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anastassius [24]

3 years ago

5

Mrs. Ramos has 37 pictures hanging up in her classroom. She has 24 pictures in the hallway. 18 of those pictures are of students

. How many are not of students?

1 answer:

Vikentia [17]3 years ago

7 0

Answer: 43

Step-by-step explanation: 37 + 24 = 61

61 - 18 = 43

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Plzzzzzzzzzzzzzzzzzzzzzz help me nowwwwwwwwwwwwwwwww

Alexus [3.1K]

Answer: C.

Step-by-step explanation:

3 0

3 years ago

Mrs. Galicia has a cupcake company. The amount of money earned is represented by

sashaice [31]

Using the profit function, it is found that:

a) She earns $36 after 25 years.

b) The function was shifted right 4 units and up 1 units.

<h3>What is the profit function?</h3>

The profit, in x years after 2015, is given by:

$p(x) = 12\sqrt[3]{x + 2}$

Item a:

She earns $36 after x years, considering x for which p(x) = 36, hence:

$p(x) = 12\sqrt[3]{x + 2}$

$36 = 12\sqrt[3]{x + 2}$

$\sqrt[3]{x + 2} = 3$

$(\sqrt[3]{x + 2})^3 = 3^3$

$x + 2 = 27$

$x = 25$

She earns $36 after 25 years.

Item b:

The new function is:

$h(x) = 12\sqrt[3]{x - 2} + 1$

At the domain, we have that $x + 2 \rightarrow x - 2$ , that is, $x \rightarrow x - 4$ , hence it was shifted right 4 units.

At the range, we have that $h(x) = h(x - 4) + 1$ , hence it was shifted up 1 unit.

You can learn more about functions at brainly.com/question/24380382

6 0

2 years ago

Rectangle ABCD is transformed to A’B’C’D’ by a dilation. What is the scale factor used for the dilation?

jekas [21]

Answer:

Scale factor = 3

Step-by-step explanation:

Scale factor used for the dilation = the ratio of the corresponding side length of the new figure to the side length of the original figure

Scale factor = B'C'/BC

B'C' = 9 units

BC = 3 units

Scale factor = 9/3 = 3

6 0

2 years ago

What is 4(-2)+(-3)(-5)?

Svetlanka [38]

I say that the answer is 7 bc it’s -8+15

8 0

2 years ago

Read 2 more answers

A^2 + b^2 + c^2 = 2(a − b − c) − 3. (1) Calculate the value of 2a − 3b + 4c.

Verdich [7]

Answer:

$2a - 3b + 4c = 1$

Step-by-step explanation:

Given

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Required

Determine $2a - 3b + 4c$

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Open bracket

$a^2 + b^2 + c^2 = 2a - 2b - 2c - 3$

Equate the equation to 0

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 3 = 0$

Express 3 as 1 + 1 + 1

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 1 + 1 + 1 = 0$

Collect like terms

$a^2 - 2a + 1 + b^2 + 2b + 1 + c^2 + 2c + 1 = 0$

Group each terms

$(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

Factorize (starting with the first bracket)

$(a^2 - a -a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$(a(a - 1) -1(a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1) (a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + b+b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b(b + 1)+1(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + c+c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c(c + 1)+1(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)^2) = 0$

Express 0 as 0 + 0 + 0

$(a - 1)^2 + (b + 1)^2 + (c + 1)^2 = 0 + 0+ 0$

By comparison

$(a - 1)^2 = 0$

$(b + 1)^2 = 0$

$(c + 1)^2 = 0$

Solving for $(a - 1)^2 = 0$

Take square root of both sides

$a - 1 = 0$

Add 1 to both sides

$a - 1 + 1 = 0 + 1$

$a = 1$

Solving for $(b + 1)^2 = 0$

Take square root of both sides

$b + 1 = 0$

Subtract 1 from both sides

$b + 1 - 1 = 0 - 1$

$b = -1$

Solving for $(c + 1)^2 = 0$

Take square root of both sides

$c + 1 = 0$

Subtract 1 from both sides

$c + 1 - 1 = 0 - 1$

$c = -1$

Substitute the values of a, b and c in $2a - 3b + 4c$

$2a - 3b + 4c = 2(1) - 3(-1) + 4(-1)$

$2a - 3b + 4c = 2 +3 -4$

$2a - 3b + 4c = 1$

7 0

3 years ago