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azamat
3 years ago
6

Humberto evaluates the expression 4t2 for t=3. He correctly substitutes 3 for t in the expression, but then says that the value

is 144. However, he is incorrect!
Mathematics
1 answer:
AveGali [126]3 years ago
4 0

Answer:

36

Step-by-step explanation:

He multiplied the four and the three first when he multiplied the equation together. However, according to the order of the operations, he was supposed to square three first. 3^2 = 9. Then multiplied 9*4 = 36.

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Miss Sarah only wants one half of a sandwich she says Mr. tom will need one 16th of a pound of cheese for her sandwich Mr. Tom s
bekas [8.4K]
Aprroximately 1 pound equals 4 cups of cheese. With that in account,

1 lb = 4 cups
1/8 lb = X
X= 4*1/8
X= 4/8
X=1/2 cup
OR
1 lb =4 cups
1/16 lb = X
X=4 * 1/16
X= 4/16
X=1/4 cup

I may go with 1/16 of a pound as that is lesser cheese than 1/8. Hope this helps.
8 0
3 years ago
Read 2 more answers
I am stuck on this question. Could someone please help?​
kkurt [141]

Step-by-step explanation:

-inf<= x <= inf

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8 0
3 years ago
Which of the following shows the expression in factored form x2 + 2x - 8
lukranit [14]

Answer:

A) (x-2)(x+4)

Step-by-step explanation:

4 0
3 years ago
Someone please pass the answers to this
arsen [322]

Answer/Step-by-step explanation:

1. The figure is composed of a triangle and a rectangle.

Area of the triangle = ½*base*height

base = 4 ft

height = 12 - 8 = 4ft

Area of triangle = ½*4*4 = 8 ft²

Area of rectangle = length * width

Length = 8 ft

Width = 4 ft

Area of rectangle = 8*4 = 32 ft²

✔️Area of the figure = 8 + 32 = 40 ft²

2. The figure is composed of a semicircle and a triangle

Area of the semicircle = ½(πr²)

radius (r) = 3 cm

π = 3

Area = ½(3*3²) = 13.5 cm²

Area of triangle = ½*base*height

base = 3*2 = 6 cm

height = 6 cm

Area = ½*6*6 = 6 cm²

✔️Area of the figure = 13.5 + 6 = 19.5 cm²

5 0
3 years ago
Prove De Morgan's law by showing that each side is a subset of the other side by considering x ∈ A⎯⎯⎯ A ¯ ∩ B⎯⎯⎯ B ¯ .
adelina 88 [10]

Solution :

We have to prove that $\overline{A \cup B} = \overline{A} \cap \overline{B}$   (De-Morgan's law)

Let  $x \in \bar{A} \cap \bar{B}, $ then $x \in \bar{A}$ and $x \in \bar{B} $

and so $x \notin \bar{A}$ and $x \notin \bar{B} $.

Thus, $x \notin A \cup B$ and so $x \in \overline{A \cup B}$

Hence, $\bar{A} \cap \bar{B} \subset \overline{A \cup B}$   .........(1)

Now we will show that $\overline{A \cup B} \subset \overline{A} \cap \overline{B}$

Let $x \in \overline{A \cup B}$ ⇒ $x \notin A \cup B$

Thus x is present neither in the set A nor in the set B, so by definition of the union of the sets, by definition of the complement.

$x \in \overline{A}$ and  $x \in \overline{B}$

Therefore, $x \in \overline{A} \cap \overline{B}$ and we have $\overline{A \cup B} \subset \overline{A} \cap \overline{B}$  .............(2)

From (1) and (2),

$\overline{A \cup B} = \overline{A} \cap \overline{B}$

Hence proved.

3 0
3 years ago
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