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3 years ago

Hi, can someone help pls! Which of the following is a counterexample for the claim that √x^2 = x ?

Mathematics

1 answer:

Leto [7]3 years ago

7 0

Answer:

Step-by-step explanation:

x=-1 is the counter example as ((-1)2)^(1/2)=1 and not -1.

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Two missiles are initially 5000 miles apart. They travel along a straight line directly towards one another, one traveling at 20

Goshia [24]

d = 5,000 mi the distance between the missiles

v1 = 2,000 mph the rate of the first missile

v2 = 1,000 mph the rate of the second missile

t = 1 min = 1/60 hr

Since distance = speed*time the distance 1 minute before the impact is:

v1t + v2t = t(v1 + v2) = (1/60)(2,000 + 1,000) = 50 miles.

7 0

3 years ago

Please help!!!

Nonamiya [84]

Answer:

c. $2a^3x^4-3a^3x^3+a^4x^2$

d. $63x^5y^3-30x^2y^4-7x^3y^2$

Step-by-step explanation:

For this exercise you need to remember the following:

1. The multiplication of signs:

$(+)(+)=+\\(-)(-)=+\\(-)(+)=-$

2. The Product of powers property, which states that:

$(a^m)(a^n)=a^{(m+n)}$

3. The Distributive property:

$a(b\±c)=ab\±ac$

Knowing that, you can solve the multiplication of the polynomials:

c. $(-2ax^2+3ax-a^2)(-a^2x^2)=\\\\=(-2ax^2)(-a^2x^2)+(3ax)(-a^2x^2)+(-a^2)(-a^2x^2)=2a^3x^4-3a^3x^3+a^4x^2$

d. $(6.3x^3y-3y^2-0.7x)(10x^2y^2)=\\\\=(6.3x^3y)(10x^2y^2)+(-3y^2)(10x^2y^2)+(-0.7x)(10x^2y^2)=63x^5y^3-30x^2y^4-7x^3y^2$

8 0

3 years ago

In △ABC , line AB is extended up to point E and line BC is extended up to D . The measure of some of the angles is given as; ∠BA

Over [174]

Answer:

$\angle ACD = 130 ^\circ$ .

Step-by-step explanation:

$\angle ABC$ and $\angle EBC$ are supplementary angles. Their sum should be $180^\circ$ . $\angle EBC = 120^\circ$ $\implies \angle ABC = 180^\circ - \angle EBC = 180^\circ - 120^\circ = 60^\circ$ .

The sum of the three angles in triangle $\triangle ABC$ : $(\angle ABC + \angle BAC + \angle ACB)$ should be equal to $180^\circ$ .

$\angle ABC = 60^\circ$ and $\angle BAC = 70^\circ$ $\implies \angle ACB = 180^\circ - \angle ABC - \angle ACB = 180^\circ - 70^\circ - 60^\circ = 50^\circ$ .

$\angle ACB$ and $\angle ACD$ form another pair of supplementary angles. Their sum should also be $180^\circ$ . $\angle ACB = 50^\circ$ $\implies \angle ACD = 180^\circ - \angle ACB = 180^\circ - 50^\circ = 130^\circ$ .

8 0

4 years ago

Solve the system using substitution and describe your steps. <br> x=y-3 <br> 2x+y=12

Misha Larkins [42]

X = y - 3 (equation 1)
2x + y = 12 (equation 2)

Substitute equation 1 into equation 2, replacing every x value with equation 1

2(y - 3) + y = 12

Now, distribute the 2 inside the brackets and simplify.
2y - 6 + y = 12
3y = 18
y = 6 (divided both sides by 3)

Substitute y back into equation 1 to find x.
x = 6 - 3
x = 3

Verify that these are correct by substituting them into equation 2.
2(3) + 6 = 6 + 6 = 12

Therefore, x = 3 and y = 6

6 0

3 years ago

Read 2 more answers

Find the missing side of each right triangle (((((Pythagorean’s Theorem))))))))

kaheart [24]

Answer:

Step-by-step explanation:

36-27 = 9

$\sqrt{9} = 3$

7 0

3 years ago