When the area of a triangle is A = 18in2

The LCM of two numbers can be found by multiplying all the prime factors of

Degger [83]

2,4,5,7 between those

6 0

3 years ago

Please help 30 points!!!

just olya [345]

Answer:

Question 1: B = 0

Question 2: 23/2 or 11-1/2

Question 3: 663/6 or 110-3/6

Step-by-step explanation:

Question 1:

(-5²)+(9+4²)=5B

-25+9+16=5B

5B=-0

B=0/5=0

Question 2:

2-7/8×2-6/9×1-1/2

=23/8×24/9×3/2

=23/2

=11-1/2

Question3:

3-1/4×6-2/3×5-1/10

=13/4×20/3×51/10

=663/6

=110-3/6

7 0

3 years ago

HELp and show ur work Which of the expressions below has a product that is greater than 100? Select all that apply. A) 5.12 × 20

Bumek [7]

A & C should be your answer

6 0

3 years ago

Read 2 more answers

Simplify each expression to a single trig function or number cosx(secx-cosx)

AlladinOne [14]

I did this test b4, yours is answer #number 12

Convert things to their basic forms.
Remember a few identities 
sin^2 + cos^2 = 1 so 
sin^2 = 1 - cos^2 and 
cos^2 = 1 - sin^2 

I'm going to skip typing the theta symbol, just to make things faster. Just assume it is there and fill it in as you work the problems. 

Follow along to see how each problem was worked out. You'll catch on to the general technique. 

====== 
1. sec θ sin θ 
1/cos * sin = sin/cos = tan 

2. cos θ tan θ 
cos * sin/cos = sin 

3. tan^2 θ- sec^2 θ 
sin^2 / cos^2 - 1/cos^2 
(sin^2 - 1)/cos^2 
-(1-sin^2)/cos^2 
-cos^2/cos^2 
-1 

4. 1- cos^2θ 
sin^2 

5. (1-cosθ)(1+cosθ) 
Remember (a+b)(a--b) = a^2 - b^2 
1-cos^2 = sin^2 

6. (secx-1) (secx+1) 
sec^2 -1 
1/cos^2 - 1 
1/cos^2 - cos^2/cos^2 
(1-cos^2)/cos^2 
sin^2/cos62 
tan^2 

7. (1/sin^2A)-(1/tan^2A) 
1/sin^2 - 1/(sin^2/cos^2) 
1/sin^2 - cos^2/sin^2 
(1-cos^2)/sin^2 
sin^2/sin^2 
1 

8. 1- (sin^2θ/tan^2θ) 
1-sin^2/(sin^2/cos^2) 
1 - sin^2*cos^2/sin^2 
1-cos^2 
sin^2 

9. (1/cos^2θ)-(1/cot^2θ) 
1/cos^2 - 1/(cos^2/sin^2) 
1/cos^2 - sin^2/cos^2 
(1-sin^2)/cos^2 
cos^2/cos^2 
1 

10. cosθ (secθ-cosθ) 
cos *(1/cos - cos) 
1-cos^2 
sin^2 

11. cos^2A (sec^2A-1) 
cos^2 * (1/cos^2 - 1) 
1 - cos^2 
sin^2 

12. (1-cosx)(1+secx)(cosx) 
(1-cos)(1+1/cos)cos 
(1-cos)(cos + 1) 
-(cos-1)(cos+1) 
-(cos^2 - 1) 
-(-sin^2) 
sin^2 

13. (sinxcosx)/(1-cos^2x) 
sin*cos/sin^2 
cos/sin 
cot 

14. (tan^2θ/secθ+1) +1 
(sin^2/cos^2)/(1/cos) + 2 
sin^2/cos + 2 
sin*tan + 2

5 0

3 years ago

You are jumping off the 12 foot diving board at the municipal pool. You bounce up at 6 feet per second and drop to the water you

NARA [144]

Answer:

When do you hit the water?

1.075 seconds after you jump.

What is your maximum height?

the maximum height is 12.5626 ft

Step-by-step explanation:

The equation:

h(t) = -16*t^2 + 6*t + 12

Is the height as a function of time.

We know that the initial height is the height when t = 0s

h(0s) = 12

and we know that the diving board is 12 foot tall.

Then the zero in h(t)

h(t) = 0

Represents the surface of the water.

When do you hit the water?

Here we just need to find the value of t such that:

h(t) = 0 = -16*t^2 + 6*t + 12

Using the Bhaskara's formula, we get:

$t = \frac{-6 \pm \sqrt{6^2 - 4*(-16)*12} }{2*(-16)} = \frac{-6 \pm 28.4}{-32}$

Then we have two solutions, and we only care for the positive solution (because the negative time happens before the jump, so that solution can be discarded)

The positive solution is:

t = (-6 - 28.4)/-32 = 1.075

So you hit the water 1.075 seconds after you jump.

What is your maximum height?

The height equation is a quadratic equation with a negative leading coefficient, then the maximum of this parabola is at the vertex.

We know that the vertex of a general quadratic:

a*x^2 + b*x + c

is at

x = -b/2a

Then in the case of our equation:

h(t) = -16*t^2 + 6*t + 12

The vertex is at:

t = -6/(2*-16) = 6/32 = 0.1875

Evaluating the height equation in that time will give us the maximum height, which is:

h(0.1875) = -16*(0.1875 )^2 + 6*(0.1875) + 12 = 12.5626

And the height is in feet, then the maximum height is 12.5626 ft

6 0

3 years ago