Study the table representing the price of different amounts of apples, in pounds, and then complete the sentences. The constant difference in the y-values is . The linear function is .

Answer:

[tex]x=-3[/tex]

[tex]y=6[/tex]

Step-by-step explanation:

Elimination method:

[tex]2x+7y=36[/tex]

[tex]6x-7y=-60[/tex]

Add these equation to eliminate y:

[tex]8x=-24[/tex]

Then solve [tex]8x=-24[/tex] for x:

[tex]8x=-24[/tex]

[tex]\frac{8x}{8} =\frac{-24}{8}[/tex]

[tex]x=-3[/tex]

Add the value of x to solve y:

[tex]2x+7y=36[/tex]

Substitute [tex]-3[/tex] for x in [tex]2x+7y=36[/tex]

[tex](2)(-3)+7y=36[/tex]

[tex]7y-6=36[/tex]

[tex]7y=36+6[/tex]

[tex]7y=42[/tex]

[tex]y=42/7\\[/tex]

[tex]y=6[/tex]

{ [tex]x=-3[/tex] and [tex]y=6[/tex] }

hope this helps....

Answer:

Step-by-step explanation:

Answer:

15*.1=1.5

so either one or two people of the 15 would be left handed

Answer:

J

Step-by-step explanation:

Answer: 748,000

Step-by-step explanation: Multiply 850,000 by 0.12 and you would get 102,000 then you would subract 102,000 from 850,000 getting 748,000

Answer:

.252

Step-by-step explanation:

[tex]{10\choose7}*.65^7*(1-.65)^3=.252219625[/tex]

Answer:

Step-by-step explanation:

simple:

2900(1+.13*30)=14210

Compounding

[tex]2900(1+.13)^{30}=113436.1041[/tex]

Answer:

113436.1041

Step-by-step explanation:

2900 ( 1 +0.13 ) ^ 30

Formula : amount ( 1 + percentage ) ^ years

Answer:

not true....

assume $100 start.

in year 1 you are at $108 (up 8%)

in year 2  $108(.98) ... that is 2% down = 105.84...

thus your profit is up only 5.84% over the two years

Step-by-step explanation:

Answer:

H1 : P1 - P2 = 0

H1 : P1 - P2 > 0

Step-by-step explanation:

The test to be performed on the given data is ; difference in proportion ;

P1 = proportion od rooms in current year

P2 = proportion of rooms

The null hypothesis ``, H0 : p1 - p2 (this onstage null hypothesis and it is the initial truth, representing the notion that no difference in proportion exists.

H1 : P1 - P2 = 0

The alternative hypothesis takes takes the side that there is an increase on proportion of rooms occupied :

H1 : P1 - P2 > 0

Answer:

11.8 yd³

Step-by-step explanation:

Answer:

50% chance

Step-by-step explanation:

lol math go brrr

Answer:

D. √2 : 1

Step-by-step explanation:

The hypotenuse = 4√2 (longest side of a right triangle)

The given side = 4

Ratio of the hypotenuse to the given side = 4√2 : 4

Simplify by dividing both numbers by 4

√2 : 1

Answer:

Q9: x = 27, Q10: x = 17

Step-by-step explanation:

Answer:

m(m – 3) = 108

The correct equation can be used to solve for m, the greater integer is,

⇒ m (m - 3) = 108

We have to given that,

Two positive integers are 3 units apart on a number line.

And, Their product is 108.

Let us assume that,

In a number line, first point is m

Then, Second point is, m - 3

So, We get;

The correct equation can be used to solve for m, the greater integer is,

⇒ m (m - 3) = 108

Learn more about the equation visit:

brainly.com/question/28871326

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Hi there!

[tex]\large\boxed{12/5}}[/tex]

tan (angle) = Opposite side / Adjacent side, so:

Tan (A) = opposite side / adjacent side

= 24 / 10

Simplify:

= 12 / 5

Answer:

cf

=41

5 f-46

Step-by-step explanation:

thiis is the answer

Answer:

To find the inverse, switch the y(F(C)) and the x(C) variables.

So this function:

[tex]y=\frac{9}{5}x+32 \\[/tex]

Will become this function:

[tex]x=\frac{9}{5}y+32 \\[/tex]

You will then solve for y:

[tex]x=\frac{9}{5}y+32 \\x-32=\frac{9}{5}y\\5(x-32)=5(\frac{9}{5}y)\\5x-160=9y\\y=\frac{5x-160}{9}\\y=\frac{5x}{9}-\frac{160}{9}[/tex]

Substitute in the variables of this problem:

[tex]C(F)=\frac{5C}{9}-\frac{160}{9}[/tex]

Answer:

8/15

Step-by-step explanation:

Answer:

8/15

Step-by-step explanation:

when you divide 8/15 its 0.53

Answer:

just keep writing down outcome on a sheet of paper then count total

Step-by-step explanation:

I can't see the diagram sorry.

Step-by-step explanation:

Is there supposed to be a picture attached?

Answer:

Step-by-step explanation:

it is a quadratic function.

Answer:

Inconsistent system

Step-by-step explanation:

Given

The attached graph

Required

The type of system

When two lines are parallel, it means they have the same slope and as such, the system has no solution.

Equations with the same slope are:

[tex]y = 2x + 6[/tex]

[tex]y = 2x- 8[/tex]

Both have a slope of 2

Such system are referred to inconsistent system.

Hence, (c) is correct.

Answer:

nr.herkyrsfdlufshfsyfs

Step-by-step explanation:

dsfsyfksutryrysyrslufzmfyzydzufmzmhfzl

hdhfuthfzhkrskyrsgj

Pentagon has sum of 540°

Answer:

2/15

Step-by-step explanation:

(-3/5) (-2/9)

Rewriting

-3/9 * -2/5

-1/3 * -2/5

A negative times a negative is a positive.

2/15

Answer:

w = 5

y = 4

Step-by-step explanation:

w+y=9

3w-y=11

4w = 20

w = 5

y = 4

Answer: P = 4s

Step-by-step explanation:

P = 4s where s = the length of each side.  

Since each side of a square is the same length, the side length is multiplied by 4.

Answer:

[tex]90[/tex] [tex]ft^3[/tex]

Step-by-step explanation:

----------------------------------------

The formula to find the volume of a rectangular prism is   [tex]V=lwh[/tex]

Let's substitute the number for the length, width, and height now.

[tex]V=(6)(5)(3)[/tex]

[tex]V=(30)(3)[/tex]

[tex]V=90[/tex]

--------------------

Hope this is helpful.

Answer:

The remainder is 3x - 4

Step-by-step explanation:

[Remember] [tex]\frac{Dividend}{Divisor} = Quotient + \frac{Remainder}{Divisor}[/tex]

So, [tex]Dividend = (Quotient)(Divisor) + Remainder[/tex]

In this case our dividend is always P(x).

Part 1

When the divisor is [tex](x - 1)[/tex], the remainder is [tex]4[/tex], so we can say [tex]P(x) = (Quotient)(x - 1) + 4[/tex]

In order to get rid of "Quotient" from our equation, we must multiply it by 0, so [tex](x - 1) = 0[/tex]

When solving for [tex]x[/tex], we get

[tex]x - 1 = 0\\x - 1 + 1 = 0 + 1\\x = 1[/tex]

When [tex]x = 1[/tex],

[tex]P(x) = (Quotient)(x - 1) + 4\\P(1) = (Quotient)(1 - 1) + 4\\P(1) = (Quotient)(0) + 4\\P(1) = 0 + 4\\P(1) = 4[/tex]

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Part 2

When the divisor is [tex](x + 3)[/tex], the remainder is [tex]104[/tex], so we can say [tex]P(x) = (Quotient)(x + 3) + 104[/tex]

In order to get rid of "Quotient" from our equation, we must multiply it by 0, so [tex](x + 3) = 0[/tex]

When solving for [tex]x[/tex], we get

[tex]x + 3 = 0\\x + 3 - 3 = 0 - 3\\x = -3[/tex]

When [tex]x = -3[/tex],

[tex]P(x) = (Quotient)(x + 3) + 104\\P(-3) = (Quotient)(-3 + 3) + 104\\P(-3) = (Quotient)(0) + 104\\P(-3) = 0 + 104\\P(-3) = 104[/tex]

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Part 3

When the divisor is [tex](x^2 - x + 1)[/tex], the quotient is [tex](x^2 + x + 3)[/tex], and the remainder is [tex](ax + b)[/tex], so we can say [tex]P(x) = (x^2 + x + 3)(x^2 - x + 1) + (ax + b)[/tex]

From Part 1, we know that [tex]P(1) = 4[/tex] , so we can substitute [tex]x = 1[/tex] and [tex]P(x) = 4[/tex] into [tex]P(x) = (x^2 + x + 3)(x^2 - x + 1) + (ax + b)[/tex]

When we do, we get:

[tex]4 = (1^2 + 1 + 3)(1^2 - 1 + 1) + a(1) + b\\4 = (1 + 1 + 3)(1 - 1 + 1) + a + b\\4 = (5)(1) + a + b\\4 = 5 + a + b\\4 - 5 = 5 - 5 + a + b\\-1 = a + b\\a + b = -1[/tex]

We will call [tex]a + b = -1[/tex] equation 1

From Part 2, we know that [tex]P(-3) = 104[/tex], so we can substitute [tex]x = -3[/tex] and [tex]P(x) = 104[/tex] into [tex]P(x) = (x^2 + x + 3)(x^2 - x + 1) + (ax + b)[/tex]

When we do, we get:

[tex]104 = ((-3)^2 + (-3) + 3)((-3)^2 - (-3) + 1) + a(-3) + b\\104 = (9 - 3 + 3)(9 + 3 + 1) - 3a + b\\104 = (9)(13) - 3a + b\\104 = 117 - 3a + b\\104 - 117 = 117 - 117 - 3a + b\\-13 = -3a + b\\(-13)(-1) = (-3a + b)(-1)\\13 = 3a - b\\3a - b = 13[/tex]

We will call [tex]3a - b = 13[/tex] equation 2

Now we can create a system of equations using equation 1 and equation 2

[tex]\left \{ {{a + b = -1} \atop {3a - b = 13}} \right.[/tex]

By adding both equations' right-hand sides together and both equations' left-hand sides together, we can eliminate [tex]b[/tex] and solve for [tex]a[/tex]

So equation 1 + equation 2:

[tex](a + b) + (3a - b) = -1 + 13\\a + b + 3a - b = -1 + 13\\a + 3a + b - b = -1 + 13\\4a = 12\\a = 3[/tex]

Now we can substitute [tex]a = 3[/tex] into either one of the equations, however, since equation 1 has less operations to deal with, we will use equation 1.

So substituting [tex]a = 3[/tex] into equation 1:

[tex]3 + b = -1\\3 - 3 + b = -1 - 3\\b = -4[/tex]

Now that we have both of the values for [tex]a[/tex] and [tex]b[/tex], we can substitute them into the expression for the remainder.

So substituting [tex]a = 3[/tex] and [tex]b = -4[/tex] into [tex]ax + b[/tex]:

[tex]ax + b\\= (3)x + (-4)\\= 3x - 4[/tex]

Therefore, the remainder is [tex]3x - 4[/tex].

Answer:

[tex]-\frac{3(\sqrt{6}-\sqrt{2})}{2}\text{ or } \frac{-3\sqrt{6}+3\sqrt{2}}{2}}\text{ or }\frac{3(\sqrt{2}-\sqrt{6})}{2}[/tex]

Step-by-step explanation:

There are multiple ways to achieve and even express the exact answer to this problem. Because the exact value of [tex]6\cos(105^{\circ}})[/tex] is a non-terminating (never-ending) decimal, it does not have a finite number of digits. Therefore, you cannot express it as an exact value as a decimal, as you'd either have to round or truncate.

Solution 1 (Cosine Addition Identity):

Nonetheless, to find the exact value we must use trigonometry identities.

Identity used:

[tex]\cos(\alpha +\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta[/tex]

Notice that [tex]45+60=105[/tex] and therefore we can easily solve this problem if we know values of [tex]\cos(45^{\circ})[/tex], [tex]\cos(60^{\circ})[/tex], [tex]\sin (45^{\circ})[/tex], and [tex]\sin(60^{\circ})[/tex], which is plausible as they are all key angles on the unit circle.

Recall from either memory or the unit circle that:

Therefore, we have:

[tex]\cos(105^{\circ})=\cos(45^{\circ}+60^{\circ}}),\\\cos(45^{\circ}+60^{\circ}})=\cos 45^{\circ}\cos 60^{\circ}-\sin 45^{\circ}\sin 60^{\circ},\\\cos(45^{\circ}+60^{\circ}})=\frac{\sqrt{2}}{2}\cdot \frac{1}{2}-\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2},\\\cos(105^{\circ})=\frac{\sqrt{2}}{4}-\frac{\sqrt{6}}{4},\\\cos(105^{\circ})={\frac{-\sqrt{6}+\sqrt{2}}{4}}[/tex]

Since we want the value of [tex]6\cos 105^{\circ}[/tex], simply multiply this by 6 to get your final answer:

[tex]6\cdot {\frac{-\sqrt{6}+\sqrt{2}}{4}}=\frac{-3\sqrt{6}+3\sqrt{2}}{2}}=\boxed{\frac{3(\sqrt{2}-\sqrt{6})}{2}}[/tex]

Solution 2 (Combination of trig. identities):

Although less plausible, you may have the following memorized:

[tex]\sin 15^{\circ}=\cos75^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4},\\\sin 75^{\circ}=\cos15^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}[/tex]

If so, we can use the following trig. identity:

[tex]\cos(\theta)=\sin(90^{\circ}-\theta)[/tex] (the cosine of angle theta is equal to the sine of the supplement of angle theta - the converse is also true)

Therefore,

[tex]\cos (105^{\circ})=\sin (90^{\circ}-105^{\circ})=\sin(-15^{\circ})[/tex]

Recall another trig. identity:

[tex]\sin(-\theta)=-\sin (\theta)[/tex] and therefore:

[tex]\sin (-15^{\circ})=-\sin (15^{\circ})[/tex]

Multiply by 6 to get:

[tex]6\cos (105^{\circ})=-6\sin (15^{\circ})=-6\cdot \frac{\sqrt{6}-\sqrt{2}}{4}=\boxed{-\frac{3(\sqrt{6}-\sqrt{2})}{2}}[/tex] (alternative final answer).