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Romashka-Z-Leto [24]
3 years ago
9

What is the equation of a line with a slope of ​ 1/2 ​ and a point (3, 1) on the line?

Mathematics
2 answers:
miv72 [106K]3 years ago
3 0
Y = mx + b

y = m1/2 +1
Pani-rosa [81]3 years ago
3 0
The slope is 1/2 and the y-intercept would be -1/2 so the answer is y = 1/2x + -1/2
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Will a 9/32 inch diameter pin fit into a reamed hole with a diameter of 0.278 inch? (Do not round off your answer and express it
ollegr [7]

Answer:

9/32 inch diameter pin will not fit into a reamed hole with a diameter of 0.278 inch

Step-by-step explanation:

Diameter of reamed hole = 0.278 inch

Radius of reamed hole = \frac{0.278}{2} =0.139

Surface area of reamed hole = \pi r^2

Surface area of reamed hole = 3.14 \times  (0.139)^2

Surface area of reamed hole = 0.06066794

Diameter of pin = \frac{9}{32}

Radius of pin =\frac{9}{64}

Surface area of pin = \pi r^2 = 3.14 \times ( \frac{9}{64})^2=0.0620

Since the surface area of pin is more than the surface area of hole .

So, a 9/32 inch diameter pin will not fit into a reamed hole with a diameter of 0.278 inch

7 0
4 years ago
108 is 36% of what number? Write and solve a proportion to solve the problem
kolezko [41]

Let the number = x

Then, it would be: (x*36)/100 = 108

x*36 = 10800

x = 10800/36

x = 300

3 0
3 years ago
Find the surface area of the cylinder. Round your answer to the nearest tenth. (Open attachment to see image of the cylinder).
Vaselesa [24]
Formula: SA = (2 × π(pie) × radius × height) + (2 × π(pie) × radius square.)
So the equation will be SA=( 2 × 3.14 × 2 × 4) + ( 2 × 3.14 × 2²)
Then, we will get                                     ↓                            ↓
                                           SA=      50.24               +         25.12
                                           <u>SA =                   75.36 in</u>
<u>²</u>
8 0
3 years ago
Calculus piecewise function. ​
Kipish [7]

Part A

The notation \lim_{x \to 2^{+}}f(x) means that we're approaching x = 2 from the right hand side (aka positive side). This is known as a right hand limit.

So we could start at say x = 2.5 and get closer to 2 by getting to x = 2.4 then to x = 2.3 then 2.2, 2.1, 2.01, 2.001, etc

We don't actually arrive at x = 2 itself. We simply move closer and closer.

Since we're on the positive or right hand side of 2, this means we go with the rule involving x > 2

Therefore f(x) = (x/2) + 1

Plug in x = 2 to find that...

f(x) = (x/2) + 1

f(2) = (2/2) + 1

f(2) = 2

This shows \lim_{x \to 2^{+}}f(x) = 2

Then for the left hand limit \lim_{x \to 2^{-}}f(x), we'll involve x < 2 and we go for the first piece. So,

f(x) = 3-x

f(2) = 3-2

f(2) = 1

Therefore, \lim_{x \to 2^{-}}f(x) = 1

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

Part B

Because \lim_{x \to 2^{+}}f(x) \ne \lim_{x \to 2^{-}}f(x) this means that the limit \lim_{x \to 2}f(x) does not exist.

If you are a visual learner, check out the graph below of the piecewise function. Notice the gap or disconnect at x = 2. This can be thought of as two roads that are disconnected. There's no way for a car to go from one road to the other. Because of this disconnect, the limit doesn't exist at x = 2.

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

Part C

You'll follow the same type of steps shown in part A.

However, keep in mind that x = 4 is above x = 2, so we'll deal with x > 2 only.

So you'd only involve the second piece f(x) = (x/2) + 1

You should find that f(4) = 3, and that both left and right hand limits equal this value. The left and right hand limits approach the same y value. The limit does exist here. There are no gaps to worry about when x = 4.

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

Part D

As mentioned earlier, since \lim_{x \to 4^{+}}f(x) = \lim_{x \to 4^{-}}f(x) = 3, this means the limit \lim_{x \to 4}f(x) does exist and it's equal to 3.

As x gets closer and closer to 4, the y values are approaching 3. This applies to both directions.

4 0
2 years ago
Cookies all about how man
KengaRu [80]
Yes cookies

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8 0
3 years ago
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