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3 years ago

B is a number larger than 1

Mathematics

2 answers:

mart [117]3 years ago

6 0

Answer:

b= infinity

Step-by-step explanation:

Because is b>1 than b can be 2, 3, 44, 100, 1000, 100000000213981623 and higher. There is no way to find the exact number though.

Ulleksa [173]3 years ago

3 0

B is greater than 1, so B>1

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2x + y = 12<br> 4x - 3y = 2

7nadin3 [17]

Answer:

The value of x for equation is 3.4

The value of y for equation is 5.2

Step-by-step explanation:

Given as :

The two linear equation are

2 x + y = 12 ........A

4 x - 3 y = 2 .........B

Now, Solving both equations , we get

3 × ( 2 x + y ) + (4 x - 3 y) = 3 × 12 - 2

Or, 6 x + 3 y + 4 x - 3 y = 36 - 2

Or, (6 x + 4 x) + (3 y - 3 y) = 34

Or, 10 x + 0 = 34

∴ x = $\dfrac{34}{10}$

i.e x = 3.4

So, The value of x for equation , x = 3.4

Put the value of x into eq A

∵ 2 x + y = 12

Or, y = 12 - 2 x

Or, y = 12 - 2 × 3.4

Or, y = 12 - 6.8

∴ y = 5.2

So, The value of y for equation , y = 5.2

Hence, The value of x for equation is 3.4 and The value of y for equation is 5.2 Answer

4 0

3 years ago

1.) -9x+8 ≤ 80 *<br><br> A. x ≥ -8<br> B. x ≤ -8<br> C. x ≥ 8<br> D. x ≤ 8

Firlakuza [10]

Answer:

$x\geq -8$

Step-by-step explanation:

$-9x+8\leq 80$

-8 -8

₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋

$\frac{-9x}{-9} \leq \frac{72}{-9}$

⇵

$x\geq -8$

7 0

2 years ago

Read 2 more answers

Determine the solubility for cuc2o4(s) in pure water. Ksp for is 2. 9 × 10^-8

Elena-2011 [213]

Based on the elements and charges in Copper (II) Oxalate, CuC₂O₄(s), the solubility in pure water is 1.7 x 10⁻⁴ M.

<h3>What is the solubility of Copper (II) Oxalate in pure water?</h3>

The solubility equilibrium (Ksp) is 2.9 x 10⁻⁸ so the solubility can be found as:

Ksp = [Cu²⁺] [C₂O₄²⁻]

Solving gives:

2.9 x 10⁻⁸ = S x S

S² = 2.9 x 10⁻⁸

S = 1.7 x 10⁻⁴ M

Find out more on solubility at brainly.com/question/23659342.

#SPJ1

3 0

2 years ago

Solve by completing the square with working out <br><br> -6x-x^2-8

Goryan [66]

-6x-x^2-8 = -(x+2) x (x+4)

(This is if you are factoring the expression)

7 0

3 years ago

Find the distance traveled by a particle with position (x, y as t varies in the given time interval. x = 3?sin2 t, y = 3?cos2 t,

ANTONII [103]

Assuming the particle's position is given by

$\begin{cases}x(t)=3-\sin2t\\y(t)=3-\cos2t\\0\le t\le4\end{cases}$

then the distance traveled over the interval is

$\displaystyle\int_{t=0}^{t=4}\sqrt{\left(\dfrac{\mathrm dx(t)}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy(t)}{\mathrm dt}\right)^2}\,\mathrm dt$
$=\displaystyle\int_0^4\sqrt{(-2\cos2t)^2+(2\sin2t)^2}\,\mathrm dt$
$=\displaystyle\int_0^4\sqrt{4\cos^22t+4\sin^22t}\,\mathrm dt$
$=\displaystyle2\int_0^4\mathrm dt$
$=2(4-0)=8$

5 0

3 years ago