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3 years ago

I WILL GIVE BRAINLIEST IF ANSWER FAST What is the measurement of this dial caliper?

Engineering

2 answers:

Galina-37 [17]3 years ago

6 0

The answer is B. 4.045 inches

Ratling [72]3 years ago

5 0

I think the answer might be A, but I’m not completely sure

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Car insurance incentives and discounts are available depending on _____. A. school attendance and driver skill B. vehicle type a

WARRIOR [948]

Answer:D. Location, vehicle type, and driving habits

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2 years ago

Read 2 more answers

Four of the minterms of the completely specified function f(a, b, c, d) are m0, m1, m4, and m5.

Sveta_85 [38]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) The required additional minterms for f so that f has eight primary implicants with two literals and no other prime implicant are $m_{2},m_{3},m_{7},m_{8},m_{11},m_{12},m_{13},m_{14}$ and $m_{15}$

b) The essential prime implicant are $c' d',a'b',ab$ and $cd$

c) The minimum sum-of-product expression for f are

$a'b' +ab +c'd'+cd+a'c',\\ a'b'+ab+c'd'+cd+a'd,\\a'b'+ab+c'd'+cd+bc' and \\ a'b'+ab+c'd' +cd+bd$

Explanation:

The explanation is shown on the second third and fourth image

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2 years ago

Which of the following describes how the author introduces dust storms in the text??

Oduvanchick [21]

Answer:

B-as deadly storms that claim lives in th great pines

Explanation:

I hope it helped if it did pls make me brainliest

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2 years ago

Read 2 more answers

X²-12x=0 ПЖ срочно реально решите помагие

nika2105 [10]

Answer:

формула pq sry i dont speak russian lol but thats the solution of the equation

4 0

2 years ago

A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte

Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

$a(t)=5(1-\frac{t}{15})$

$a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})$

Integrating both sides we get

$\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c$

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

$\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D$

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

$x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m$

Thus the remaining 125 meters will be covered with a constant speed of

$v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s$

in time equalling $t_{2}=\frac{125}{37.5}=3.33seconds$

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0

2 years ago