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4 years ago

I estimated my bag of chips would have 20 chips, but it really had 16. Was was my percent error ? Round to the nearest percent

1 answer:

3 0

Ok so............ you take 16 and 20 and divide them both into 4.. .....you get 4 and 5 right? Then, you take 4 and 5 and put them into a fraction....4/5! Ok so then you you calculate 4/5 into a percent which is 80% !!!

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Find the coordinates of the midpoint of TU for T(3,1) and U(5,3)

bixtya [17]

<h3>Answer: (4,2)</h3>

==========================================================

The given points are T(3,1) and U(5,3)

The x coordinates of these given points are 3 and 5. Add them up to get 3+5 = 8. Then cut this in half getting 8/2 = 4. This is the x coordinate of the midpoint.

The y coordinates are 1 and 3. They add to 1+3 = 4. Then that cuts in half to get 4/2 = 2. This is the y coordinate of the midpoint.

The midpoint is (4,2)

Effectively, we just used the midpoint formula

$(x_m, y_m) = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)$

In short: add the corresponding coordinates, then divide by 2.

3 0

3 years ago

How many sig figs does 602 200 000 000 000 000 000 000 molecules have ? Explain

lapo4ka [179]

Answer:

Step-by-step explanation:because 6022 is the sig figs, and 00000000000000000000 doesnt work because it is zero

6 0

3 years ago

Divide:(10x²-3x+4)÷(2x-5)

Helen [10]

Answer:

$\frac{10x^2-3x+4}{2x-5}=5x+11+\frac{59}{2x-5}$

Quotient: 5x+11

Remainder: 59

Step-by-step explanation:

I'm going to do long division.

The bottom goes on the outside and the top goes in the inside.

Setup:

---------------------------------

2x-5 | 10x^2 -3x +4

Starting the problem from the setup:

5x +11 (I put 5x on top because 5x(2x)=10x^2)

--------------------------------- (We are going to distribute 5x to the divisor)

2x-5 | 10x^2 -3x +4

-(10x^2 -25x) (We are now going to subtract to see what's left.)

-----------------------------------

22x +4 (I know 2x goes into 22x, 11 times.)

( I have put +11 on top as a result.)

-(22x -55) (I distribute 11 to the divisor.)

-----------------------

59 (We are done since the divisor is higher degree.)

The quotient is 5x+11.

The remainder is 59.

The result of the division is equal to:

$5x+11+\frac{59}{2x-5}$ .

We can actually use synthetic division as well since the denominator is linear.

Let's solve 2x-5=0 to find what to put on the outside of the synthetic division setup:

2x-5=0

Add 5 on both:

2x=5

Divide both sides by 2:

x=5/2

Or realize that 2x-5 is the same as 2(x-(5/2)) which you will have to do anyways if you choose this route:

So 5/2 will go on the outside:

5/2 | 10 -3 4

| 25 55

------------------------------

10 22 59

So we have:

$\frac{10x^2-3x+4}{2x-5}$

$=\frac{10x^2-3x+4}{2(x-\frac{5}{2})}=\frac{1}{2} \cdot \frac{10x^2-3x+4}{x-\frac{5}{2}}=\frac{1}{2}(10x+22+\frac{59}{x-\frac{5}{2}})$

Distribute the 1/2 back:

$\frac{10x^2-3x+4}{2x-5}=\frac{10x+22}{2}+\frac{59}{2(x-\frac{5}{2})}$

$\frac{10x^2-3x+4}{2x-5}=5x+11+\frac{59}{2x-5}$

3 0

4 years ago

Given a function I and a subset A of its domain, let I(A) represent the range of lover the set A; that is, I(A) = {I(x) : x E A}

Ede4ka [16]

Step-by-step explanation:

(a)

Using the definition given from the problem

$f(A) = \{x^2 \, : \, x \in [0,2]\} = [0,4]\\f(B) = \{x^2 \, : \, x \in [1,4]\} = [1,16]\\f(A) \cap f(B) = [1,4] = f(A \cap B)\\$

Therefore it is true for intersection. Now for union, we have that

$A \cup B = [0,4]\\f(A\cup B ) = [0,16]\\f(A) = [0,4]\\f(B)= [1,16]\\f(A) \cup f(B) = [0,16]$

Therefore, for this case, it would be true that $f(A\cup B) = f(A)\cup f(B)$ .

(b)

1 is not a set.

(c)

To begin with

$A\cap B \subset A,B$

Therefore

$g(A\cap B) \subset g(A) \cap g(B)$

Now, given an element of $g(A) \cap g(B)$ it will belong to both sets, therefore it also belongs to $g(A\cap B)$ , and you would have that

$g(A)\cap g(B) \subset g(A \cap B)$ , therefore $g(A)\cap g(B) = g(A \cap B)$ .

(d)

To begin with $A,B \subset A \cup B$ , therefore

$g(A) \cup g(b) \subset g(A\cup B)$

7 0

4 years ago

Picture is shown above<br>suppose that y varies directly with x, and y=12 when x=15

a_sh-v [17]

Step-by-step explanation:

Since y is directly proportional to x,

we have y = kx, where k is a real constant.

When x = 15, y = 12.

=> (12) = k(15), k = 0.8.

Therefore we have y = 0.8x.

When x = 2, y = 0.8(2) = 1.6.

The answer is y = 1.6 when x = 2.

7 0

3 years ago

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