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2 years ago

Please answer number 3.

Mathematics

1 answer:

Alenkasestr [34]2 years ago

6 0

I can't exactly SHOW you where to put the numbers, but I can teach you the process of how you'd do it.

First off, label your number line from 0-15, as it is the simplest. (You'd be counting by 1 per each line). Then, follow this process:

1) Look at the first digit of your value. Place your number according to your first digit. (So, you'd put 0.365 at the 0 line and 3.521 at the 3 line)

2) Look at the second digit of your value. Imagine that between the two main lines (0-1 and 3-4) that there is 10 smaller lines. Then, you can place your number according to your second digit. (So, you'd put 0.365 at the 0.3 line and 3.521 at the 3.5 line).

3) Look at the third digit of your value. Imagine that between the two smaller lines (0.3-0.4 and 3.5-3.6) that there is 10 smaller lines. Then, you can place your number according to your third digit. (So, you'd put 0.365 at the 0.36 line and 3.521 at the 3.52 line).

4) Look at the fourth digit of your value. Imagine that between the two even smaller lines (0.36-0.37 and 3.52-3.53) that there is 10 smaller lines. Then, you can place your number according to your fourth digit. (So you'd place 0.365 at the 0.365 line and 3.521 at the 3.521 line)

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Find the equation of a line that contains the points (6,3) and (6,−2).

RSB [31]

Answer:

x = 6

Step-by-step explanation:

First find the gradient,

$Gradient = \frac{y2-y1}{x2-x1} = \frac{-2-3}{6-6} =\frac{-5}{0} =0$

Since this line has no gradient, it means that it's a horizontal or vertical line. We cannot substitute in the values because the gradient is 0 so this line will only be x = 6

3 0

3 years ago

Read 2 more answers

Iestion 10 (5 points)

marta [7]

Answer:

Jake is 9, and Jake's dad is 43.

Step-by-step explanation:

Ok, let's start off by putting a variable for Jack's age, A.

We know that Jack's father is 2 less than 5 times, so we can write his as 5x -2.

Now let's make an equation:

52 = x + 5x - 2

We need to isolate the x-values so we need to add 2 on both sides:

54= x + 5x

No we need to add the like terms:

54 = 6x

Finally, we just divide by 6 on each side.

Now we know X = 9 but we're not done yet :)

So because Jake's dad is almost 5 times as much, we first multiply Jack's age by 5. We get 45 but we need to subtract 2 which leaves us with 43.

3 0

3 years ago

Which equation can be used to solve for in the following diagram?<br> (8x - 20)°<br> 3xº

gayaneshka [121]

Answer:

A. 3x° + (8x - 20)° = 90°

Step-by-step explanation:

The two given angles are complementary and sum to 90°

3x° + (8x - 20)° = 90°

Therefore correct choice is A.

7 0

2 years ago

A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.

Usimov [2.4K]

Answer:

a) $\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

b) $v(t=3) = 3(3)^2 -18(3) +15=-12$

c) $t =1s, t=5s$

d) $[0,1) \cup (5,\infty)$

e) $D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

f) $a(t) = \frac{dv}{dt}= 6t -18$

g) The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

Step-by-step explanation:

For this case we have the following function given:

$f(t) = s = t^3 -9t^2 +15 t$

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

$\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

$v(t=3) = 3(3)^2 -18(3) +15=-12$

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

$3t^2 -18 t +15 =0$

We can divide both sides of the equation by 3 and we got:

$t^2 -6t +5=0$

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

$(t-5)*(t-1) =0$

And for this case we got $t =1s, t=5s$

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

$t^2 -6t +5 >0$

$(t-5) *(t-1)>0$

So then the intervals positive are $[0,1) \cup (5,\infty)$

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

$D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6$

And if we replace we got:

$D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

$a(t) = \frac{dv}{dt}= 6t -18$

Part g and h

The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

5 0

3 years ago

How many different 4-question geometry quizzes can a teacher make if there are 7 different problems to test on?.

stich3 [128]

Answer:

35 quizzes

Step-by-step explanation:

We need to determine how many ways we can choose 4 questions out of 7 to make the quizzes.

We do this by using the formula $\displaystyle nCr=\frac{n!}{r!(n-r)!}$ which describes the number of ways we can choose $r$ objects given $n$ possible choices. So, if $n=7$ and $r=4$ , then:

$_4C_7=\frac{7!}{4!(7-4)!}\\\\_4C_7=\frac{7*6*5*4!}{4!*3!}\\\\_4C_7=\frac{210}{6}\\\\_4C_7=35$

Hence, the teacher can make 35 different geometry quizzes.

3 0

2 years ago