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2 years ago

Please solve both of these with how you solved them.

Mathematics

1 answer:

Paladinen [302]2 years ago

6 0

Answer:

a. (2,5)

b. (14,15.5)

Step-by-step explanation:

y=4x-3

y=-2x+9

Set both equations equal to each other to solve for x.

4x-3=-2x+9

6x=12

x=2

Plug in x to solve for y.

y=4x-3

y=4(2)-3

y=8-3

y=5

(2,5)

y=(5/4x)-2

y=(-1/4x)+19

Set both equations equal to each other to solve for x.

(5/4x)-2=(-1/4x)+19

4((5/4x)-2=(-1/4x)+19)

5x-8=-x+76

6x=84

x=14

Plug in x to solve for y.

y=(5/4x)-2

y=(5/4(14))-2

y=17.5-2

y=15.5

(14,15.5)

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A cell phone plan has a monthly cost that is shown in the table below. What is the correct statement regarding the average rate

BARSIC [14]

Answer:

Second from top...

The average rate of change is $0.07, meaning that for each minute of talk time, the monthly bill increases by $0.07.

Step-by-step explanation:

Pick any two points and calculate the slope which will represent the rate of change:

(Minutes, Cost)

(0, 9.95)

(10, 10.65)

Rate of change:

slope m = (y₂ - y₁) / (x₂ - x₁)

= (10.65 - 9.95) / (10 - 0)

= 0.7 / 10

= 0.07

y is cost

m is rate of change

x is minute(s) use

b is fixed monthly cost regardless if any minutes is used

y = mx + b

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4 0

3 years ago

The sum of the first term of an ap is 240 and the sum of the next 4 term is 220 find the first term of the ap

Sindrei [870]

Answer:

The common difference is -5/4

T(n) = T(0) - 5n/4,

where T(0) can be any number. d = -5/4

Assuming T(0) = 0, then first term

T(1) = 0 -5/4 = -5/4

Step-by-step explanation:

T(n) = T(0) + n*d

Let

S1 = T(x) + T(x+1) + T(x+2) + T(x+3) = 4*T(0) + (x + x+1 + x+2 + x+3)d = 240

S2 = T(x+4) + T(x+5) + T(x+6) + T(x+7) = 4*T(0) + (x+5 + x+6 + x+7 + x+8)d = 220

S2 - S1

= 4*T(0) + (x+5 + x+6 + x+7 + x+8)d - (4*T(0) + (x+1 + x+2 + x+3 + x+4)d)

= (5+6+7+8 - 1 -2-3-4)d

= 4(4)d

= 16d

Since S2=220, S1 = 240

220-240 = 16d

d = -20/16 = -5/4

Since T(0) has not been defined, it could be any number.

7 0

3 years ago

Find the exact value of sin2 theta given sin theta =5/13, 90°<theta <180°

Zepler [3.9K]

Now, we know that 90°< θ <180°, that simply means the angle θ is in the II quadrant, where sine is positive and cosine is negative.

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{13}}\impliedby \textit{now let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}$

$\bf \pm\sqrt{13^2-5^2}=a\implies \pm\sqrt{144}=a\implies \pm 12 =a \implies \stackrel{II~quadrant}{-12=a}
\\\\\\
therefore \qquad cos(\theta )=\cfrac{\stackrel{adjacent}{-12}}{\stackrel{hypotenuse}{13}}
\\\\
-------------------------------\\\\
sin(2\theta )\implies 2sin(\theta )cos(\theta )\implies 2\left(\frac{5}{13} \right)\left( \frac{-12}{13} \right)\implies -\cfrac{120}{169}$

3 0

4 years ago

The vertex form of the equation of a parabola is y = 2(x + 3)2 - 5. What is the standard form of the equation?

svet-max [94.6K]

Hello :
<span>y = 2(x + 3)² - 5
y = 2(x²+6x+9) -5
y = 2x² +12x +13...(answer : </span><span>A) y=2x^2+12x+13 )</span>

7 0

3 years ago

Read 2 more answers

What is the quotient when 4x3 + 2x + 7 is divided by x + 3?

Arte-miy333 [17]

Answer:

The quotient of this division is $(4x^2 -12x + 38)$ . The remainder here would be $-26$ .

Step-by-step explanation:

The numerator $4x^3 + 2x + 7$ is a polynomial about $x$ with degree $3$ .

The divisor $x + 3$ is a polynomial, also about $x$ , but with degree $1$ .

By the division algorithm, the quotient should be of degree $3 - 1 = 2$ , while the remainder shall be of degree $1 - 1 = 0$ (i.e., the remainder would be a constant.) Let the quotient be $a\,x^2 + b\, x + c$ with coefficients $a$ , $b$ , and $c$ .

$4x^3 + 2x + 7 = \left(a\,x^2 + b\, x + c\right)(x + 3)$ .

Start by finding the first coefficient of the quotient.

The degree-three term on the left-hand side is $4 x^3$ . On the right-hand side, that would be $a\, x^3$ . Hence $a = 4$ .

Now, given that $a = 4$ , rewrite the right-hand side:

$\begin{aligned}&\left(4\,x^2 + b\, x + c\right)(x + 3) \cr =& \left(4x^2 + (b\, x + c)\right)(x + 3) \cr =& 4x^2(x + 3) + (bx + c)(x + 3) \cr =& 4x^3 + 12x^2 + (bx + c)(x + 3)\end{aligned}$ .

Hence:

$4x^3 + 2x + 7 = 4x^3 + 12x^2 + (b\,x + c)(x + 3)$

Subtract $\left(4x^3 + 12x^2\right$ from both sides of the equation:

$-12x^2 + 2x + 7 = (b\,x + c)(x + 3)$ .

The term with a degree of two on the left-hand side has coefficient $(-12)$ . Since the only term on the right hand side with degree two would have coefficient $b$ , $b = -12$ .

Again, rewrite the right-hand side:

$\begin{aligned}&\left(-12 x + c\right)(x + 3) \cr =& \left(-12 x+ c\right)(x + 3) \cr =& (-12x)(x + 3) + c(x + 3) \cr =& -12x^2 -36x + (bx + c)(x + 3)\end{aligned}$ .

Subtract $-12x^2 -36x$ from both sides of the equation:

$38x + 7 = c(x + 3)$ .

By the same logic, $c = 38$ .

Hence the quotient would be $(4x^2 - 12x + 38)$ .

6 0

3 years ago