What equation do I use for this? If anyone can help, I’ll appreciate it!

Ulleksa [173]

Answer:

Step 1 Eliminate fractions by multiplying all terms by the least common denominator of all fractions.

Step 2 Simplify by combining like terms on each side of the inequality.

Step 3 Add or subtract quantities to obtain the unknown on one side and the numbers on the other.

Step 4 Divide each term of the inequality by the coefficient of the unknown. If the coefficient is positive, the inequality will remain the same. If the coefficient is negative, the inequality will be reversed.

Step 5 Check your answer.

3 0

3 years ago

Read 2 more answers

Answer these questions please.

Ivenika [448]

ABCDEFGH = 22.
HEFG = 12
ABCD = 10

8 0

3 years ago

These are questions 2-5 out of 20. are these correct!?

dusya [7]

I believe so, but It is kind of blurry for some reason on my screen but from what I can make out yes.

3 0

4 years ago

A ball is thrown into the air from a height of 4 feet at time t = 0. The function that models this situation is h(t) = -16t2 + 6

katrin2010 [14]

Answer:

Part a) The height of the ball after 3 seconds is $49\ ft$

Part b) The maximum height is 66 ft

Part c) The ball hit the ground for t=4 sec

Part d) The domain of the function that makes sense is the interval

[0,4]

Step-by-step explanation:

we have

$h(t)=-16t^{2} +63t+4$

Part a) What is the height of the ball after 3 seconds?

For t=3 sec

Substitute in the function and solve for h

$h(3)=-16(3)^{2} +63(3)+4=49\ ft$

Part b) What is the maximum height of the ball? Round to the nearest foot.

we know that

The maximum height of the ball is the vertex of the quadratic equation

so

Convert the function into a vertex form

$h(t)=-16t^{2} +63t+4$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$h(t)-4=-16t^{2} +63t$

Factor the leading coefficient

$h(t)-4=-16(t^{2} -(63/16)t)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$h(t)-4-16(63/32)^{2}=-16(t^{2} -(63/16)t+(63/32)^{2})$

$h(t)-(67,600/1,024)=-16(t^{2} -(63/16)t+(63/32)^{2})$

Rewrite as perfect squares

$h(t)-(67,600/1,024)=-16(t-(63/32))^{2}$

$h(t)=-16(t-(63/32))^{2}+(67,600/1,024)$

the vertex is the point (1.97,66.02)

therefore

The maximum height is 66 ft

Part c) When will the ball hit the ground?

we know that

The ball hit the ground when h(t)=0 (the x-intercepts of the function)

so

$h(t)=-16t^{2} +63t+4$

For h(t)=0

$0=-16t^{2} +63t+4$

using a graphing tool

The solution is t=4 sec

see the attached figure

Part d) What domain makes sense for the function?

The domain of the function that makes sense is the interval

[0,4]

All real numbers greater than or equal to 0 seconds and less than or equal to 4 seconds

Remember that the time can not be a negative number

6 0

3 years ago

Can someone help me with these questions? i’m so lost

spayn [35]

Answer:

Q.5 ab=cd

Q.6 ad=bc

Q.7 ce=ae

Q.8 eb=ed

Q.9 angle D=angle B (opposite angle of parallelogram are equal)

let other angle of parallelogram be x.

angle A+angle B +angle C + angle D= 360° (sum of quadrilateral is 360°)

x+130°+x+130°=360°

2x+260°=360°

2x=360°-260°

2x=100°

x=100/2

x=50°

Q.10 similarly, angle b= angle d

let other angle be x.

x+61°+ x+61°=360°

2x+122°=360°

2x=360°+122°

2x=238°

x=238°/2

x=119°

Q.11 in quadrilateral opposite angles are equal and opposite angle of parallelogram are equal.

Q.12 in quadrilateral opposite angle are equal and opposite angle of parallelogram are equal.

Q.13 in quadrilateral opposite sides are equal and opposite sides are parellel and this property is also present in parallelogram.

q.14 in quadrilateral diagonal bisected each other and diagonal of parallelogram also bisect each other.

8 0

3 years ago