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2 years ago

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Three less than three times a number is three more than five times that same number . What is the number

1 answer:

Pachacha [2.7K]2 years ago

4 0

Answer: -3

Solution:

Let n be that number.

So,

Three less than 3*no = 3 more than 5*no

3*n-3=5*n+3

5n-3n=-3-3

2n=-6

n=-3

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Write a quotient that gives the same answer as 4.65 divided by 0.12

jeyben [28]

4.65/.12= 465/12
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2 years ago

Read 2 more answers

the frame of a picture is 28 cm by 32 cm outside and is of uniform width.What is the width of the frame if 192 cm2 of the pictur

natita [175]

Answer:

The width of the frame is equal to $8\ cm$

Step-by-step explanation:

Let

x-------> the width of the frame picture

we know that

$(28-2x)(32-2x)=192$

$896-56x-64x+4x^{2} =192$

$4x^{2}-120x+704=0$

Simplify

$x^{2}-30x+176=0$

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^{2}-30x+176=0$

so

$a=1\\b=-30\\c=176$

substitute in the formula

$x=\frac{30(+/-)\sqrt{(-30)^{2}-4(1)(176)}} {2(1)}$

$x=\frac{30(+/-)\sqrt{196}} {2}$

$x=\frac{30(+/-)14} {2}$

$x=\frac{30+14} {2}=22\ cm$ ----> this solution is not logic

$x=\frac{30-14} {2}=8\ cm$

The width of the frame is equal to $8\ cm$

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3 years ago

If tan 0=1/3<br> find cot 0.

Karolina [17]

Answer: cot 0 = 3

Cotangent is the reciprocal of tangent, so it would be 3/1, which simplified to 3.

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2 years ago

Please help me with the below question.

VMariaS [17]

By letting

$y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}$

we get derivatives

$y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}$

$y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}$

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

$5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0$

Examine the lowest degree term $\left(x^{r-1}\right)$ , which gives rise to the indicial equation,

$5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0$

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients $c_k$ is

$(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}$

so that with r = 4/5, the coefficients are governed by

$c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}$

c) Starting with $c_0=1$ , we find

$c_1 = -\dfrac{c_0}5 = -\dfrac15$

$c_2 = -\dfrac{c_1}{10} = \dfrac1{50}$

so that the first three terms of the solution are

$\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}$

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2 years ago

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VLD [36.1K]

Answer:

t greater than or less than -10

Step-by-step explanation:

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2 years ago