Twice the sum of a number and negative nine

PLS HELP!!! FIRST ANSWER GETS BRAINLIEST!!!

olga55 [171]

Answer:C

Step-by-step explanation:11m is the radius and to find the diameter u multiply 11 x 2 which is radius x 2 which = 22

5 0

3 years ago

If

Tju [1.3M]

Answer:

Step-by-step explanation:

Rationalize the denominator of b. So, multiply the numerator and denominator by $\sqrt{x}$

$b = \frac{(1-2\sqrt{x}) *\sqrt{x}}{\sqrt{x}*\sqrt{x} }=\frac{1*\sqrt{x} -2\sqrt{x} *\sqrt{x} }{\sqrt{x} *\sqrt{x} }\\\\=\frac{\sqrt{x} -2x}{x}\\$

Now, find a +b

$a +b = \frac{2x+\sqrt{x} }{x}+\frac{\sqrt{x} -2x}{x}\\\\=\frac{2x+\sqrt{x} +\sqrt{x} -2x}{x}$

Combine like terms

$= \frac{2x-2x+\sqrt{x} +\sqrt{x} }{x}\\\\=\frac{2\sqrt{x} }{x}$

Now find (a + b)²

(a +b)² = $(\frac{2\sqrt{x} }{x})^{2}$

$= \frac{2^{2}*(\sqrt{x} )^{2}}{x^{2}}\\\\= \frac{4* x}{x^{2}}\\\\= \frac{4}{x}$

Hint: $\sqrt{x} *\sqrt{x} =\sqrt{x*x}=x$

5 0

3 years ago

Kelly runs a restaurant that sells two kinds of desserts. Kelly knows the restaurant must make at least 25 dozen and at most 45

MissTica

Answer:

(c, m) = (45, 10)

Step-by-step explanation:

A dozen White Chocolate Blizzards generate more income and take less flour than a dozen Mint Breezes, so production of those should clearly be maximized. Making 45 dozen Blizzards does not use all the flour, so the remaining flour can be used to make Breezes.

Maximum Blizzards that can be made: 45 dz. Flour used: 45×5 oz = 225 oz.

The remaining flour is ...

315 oz -225 oz = 90 oz

This is enough for (90 oz)/(9 oz/dz) = 10 dozen Mint Breezes. This is in the required range of 2 to 15 dozen.

Kelly should make 45 dozen White Chocolate Blizzards and 10 dozen Mint Breezes: (c, m) = (45, 10).

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In the attached graph, we have reversed the applicable inequalities so the feasible region shows up white, instead of shaded with 5 different colors. The objective function is the green line, shown at the point that maximizes income. (c, m) ⇔ (x, y)

8 0

3 years ago

A golfer keeps track of his score for playing nine holes of golf (half a normal golf round). His mean score is 8080 with a sta

solniwko [45]

Answer:

Therefore the mean and standard deviation of his total score if he plays a full 18 holes are 160 and $11\sqrt2$ respectively.

Step-by-step explanation:

Given that,

For the first 9 holes X:

E(X) = 80

SD(X)=13

For the second 9 holes Y:

E(Y) = 80

SD(Y)=13

For the sum W=X+Y, the following properties holds for means , variance and standard deviation :

E(W)=E(X)+E(Y)

and

V(W)=V(X)+V(Y)

⇒SD²(W)=SD²(X)+SD²(Y) [ Variance = (standard deviation)²]

$\Rightarrow SD(W)=\sqrt{SD^2(X)+SD^2(Y)}$

∴E(W)=E(X)+E(Y) = 80 +80=160

and

∴ $SD(W)=\sqrt{SD^2(X)+SD^2(Y)}$

$=\sqrt{11^2+11^2}$

$=\sqrt{2.11^2}$

$=11\sqrt2$

Therefore the mean and standard deviation of his total score if he plays a full 18 holes are 160 and $11\sqrt2$ respectively.

4 0

3 years ago

Consider the following functions. f1(x) = x, f2(x) = x2, f3(x) = 2x − 4x2 g(x) = c1f1(x) + c2f2(x) + c3f3(x) Solve for c1, c2, a

bekas [8.4K]

9514 1404 393

Answer:

(c1, c2, c3) = (-2t, 4t, t) . . . . for any value of t
NOT linearly independent

Step-by-step explanation:

We want ...

c1·f1(x) +c2·f2(x) +c3·f3(x) = g(x) ≡ 0

Substituting for the fn function values, we have ...

c1·x +c2·x² +c3·(2x -4x²) ≡ 0

This resolves to two equations:

x(c1 +2c3) = 0

x²(c2 -4c3) = 0

These have an infinite set of solutions:

c1 = -2c3

c2 = 4c3

Then for any parameter t, including the "trivial" t=0, ...

(c1, c2, c3) = (-2t, 4t, t)

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f1, f2, f3 are NOT linearly independent. (If they were, there would be only one solution making g(x) ≡ 0.)

7 0

2 years ago