1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
bagirrra123 [75]
2 years ago
8

A- 6=9. Please help

Mathematics
2 answers:
posledela2 years ago
8 0

Answer:

a=15

Step-by-step explanation:

Move all terms not containing  a  to the right side of the equation.

15-6=9

olchik [2.2K]2 years ago
3 0

Answer:

a = 15

Step-by-step explanation:

6 + 9 = 15

a = 15

15 - 6 = 9

yes

You might be interested in
Determine the maximum or minimum value of the following quadratic function. LaTeX: f\left(x\right)=-3\left(x+5\right)^2-1f ( x )
Genrish500 [490]

Answer:

Step-by-step explanation:

y=-3(x^2+10x+25)-1\\ \\ y=-3x^2-30x-76\\ \\ dy=-6x-30\\ \\ d^2y=-6\\ \\ \text{Since the second derivative is always negative, when the first derivative is equal to zero the function will be at a global maximum.}\\ \\ dy=0=-6x-30\\ \\ x=-5\\ \\ f(-5)=-1\\ \\ \text{So the global maximum occurs at the point (-5,-1)

3 0
3 years ago
Please help me with this question!
alekssr [168]

Answer: C) 12.2

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

Explanation:

We have a known adjacent side (10) and an unknown hypotenuse (x). The cosine rule ties the two sides together.

cos(angle) = adjacent/hypotenuse

cos(35) = 10/x

x*cos(35) = 10

x = 10/cos(35)

x = 12.2077458876146 approximately

x = 12.2

Make sure your calculator is in degree mode.

8 0
2 years ago
Read 2 more answers
Find the missing measure in the triangle below.
denis23 [38]

Answer:

C

Step-by-step explanation:

To find x take 180 and subtract 63 and 77

7 0
3 years ago
Read 2 more answers
Write an equation point slope form for the line that goes through the given point with the given slope.
Aleksandr-060686 [28]
Y - y1 = m(x - x1)
slope(m) = -1/2
(-4,-2)...x1 = -4 and y1 = -2
now we sub
y - (-2) = -1/2(x - (-4) =
y + 2 = -1/2(x + 4) <=====
3 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
Other questions:
  • A calzone is divided into 24 equal pieces. Glenn and Ben each ate one eighth of the calzone on Thursday. The next​ day, Ben ate
    6·1 answer
  • true or false? the domain of the function in the box is the set of all positive real numbers f(x)=(3/4)^x
    15·1 answer
  • Kelly Jackson is an administrative assistant. She earns a weekly salary of $412.00. She applied and wad hired for an administrat
    6·1 answer
  • How long will it take you to drive 175 miles at a speed of 25 miles per hour
    6·1 answer
  • Ross needs to buy a countertop for a laundry room. He calculated the area to be 12 square feet. The actual area is 11.8 square f
    5·2 answers
  • Which rational function has zeros at x = 1 and x = 3?
    6·1 answer
  • Find the perimeter of M
    12·1 answer
  • 3y - 15 =y + 1 solve for y
    10·1 answer
  • Which of the following is equivalent to 16 3/4 <br>6<br>8 <br>12 <br>64​
    5·1 answer
  • Mrs. Taylor is making new foot all uniforms for Friday nights homecoming game. The trim for the sleeves of one uniforms requires
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!