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Degger [83]
2 years ago
9

Evaluate ∣∣256+y∣∣ for y=74. A. 225 B. 315 C. 345 D. 4712

Mathematics
1 answer:
aleksklad [387]2 years ago
5 0

Answer:

678

Step-by-step explanation:

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16.A guidance counselor is planning schedules for 30 students. Sixteen students say they want to take French, 16 want to take Sp
MariettaO [177]

Answer:7

Step-by-step explanation:

This can be solved by Venn-diagram

Given there  are total 5 students who want french and Latin

also 3 among them want Spanish,french & Latin

i.e. only 2 students wants both french and Latin only.

Also Student who want only Latin is 5

Thus Student who wants Latin and Spanish both only is 11-5-3-2=1

Students who want only Spanish is 8 Thus students who wants Spanish and French is 4

Similarly Students who wants Only French is 16-4-3-2=7

8 0
2 years ago
Read 2 more answers
In each of Problems 5 through 10, verify that each given function is a solution of the differential equation.
WARRIOR [948]

Answer:

For First Solution: y_1(t)=e^t

y_1(t)=e^t is the solution of equation y''-y=0.

For 2nd Solution:y_2(t)=cosht

y_2(t)=cosht  is the solution of equation y''-y=0.

Step-by-step explanation:

For First Solution: y_1(t)=e^t

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

y_1(t)=e^t

First order derivative:

y'_1(t)=e^t

2nd order Derivative:

y''_1(t)=e^t

Put Them in equation y''-y=0

e^t-e^t=0

0=0

Hence y_1(t)=e^t is the solution of equation y''-y=0.

For 2nd Solution:

y_2(t)=cosht

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

y_2(t)=cosht

First order derivative:

y'_2(t)=sinht

2nd order Derivative:

y''_2(t)=cosht

Put Them in equation y''-y=0

cosht-cosht=0

0=0

Hence y_2(t)=cosht  is the solution of equation y''-y=0.

3 0
2 years ago
In ADEF, f = 610 inches, e = 590 inches and ZE=70°. Find all possible values of ZF,
Elina [12.6K]

Answer:

"76°" is the appropriate solution.

Step-by-step explanation:

Please find attachment of the diagram according to the given query.

The given values are:

In ΔDEF,

f = 610 inches

e = 590

∠E = 70°

∠F = ?

By using the law of sines, we get

⇒  \frac{Sin E}{e} =\frac{Sin  F}{f}

On substituting the values, we get

⇒  \frac{Sin70^{\circ}}{590} =\frac{SinF}{610}

On applying cross multiplication, we get

⇒   SinF=\frac{610\times Sin 70^{\circ}}{590}

On substituting the values, we get

⇒            =\frac{61\times 0.95969262}{59}

⇒            =\frac{57.3212499}{59}

⇒            =0.971546608

now,

⇒  F=Sin^{-1}(0.971546608)

⇒      =76^{\circ}

4 0
2 years ago
Kaylee has 16 pairs of flip-flops in her closet. She discovers that she has 1 red, 3 blue, 2 brown, 5 black, 1 purple, and 4 pai
Natalija [7]

Answer:3/16 and 4/16

Step-by-step explanation:

4 0
2 years ago
What are the domain, range, and asymptote of h(x) = 6x – 4? domain: {x | x is a real number}; range: {y | y > 4}; asymptote:
Mkey [24]

Answer:

Step-by-step explanation:

The domain of a function is the set for which the function is defined. Our function is the function h(x) = 6x-4. This function is defined regardless of the value of x, so it is defined for every real value of x. That is, it's domain is the set {x|x is a real number}.

The range of the function is the set of all possible values that the function might take, that is {y|y=6x-4}. Recall that every real number y could be written of the form y=6x-4 for a particular x. So the range of the function is the set {y|y is a real number}.

Note that as x gets bigger, the value of 6x-4 gets also bigger, then it doesn't approach any particular number. Note also that as x approaches - infinity, the value of 6x-4 approaches also - infinity. In this case, we don't have any horizontal asymptote. Since the function is defined for every real number, it doesn't have any vertical asymptote. Since h is a linear function, it cannot have any oblique asymptote, then h doesn't have any asymptote.

4 0
2 years ago
Read 2 more answers
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