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2 years ago

If A = 4x2 + 9x - 5 and B = 6x2 – 3x - 9, then A - B equals

Mathematics

1 answer:

SSSSS [86.1K]2 years ago

3 0

Answer:

1 I think I'm not 100% sure though

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tia_tia [17]

Answer:

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7 0

2 years ago

Rico bought 3 adult tickets and 5 child tickets for a total of $50.

aliina [53]

Step-by-step explanation:

3x + 5y = 50

x + 5y = 31

use substitution

x = 31-5y

3(31-5y) +5y = 50

93 -15y + 5y = 50

93 - 10y = 50

--10y = -43

y = 43/10 = $4.30 for kids

now solve for x

x + 5(4.30) = 31

x + 21.5 = 31

x = $9.5 for adults

4 0

2 years ago

Which proportion can you use to find the value of a?

Ugo [173]

<u>Given</u>:

The given triangle is a similar triangle.

The length of the hypotenuse is 18 units.

The length of the leg is a.

The length of the part of the hypotenuse is 16 units.

We need to determine the proportion used to find the value of a.

<u>Proportion to find the value of a:</u>

We shall find the proportion to determine the value of a using the geometric mean leg rule.

Applying the leg rule, we have;

$\frac{\text { hypotenuse }}{\text { leg }}=\frac{\text { leg }}{\text { part }}$

Substituting the values of hypotenuse, leg and part, we get;

$\frac{18}{a}=\frac{a}{16}$

Thus, the proportion used to find the value of a is $\frac{18}{a}=\frac{a}{16}$

Hence, Option D is the correct answer.

6 0

3 years ago

Which one of these answers is the actual one

Arturiano [62]

The answer is: 3 3/8 sq ft so C

3 0

2 years ago

Read 2 more answers

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$