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3 years ago

F(x)=2(x+6)-4, f(x)=6

Mathematics

2 answers:

Aleks04 [339]3 years ago

6 0

F(X) = 6

6 = 2(X+6) - 4

10 = 2(X+6)

5 = X+6

X = 5 - 6

X = -1.

IRINA_888 [86]3 years ago

6 0

5=x+6 gllllllllllllll

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3 years ago

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Talja [164]

Answer:

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Step-by-step explanation:

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3 years ago

If one basket contains five bottles each weighing 7 1/4 ANSYS and the other basket has six apples each weighing 6 1/2 ounces woo

Tems11 [23]

<em>Question:</em>

<em>If one basket contains five bottles each weighing 7 1/4 and the other basket has six apples each weighing 6 1/2 ounces. Which basket is heavier and by how much ?</em>

Answer:

The basket of bottles weighs more than the basket of apples by $3\frac{3}{4}$ ounces

Step-by-step explanation:

Given

$Bottles = 5$

$Weight(Bottle) = 7\frac{1}{4}$

$Apples = 6$

$Weight(Apple) = 6\frac{1}{2}$

Required

Determine which basket is heavy

and by how much?

First, we need the total weight of each basket.

For the bottles' basket

$Total\ Weight = Weight * Bottles$

$Total\ Weight = 7\frac{1}{4} * 5$

$Total\ Weight = \frac{29}{4} * 5$

$Total\ Weight = \frac{29 * 5}{4}$

$Total\ Weight = \frac{145}{4}$

$Total\ Weight = 36\frac{1}{4}$

For the bottles' basket

$Total\ Weight = Weight * Apples$

$Total\ Weight = 6\frac{1}{2} * 5$

$Total\ Weight = \frac{13}{2} * 5$

$Total\ Weight = \frac{13* 5}{2}$

$Total\ Weight = \frac{65}{2}$

$Total\ Weight = 32\frac{1}{2}$

By comparing both total weights, $36\frac{1}{4} > 32\frac{1}{2}$

Hence:

The basket of bottles weighs more than the basket of apples

Calculating the difference:

$Difference = 36\frac{1}{4} - 32\frac{1}{2}$

$Difference = \frac{145}{4} - \frac{65}{2}$

$Difference = \frac{145 - 130}{4}$

$Difference = \frac{15}{4}$

$Difference = 3\frac{3}{4}$

Hence:

It weighs more by $3\frac{3}{4}$ ounces

8 0

3 years ago

The manufacturer of a CD player has found that the revenue R (in dollars) is Upper R (p )equals negative 5 p squared plus 1 com

AleksAgata [21]

Answer:

The maximum revenue is $1,20,125 that occurs when the unit price is $155.

Step-by-step explanation:

The revenue function is given as:

$R(p) = -5p^2 + 1550p$

where p is unit price in dollars.

First, we differentiate R(p) with respect to p, to get,

$\dfrac{d(R(p))}{dp} = \dfrac{d(-5p^2 + 1550p)}{dp} = -10p + 1550$

Equating the first derivative to zero, we get,

$\dfrac{d(R(p))}{dp} = 0\\\\-10p + 1550 = 0\\\\p = \dfrac{-1550}{-10} = 155$

Again differentiation R(p), with respect to p, we get,

$\dfrac{d^2(R(p))}{dp^2} = -10$

At p = 155

$\dfrac{d^2(R(p))}{dp^2} < 0$

Thus by double derivative test, maxima occurs at p = 155 for R(p).

Thus, maximum revenue occurs when p = $155.

Maximum revenue

$R(155) = -5(155)^2 + 1550(155) = 120125$

Thus, maximum revenue is $120125 that occurs when the unit price is $155.

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3 years ago

In what quadrant will θ terminate, if sin θ is positive and sec θ is negative?

kow [346]

Sin x is possitive in two quadrants :

quadrant I and quadrant II

in quadrant I, cos x is positive

in quadrant II , cos x is negative

so i think the answer is : quadrant II

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3 years ago