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3 years ago

A.3/4-3/10 Do this for me please and this B.3 1/2- 1 1/3

Mathematics

1 answer:

rosijanka [135]3 years ago

7 0

Answer:

A: 3/4 - 3/10=9/20

B: 3 1/2 - 1 1/3=2 1/6

Decimal form

A: 0.45

B: 2.166667

Step-by-step explanation:

Please mark brainlliest if you want but please do. Also mark 5 stars if correct.

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If fixed cost of 20 article is 500 and variable cost for each article is 40. Find the total cost function? Also ,find the total

Deffense [45]

Answer:

Cost function = 40a + 500

Cost of 90 articles = $4,100

Step-by-step explanation:

The fixed cost is $500 and it will.not change regardless of production level.

The Variable cost is $40 and increases by every additional unit produced.

Assume the number of articles produced is a.

Cost function would be:

Total cost = Variable cost * Number of articles + Fixed cost

TC = 40a + 500

Using this, the cost of 90 articles is:

= 40 * 90 + 500

= $4,100

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3 years ago

What is the greatest common factor of 30, 48, and 2?

arlik [135]

Answer:

<h3>The greatest common factor of 30, 48 and 2 is 2</h3>

Step-by-step explanation:

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3 years ago

Find all solutions by factoring.<br> c? + 13c + 40 = 0

Karolina [17]

Answer: c = -5, -8

Step-by-step explanation:

c^2 +13c+40=0

(c+5)(c+8)=40

c = -5, -8

3 0

3 years ago

Someone help me please?

gavmur [86]

The answer would be Acute angles

7 0

3 years ago

Let A = { x ∈ R : x 2 − 5 x + 4 ≤ 0 } , B = (3 , 5), and C = (3 , 4]. Show that A ∩ B = C . (Recall, you must show two separate

kolbaska11 [484]

Answer:

You can proceed as follows:

Step-by-step explanation:

First solve the quadratic inequality $x^{2}-5x+4\leq 0$ . To do that, factorize, then we have that $(x-4)(x-1)\leq 0$ . This implies that

$x-1\leq 0\, \text{and}\, x-4\geq 0$

$x-1 \geq 0\, \text{and}\, x-4\leq 0$

In the first case the solution is the empty set $\emptyset$ . In the second case the solution is the interval $1\leq x \leq 4$ . Now we have that

$A=[1,4]$

$B=(3,5)$

$C=(3,4]$ .

To show that $A\cap B\subseteq C$ consider $x\in A\cap B$ . Then $1\leq x \leq 4\, \text{and}\, 1$ , this implies that $3$ , then $x\in C$ . Now, to show that $C\subseteq A\cap B$ consider $x\in C$ , then $3$ , then $1\leq x \leq 4\, \text{and}\, 3$ , then $x\in [1,4] \, \text{and}\, x\in (3,5)$ , this implies that $x\in A\cap B$ .

Observe the image below.

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3 years ago