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3 years ago

-10 + (-15) =+ what do i do first

Mathematics

2 answers:

Rus_ich [418]3 years ago

8 0

Answer:

-10 + (-15) is -25 and you do subtraction first.

Step-by-step explanation:

Since negative + negative is negative, you ignore the plus sign and continue with the equation. So -10 minus -15 is -25.

LuckyWell [14K]3 years ago

3 0

Answer:

-15 then -10 = -25 we can only take one sign that accompanies the -10 and put in front of the brackets.

Step-by-step explanation:

-10 + (-15) = -25

just like (-15) -10 = 25

This is because ( ) is order of parenthesis Brackets first.

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skelet666 [1.2K]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Can someone please help me I posted this multiple times but no one answers pls help me quick

mario62 [17]

You would have to make 465=1550-7 1/2 x and find x
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2 years ago

Lila flies a drone in the direction of 90° towards Avery. Once she gets to Avery she changes course and flies in the direction o

Nikitich [7]

Answer:

X = 13.13 feet

Step-by-step explanation:

Let the distance between Lila and Mark be x.

Let the distance between Avery and lila be A= 20

Let the distance between Avery and Mark be B= 30

Angle x = 200-180=20°

To solve this question we will use the cosine formula which is

X² = A²+ B² -2ABcosx

X²= 20² + 30² -2(20)(30)cos20

X²=1300-1127.6

X²= 172.4

X= 13.13

8 0

3 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

What is the conjugate of -2 + 8i

topjm [15]

Answer:

-2 - 8i

Step-by-step explanation:

To find the conjugate of a complex number, change the sign of the imaginary part.

The conjugate of -2 + 8i is -2 - 8i.

3 0

3 years ago

-10 + (-15) =+ what do i do first​

-10 + (-15) =+ what do i do first