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3 years ago

Hi I NEED HELP NOW

Mathematics

1 answer:

MrMuchimi3 years ago

5 0

1. 2/2 - 1/2=1/2
2. 3/4-2/4=1/4
3. 2/4-1/4=1/4
4. 4/6-2/6=2/6

Are you you learning about subtracting fractions?
I’m pretty sure this is it, and you would shade in the parts that are left in the other circles.

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Find the limit of the function by using direct substitution. limit as x approaches zero of quantity x squared minus one.

Allisa [31]

The given function is:
x^2 - 1
We want to calculate the limit of this function as x approaches zero. To do so, we will use direct substitution.
We will substitute the x with 0 in the given function to calculate its limit as follows:
Limit as x approaches 0 = (x)^2 - 1 = -1

Therefore, the correct choice is:
-1

4 0

4 years ago

The manufacturing cost of a cell phone was $136. Sellers first marked up the cost by 25 percent. Sales were poor, so sellers dec

Helen [10]

Manufacturing cost = $136
sellers added 25% = $ 136 * 25/100 = $34
manufacturers cost + 25% = $136 + $34 = $170.
old selling price = $170

Sellers decreased the markup to 17% (of manufacturing cost): $136 x 17/100 = $23.12
New selling price = $136 + $23.12 = $153.12

both of these answers show: original cost + markup value

6 0

3 years ago

Pls answer! I don’t know math @ all

Lana71 [14]

The probability that the outcome is a sum that is a multiple of 6 or a sum that is a multiple of 4 is $\frac{5}{12}$ .

Solution:

Total number of outcomes N(S) = 36

Let A be the sum that is a multiple of 6 and

B be the sum that is a multiple of 4.

Sum that is a multiple of 6 = (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6)

N(A) = 6

Sum that is a multiple of 4 = (1, 3), (2, 2), (2, 6), (3, 1), (3, 5),

(4, 4), (5, 3), (6, 2), (6, 6)

N(B) = 9

$$P(A)=\frac{N(A)}{N(S)}$

$$P(A)=\frac{6}{36}=\frac{1}{6}$

$$P(B)=\frac{N(B)}{N(S)}$

$$P(B)=\frac{9}{36}=\frac{1}{4}$

Probability that the outcome is a sum that is a multiple of 6 or a sum that is a multiple of 4:

$P(A \cup B)=P(A)+P(B)$

$$P(A \cup B)=\frac{1}{6} +\frac{1}{4}$

$$=\frac{2+3}{12}$ (Make the denominator same)

$$=\frac{5}{12}$

Hence the probability that the outcome is a sum that is a multiple of 6 or a sum that is a multiple of 4 is $\frac{5}{12}$ .

8 0

3 years ago

2/3÷2⁴+(3/4+1/6)÷1/3<br> How do you figure this out

Alecsey [184]

Answer: $\frac{67}{24}$

Step-by-step explanation:

$\frac{2}{3} /2^{4}+(\frac{3}{4} +\frac{1}{6})/\frac{1}{3} \\\\\frac{2}{3} /2^{4}+ \frac{11}{12} /\frac{1}{3} = \frac{2}{3} /16+ \frac{11}{12} /\frac{1}{3}$