.

Let s= small cups and g= large cups

Next, make 2 equations. 1 for the money and 1 for the number of cups.

<u>Money</u>

A small cup costs 1.25 and a large cup costs 2.50. The total money made is 265.
1.25s+2.50g=265

<u>Number</u>

A total of 155 cups of lemonade were sold. The sum of small cups and large cups sold will equal 155.
s+g=155

Here is our system of equations:

$1.25s+2.50g=265 \\s+g=155$

Now, solve. Let's isolate a variable in the 2nd equation, so we can plug it in after.

Subtract g from both sides of the equation.
s+g=155
s+g-g=155-g
s=155-g

Now we know that s is equal to 155-g. We can substitute (155-g) in for s in the first equation.

$1.25s +2.50g=265$

$1.25(155-g)+2.50g=265$

Distribute the 1.25

1.25(155-g)= (1.25*155)+ (1.25*-g)= 193.75-1.25g

$193.75-1.25g+2.50g=265$

Work to isolate the variable. Subtract 193.75 from both sides of the equation and combine like terms on the left.

$193.75-1.25g+2.50g=265-193.75$ (Subtract 193.75)

$-1.25g+2.50g=265-193.75$

$1.25g=71.25$ (Combine like terms).

1.25 and g are being multiplied. The inverse of multiplication is division. Divide both sides by 1.25

$1.25g/1.25=71.25/1.25\\g=71.25/1.25\\g=57$

Now we know that 57 large cups were sold. Plug 57 back into the original second equation to find the small cups.

$s+g=155$

$s+57=155$

Subtract 57 from both sides of the equation to isolate the variable, s.

$s+57-57=155-57\\s=155-57\\s=98$

Ashley sold 57 large cups and 98 small cups.

5 0

3 years ago

Read 2 more answers

Find the slope and y-intercept of equation y= 10x + 18

pickupchik [31]

Answer:

slope is 10

y-intercept is 18

Step-by-step explanation:

3 0

2 years ago

Find the perimeter of quadrilateral ABCD with vertices A=(1,1) B=(4,4) C=(7,1) D=(4,-2). Explain your answer.

Vanyuwa [196]

we have the coordinates of ABCD

there are 2 ways of doing this one is you find the distance of ab, bc, cd and da and then add the answers together. there is another way but i dont remember it properly so i am just gonna do it this way

distance between 2 point = $\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}$

$d = \sqrt {(4 - 1)^2 + (4 - 1)^2}$

$d = \sqrt {(3)^2 + (3)^2}$

$d = \sqrt {9 + 9}$

$d = \sqrt 18$

AB = 4.24

now repeat the same with all

BC= 4.24

CD= 4.24

DA= 4.24

so its a square with all the same sides so perimeter is 4 x 4.24

perimeter is 16.96

hope this helps

mark me as brainliest please

3 0

2 years ago

All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be

Aneli [31]

Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a student is proficient in reading but not mathematics and $A \cap B$ is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

$B = b + (A \cap B)$

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

$(A \cup B) + C = 1$