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3 years ago

F. ALGEBRA Find x. a. 25.6 b. 22.5 c. 38 d. 40

Mathematics

1 answer:

LenaWriter [7]3 years ago

5 0

Answer:

B. 22.5

Step-by-step explanation:

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The length of atrain car is 50.6 feet. This is 5.8 feet less than 6 times the width. What is the width?

hichkok12 [17]

∗ l = 50.6 ft
∗ w = ?

∗ The length is 50.6 feet. The question is saying that 50.6 is 5.8 feet less than 6 times the width.

∗ 6w - 5.8 = 50.6

∗ 6w = 56.4

∗ w = 9.4

Now lets check.

∗ If the length is 5.8 feet less than 6 times the width, and we found 9.4 as the answer, then;
6 times 9.4 - 5.8 should equal 50.6

∗ 6 • 9.4 - 5.8 = l

∗ 56.4 - 5.8 = l

∗ 50.6 = l

⋆The width is 9.4 feet⋆

♡´･ᴗ･`♡вяαιиℓιєѕт ρℓєαѕє♡´･ᴗ･`♡

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2 years ago

If (fg)(x) = h(x) such that h of x is equal to the square root of the quantity 8 times x plus 6 end quantity which of the follow

Sladkaya [172]

The value of the functions f(x) and g(x) will be √(4x + 3) and √2. Then the correct option is B.

<h3>What is a function?</h3>

A statement, principle, or policy that creates the link between two variables is known as a function. Functions are found all across mathematics and are required for the creation of complex relationships.

If (f g)(x) = h(x) such that h(x) = √(8x + 6). Then we have

(f g)(x) = h(x)

f(x) · g(x) = h(x)

Then put the value of h(x), then we have

f(x) · g(x) = √(8x + 6)

f(x) · g(x) = √2(4x + 3)

f(x) · g(x) = √(4x + 3) × √2

Thus, the value of the functions f(x) and g(x) will be √(4x + 3) and √2.

Then the correct option is B.

More about the function link is given below.

brainly.com/question/5245372

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1 year ago

Question 13<br> If the nth term of a sequence is<br> 6n + 4 what is the 9th term?

GarryVolchara [31]

Answer:

Step-by-step explanation:

hello :

the 9th term when : n=9

6(9) + 4 =58

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3 years ago

Bowl A contains three red and two white chips, and bowl B contains four red and three white chips. A chip is drawn at random fro

____ [38]

$\Large \text {Answer:}$

$\large \boxed {\dfrac{23}{40}}$

$\Large \text {Step-by-step explanation:}$

$\text{P}_{B}(r) = \text{P}_{A}(r) \cdot \text{P}_{B}(r) + \text{P}_{A}(w) \cdot \text{P}_{B}(r)$

$\text{P}_{B}(r) = \left (\dfrac{3}{5} \right ) \negthickspace \left(\dfrac{5}{8}\right ) + \left (\dfrac{2}{5} \right ) \negthickspace \left (\dfrac{4}{8} \right )$

$\text{P}_{B}(r) = \dfrac{23}{40}$

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2 years ago

Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E

Ipatiy [6.2K]

Answer:

(A) $y=ke^{2t}$ with $k\in\mathbb{R}$ .

(B) $y=ke^{2t}/2-1/2$ with $k\in\mathbb{R}$

(D) $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ ,

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then $0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}$ . With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form $e^{2t}$ with $t$ real.

(B) Proceeding and the previous item, we obtain $1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2$ . Which is not a vector space with the usual operations (this is because $-1/2$ ), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set $y=e^{mt} \Rightarrow y''=m^2e^{mt}$ and we obtain the characteristic equation $0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$ , and as in the first items the set of solutions form a vector space.

(D) Using C, let be $y=me^{3t}$ we obtain that it must satisfies $3^2m-4m=1\Rightarrow m=1/5$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ , and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).

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3 years ago