MATHEMATICS HIGH SCHOOL

Get this question correct image attached

Answers

Answer 1

Answer:

Answer: x ≤ 5 or x > 7

Step-by-step explanation:

First, we will write an equation for the left side. It's a closed dot going to the left (less than) of 5.

x ≤ 5

Next, we will write the right side. It's an open dot going to the right (greater than) of 7.

x > 7

Lastly, we will combine these to create our compound inequality.

x ≤ 5 or x > 7

Answer 2

Answer:

Answer:

-infinity less than or equal to 5 and 7 less than infinity

Step-by-step explanation:

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4. Solve the quadratic by factoring, isolating x^2, or using the quadratic formula:a. 2x^2 – 7x = 4

b. 81x^4 – 1

c. 2x^2 – 32 = 0

d. X^2 – 7x + 12

Answers

Hello,

A:
2x²-7x-4=0
==>2x²-8x+x-4=0
==>2x(x-4)+1(x-4)=0
==>(x-4)(2x+1)=0
==>(x=4) or (x=-1/2)

B: where is the sign "=" !!!!!!!!!!!!!!!!!!

81x^-1=(9x²+1)(3x+1)(3x-1) I can not go any further.

C)
2x²-32=0
==>x²-16=0
==>(x+4)(x-4)0
==>x=-4 or x=4

D:
where is the sign "="!!!!!!!!!!!!!!!!!!!!!!!!

x^2-7x+12=(x-4)(x-3) but I can not go ant further....

Solve for x in this equation log9 1/3=x

Answers

The work is shown in the attachment

Would you please help me to solve this 9.1x10^-5)^-4

Answers

[( 91 * 10^-1 ) *10^ -5]^-4 = (91 * 10^-6)^-4 = 91^-4 * 10^24 = ( 10^24 )/ ( 91^4 ) =
10000000 ≈ 1458258211769

Which type of transformation does NOT result in a figure that is congruent to the original one?

Answers

Dilation never does.
Reflection often doesn't.

Which of the following is equivalent to -(5/12) ?A. -5/-12
B. 5/12
C. 12/-5
D. -5/12

Answers

The answer is D. If you divide -5 by -12 you will get the same answer

8. The base of a circular cone has a diameter of 10 cm and an altitude of 10 cm. The cone is filled with water. A sphere is lowered into the cone until it just fits. Exactly one-half of the sphere remains out of the water. Once the sphere is removed, how much water remains in the cone?

Answers

Answer:

The volume of water that remains on the cone is 523.6 cm³

Step-by-step explanation:

To solve this problem you have to keep in mind the formules that describes the volume of a cone and the volume of a sphere.

Volume of a cone = (πr²h)/3

Volume of a sphere = (4/3)πr³

So, if the base of the cone has a diameter of 10 cm, its radius is 5 cm. Its altitude is 10 cm. ⇒Volume = (πr²h)/3 ⇒ Volume = [π(5²)10) ⇒

Volume = 785.4 cm³. This is the initial volume of water.

Now if the sphere fits in the cone and half of it remains out of the water, the other half is inside the cone. Estimating the volume of the sphere and dividing it by two, you find the volume of water that was displaced.

Volume of a sphere = (4/3)πr³, here the radius is the same of the base of the cone (5 cm).

⇒ Volume = (4/3)π(5³) ⇒ Volume = 523.6 cm³ ⇒ The half of this volume is 261.8 cm³. This is the volume of water displaced.

⇒ The volume of water that remains on the cone is 523.6 cm³ (785.4 cm³- 261.8 cm³)

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