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3 years ago

What are the terms in the expression 4 + 9V +6w?

Mathematics

1 answer:

spayn [35]3 years ago

4 0

Answer:

4, 9v, and 6w

Step-by-step explanation:

I'm pretty sure it's right

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Step 1 5x- y= 4

valkas [14]

Answer: C, In Step 2 the expression for y should be subsituted in the other equation.

5 0

3 years ago

In the coordinate plane, draw quadrilateral ABCD with A(–5, 0), B(2, –6), C(8, 1), and D(1, 7), then demonstrate that ABCD is a

Len [333]

Answer:

ABCD is a rectangle

Step-by-step explanation:

∵ A = (-5 , 0) , B = (2 , -6) , C = (8 , 1) , D = (1 , 7)

∵ The x-coordinate of the mid-point of AC = (-5 + 8)/2 =3/2

∵ The y-coordinate of the mid-point of AC = (0 + 1)/2 = 1/2

∴ The mid-point of AC = (3/2 , 1/2)

∵ The x-coordinate of the mid-point of BD = (2 + 1)/2 =3/2

∵ The y-coordinate of the mid-point of BD = (-6 + 7)/2 = 1/2

∴ The mid-point of BD = (3/2 , 1/2)

∴ The mid-point of AC = The mid-point of BD ⇒ (1)

∵ AC = √[(8 - -5)²+(1 - 0)² = √170

∵ BD = √[(1 - 2)²+(7 - -6)² = √170

∴ AC = BD ⇒ (2)

From (1) and (2)

AC and BD equal each other and bisects each other

∴ ABCD is a rectangle

7 0

3 years ago

Solve the equation by completing the square

Nana76 [90]

Answer:

s = - 1 ± $\sqrt{7}$

Step-by-step explanation:

Given

s² + 2s - 6 = 0 ( add 6 to both sides )

s² + 2s = 6

To complete the square

add ( half the coefficient of the s- term )² to both sides

s² + 2(1)s + 1 = 6 + 1

(s + 1)² = 7 ( take the square root of both sides )

s + 1 = ± $\sqrt{7}$ ( subtract 1 from both sides )

s = - 1 ± $\sqrt{7}$

Thus

s = - 1 - $\sqrt{7}$ , s = - 1 + $\sqrt{7}$

4 0

3 years ago

Is it C or no? Can u help please

Westkost [7]

the answer is Probably C.

7 0

3 years ago

Determine whether the improper integral converges or diverges, and find the value of each that converges.

Brrunno [24]

Answer:

Converges at -1

Step-by-step explanation:

The integral converges if the limit exists, if the limit does not exist or if the limit is infinity it diverges.

We will make use of integral by parts to determine:

let:

$u=x$ $dv=e^(2x)\cdot{dx}$

$du=dx$ $v=2\cdot{e^(2x)}$

$\int\limits^a_b {u} \, dv = uv -\int\limits^a_b {v} \, du$

$\int\limits^a_b {x\cdot{e^2^x} \, dx =2xe^2^x- \int\limits^a_b {2e^2^x} \, dx$

$\int\limits^a_b {xe^2^x} \, dx = 2xe^2^x-2e^2^x-C$

We can therefore determine that if x tends to 0 the limit is -1

$\lim_{x \to \0} 2xe^2^x-2e^2^x=0-1=-1$

5 0

3 years ago