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3 years ago

15

If f(1) = 10 and f(n) = f(n = 1) – 4 then find the value of f(5).

1 answer:

svetoff [14.1K]3 years ago

5 0

Given:

$f(1)=10$ and $f(n)=f(n-1)-4$ .

To find:

The value of f(5).

Solution:

We have,

$f(n)=f(n-1)-4$

For $n=2$ ,

$f(2)=f(2-1)-4$

$f(2)=f(1)-4$

$f(2)=10-4$

$f(2)=6$

For $n=3$ ,

$f(3)=f(3-1)-4$

$f(3)=f(2)-4$

$f(3)=6-4$

$f(3)=2$

For $n=4$ ,

$f(4)=f(4-1)-4$

$f(4)=f(3)-4$

$f(4)=2-4$

$f(4)=-2$

For $n=5$ ,

$f(5)=f(5-1)-4$

$f(5)=f(4)-4$

$f(5)=-2-4$

$f(5)=-6$

Therefore, the value of $f(5)$ is $-6$ .

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If the original square had a side length of

irina [24]

Answer:

Part a) The new rectangle labeled in the attached figure N 2

Part b) The diagram of the new rectangle with their areas in the attached figure N 3, and the trinomial is $x^{2} +11x+28$

Part c) The area of the second rectangle is 54 in^2

Part d) see the explanation

Step-by-step explanation:

The complete question in the attached figure N 1

Part a) If the original square is shown below with side lengths marked with x, label the second diagram to represent the new rectangle constructed by increasing the sides as described above

we know that

The dimensions of the new rectangle will be

$Length=(x+4)\ in$

$width=(x+7)\ in$

The diagram of the new rectangle in the attached figure N 2

Part b) Label each portion of the second diagram with their areas in terms of x (when applicable) State the product of (x+4) and (x+7) as a trinomial

The diagram of the new rectangle with their areas in the attached figure N 3

we have that

To find out the area of each portion, multiply its length by its width

$A1=(x)(x)=x^{2}\ in^2$

$A2=(4)(x)=4x\ in^2$

$A3=(x)(7)=7x\ in^2$

$A4=(4)(7)=28\ in^2$

The total area of the second rectangle is the sum of the four areas

$A=A1+A2+A3+A4$

State the product of (x+4) and (x+7) as a trinomial

$(x+4)(x+7)=x^{2}+7x+4x+28=x^{2} +11x+28$

Part c) If the original square had a side length of x = 2 inches, then what is the area of the second rectangle?

we know that

The area of the second rectangle is equal to

$A=A1+A2+A3+A4$

For x=2 in

substitute the value of x in the area of each portion

$A1=(2)(2)=4\ in^2$

$A2=(4)(2)=8\ in^2$

$A3=(2)(7)=14\ in^2$

$A4=(4)(7)=28\ in^2$

$A=4+8+14+28$

$A=54\ in^2$

Part d) Verify that the trinomial you found in Part b) has the same value as Part c) for x=2 in

We have that

The trinomial is

$A(x)=x^{2} +11x+28$

For x=2 in

substitute and solve for A(x)

$A(2)=2^{2} +11(2)+28$

$A(2)=4 +22+28$

$A(2)=54\ in^2$ ----> verified

therefore

The trinomial represent the total area of the second rectangle

7 0

3 years ago

In ΔOPQ, o = 58 cm, q = 13 cm and ∠Q=52°. Find all possible values of ∠O, to the nearest degree.

PSYCHO15rus [73]

Answer:

There is no possible triangles

Step-by-step explanation:

\frac{\sin A}{a}=\frac{\sin B}{b}

a

sinA

=

b

sinB

From the reference sheet (reciprocal version).

\frac{\sin O}{58}=\frac{\sin 52}{13}

58

sinO

=

13

sin52

Plug in values.

\sin O=\frac{58\sin 52}{13}\approx 3.5157403

sinO=

13

58sin52

≈3.5157403O=\sin^{-1}(3.5157403)= \text{ERROR}

O=sin

−1

(3.5157403)=ERROR

Sine cannot be greater than 1.

{No Possible Triangles}

No Possible Triangles

{}

{}

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tangare [24]

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</span><span>
78: 1,2,3,6,13, 26, 39, 78
</span>
The answer is 6 :)

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BigorU [14]

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Answer:

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Step-by-step explanation:

Just take my word for it please.

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