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AlexFokin [52]
1 year ago
7

The area on a wall covered by a rectangular poster is 294 square inches. The length of the poster is 1.5 times longer than the w

idth of the poster. What are the dimensions of the poster?
Mathematics
2 answers:
Dafna1 [17]1 year ago
7 0
A= l×w
296 = 1.5l × w

You should get 20 at the end
Alex787 [66]1 year ago
4 0
Hey there,

Your question states: <span>The area on a wall covered by a rectangular poster is 294 square inches. The length of the poster is 1.5 times longer than the width of the poster. What are the dimensions of the poster? </span>

l= 16 \ x  \ 1.25 = 20  in

Your correct answer would be \boxed{20}

Hope this helps.
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Identify the rule of inference that is used to derive the conclusion "You do not eat tofu" from the statements "For all x, if x
Arturiano [62]

Answer:

1(b) ∀ (A(x) ⇒ B(x) )

2(b) ∀ (B(x) ⇒ C(x) )

3(b) ∀ (B(x) ⇒ E(x) )

Step-by-step explanation:

1) Tofu is healthy

2) Tofu is healthy to eat

3) Tofu eats what taste good

1a) For all x, if x is healthy to eat

2a) For all x, if x is not healthy to eat, then x does not taste good.

3a) For all x, if x is healthy to eat, then x is healthy to eat what tastes good

For all x in order to symbolize the statement

1(a)   2(a)  3(a)

If we use:

A(x): Tofu is healthy

B(x): Tofu is healthy to eat

C(x): Tofu eats what taste good

E(x): Tofu only eat what tastes good

If we symbolize "For all x" by the symbol ∀ then then the propositions 1(a), 2(a) and 3(a) can be written as:

1(b) ∀ (A(x) ⇒ B(x) )

2(b) ∀ (B(x) ⇒ C(x) )

3(b) ∀ (B(x) ⇒ E(x) )

6 0
2 years ago
The admission fee at a movie theater is $5 for children and $9 for adults. If 3200 people go to the movies and $24000 is collect
Usimov [2.4K]
For this problem, let x be the number of children and y for adults. Formulate the equations: 1st equation, x + y = 3,200 and 2nd equation 5x + 9y = 24,000. Re-arrange 1st equation into x = 3200 - y. Then, substitute into 2nd equation, 5(3,200-y) + 9y = 24,000. Then, solve for y. The 16,000 - 5y + 9y = 24000. Final answer is, y = 2000 adults went to watch the movie.
6 0
2 years ago
The earth has a mass of approximately 6\cdot 10^{24}6⋅10 24 6, dot, 10, start superscript, 24, end superscript kilograms (\text{
Alex_Xolod [135]

Answer:

0.02

Step-by-step explanation:

The volume of the earth's oceans is approximately 1.34\cdot 10^{9}1.34⋅10

9

1, point, 34, dot, 10, start superscript, 9, end superscript cubic kilometers (\text{km}^3)(km

3

)left parenthesis, start text, k, m, end text, cubed, right parenthesis, and ocean water has a mass of about 1.03 \cdot 10^{12}\,\dfrac{\text{kg}}{\text{km}^3}1.03⋅10

12

 

km

3

kg

​

1, point, 03, dot, 10, start superscript, 12, end superscript, start fraction, start text, k, g, end text, divided by, start text, k, m, end text, cubed, end fraction .

To simplify, we will use the product of powers property of exponents that says that x^a\cdot x^b = x^{a+b}x

a

⋅x

b

=x

a+b

x, start superscript, a, end superscript, dot, x, start superscript, b, end superscript, equals, x, start superscript, a, plus, b, end superscript.

\qquad 1.34\cdot 10^{9}\,\cancel{\text{km}^3} \cdot 1.03 \cdot 10^{12}\,\dfrac{\text{kg}}{\cancel{\text{km}^3}} = 1.3802 \cdot 10^{21}\,\text{kg}1.34⋅10

9

 

km

3

⋅1.03⋅10

12

 

km

3

kg

​

=1.3802⋅10

21

kg1, point, 34, dot, 10, start superscript, 9, end superscript, start cancel, start text, k, m, end text, cubed, end cancel, dot, 1, point, 03, dot, 10, start superscript, 12, end superscript, start fraction, start text, k, g, end text, divided by, start cancel, start text, k, m, end text, cubed, end cancel, end fraction, equals, 1, point, 3802, dot, 10, start superscript, 21, end superscript, start text, k, g, end text

Hint #2

Next we want to know what portion of the earth's mass this represents. We have:

\qquad \begin{aligned} \dfrac{\text{mass of the oceans}}{\text{total mass of the earth}} &= \dfrac{1.3802 \cdot 10^{21}\,\text{kg}}{6\cdot 10^{24}\,\text{kg}} \\\\ &= \dfrac{1.3802}{6\cdot 10^{3}} \\\\ &= \dfrac{1.3802}{6000} \\\\ &= 0.0002300\overline{3} \end{aligned}

total mass of the earth

mass of the oceans

​

​

 

=

6⋅10

24

kg

1.3802⋅10

21

kg

​

=

6⋅10

3

1.3802

​

=

6000

1.3802

​

=0.0002300

3

​

To convert this to a percent, we multiply by 100100100, so the oceans represent 0.02300\overline{3}\%0.02300

3

%0, point, 02300, start overline, 3, end overline, percent of the earth's total mass, according to these figures.

Hint #3

To the nearest hundredth of a percent, 0.020.020, point, 02 percent of the earth's mass is from oceans.

6 0
1 year ago
A new drug on the market is known to cure 30% of patients with cervical cancer. If a group of 18 patients is randomly selected,
Dima020 [189]

Answer:

Probability cured of cervical cancer =  18C0 (0.30)⁰(0.70)¹⁸ +  18C1(0.30)(0.70)¹⁷

Step-by-step explanation:

Given:

Patients cured = 30% = 0.30

Number of patients (n) = 18

Probability cured of cervical cancer = P(X≤1)

Probability cured of cervical cancer  = P(X=0) + P(X=1)

Probability cured of cervical cancer =  18C0 (0.30)⁰(0.70)¹⁸ +  18C1(0.30)(0.70)¹⁷

5 0
1 year ago
Consider the initial value problem: 2ty′=8y, y(−1)=1. Find the value of the constant C and the exponent r so that y=Ctr is the s
VikaD [51]

The correct question is:

Consider the initial value problem

2ty' = 8y, y(-1) = 1

(a) Find the value of the constant C and the exponent r such that y = Ct^r is the solution of this initial value problem.

b) Determine the largest interval of the form a < t < b on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution.

c) What is the actual interval of existence for the solution obtained in part (a) ?

Step-by-step explanation:

Given the differential equation

2ty' = 8y

a) We need to find the value of the constant C and r, such that y = Ct^r is a solution to the differential equation together with the initial condition y(-1) = 1.

Since Ct^r is a solution to the initial value problem, it means that y = Ct^r satisfies the said problem. That is

2tdy/dt - 8y = 0

Implies

2td(Ct^r)/dt - 8(Ct^r) = 0

2tCrt^(r - 1) - 8Ct^r = 0

2Crt^r - 8Ct^r = 0

(2r - 8)Ct^r = 0

But Ct^r ≠ 0

=> 2r - 8 = 0 or r = 8/2 = 4

Now, we have r = 4, which implies that

y = Ct^4

Applying the initial condition y(-1) = 1, we put y = 1 when t = -1

1 = C(-1)^4

C = 1

So, y = t^4

b) Let y = F(x,y)................(1)

Suppose F(x, y) is continuous on some region, R = {(x, y) : x_0 − δ < x < x_0 + δ, y_0 −ę < y < y_0 + ę} containing the point (x_0, y_0). Then there exists a number δ1 (possibly smaller than δ) so that a solution y = f(x) to (1) is defined for x_0 − δ1 < x < x_0 + δ1.

Now, suppose that both F(x, y)

and ∂F/∂y are continuous functions defined on a region R. Then there exists a number δ2

(possibly smaller than δ1) so that the solution y = f(x) to (1) is

the unique solution to (1) for x_0 − δ2 < x < x_0 + δ2.

c) Firstly, we write the differential equation 2ty' = 8y in standard form as

y' - (4/t)y = 0

0 is always continuous, but -4/t has discontinuity at t = 0

So, the solution to differential equation exists everywhere, apart from t = 0.

The interval is (-infinity, 0) n (0, infinity)

n - means intersection.

7 0
1 year ago
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