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2 years ago

5

Log3(2x-1)=3 help me out:(

1 answer:

aleksley [76]2 years ago

5 0

Answer:

x = 2.

Step-by-step explanation:

(Brainliest?)

So log3(2x-1)+log3(x+7)=3 = log3(3^3) So (2x- 1)(x+7) = 27..... 2x^2+13x - 34 =0 (2x+17)(x - 2) = 0 ..so x = 2 or - 17/2. The solution x = - 17/2 is rejected since we cannot have log 3(2*- 17/2 - 1),a negative number. So the answer is x = 2.

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2. In circle 0, points A and B lie on the circle such that OA = 2x + 9 and OB = 27 – x. Which of the following

malfutka [58]

6 because you set them equal and solve

4 0

2 years ago

Please help me with this question please!!!

lozanna [386]

A = x * x = x^2

B = 6 * x = 6x

C = 8 * x = 8x

D = 6 * 8 = 48

So the area of the whole shape is each side added up then multiplied to give (x+6)(x+8) =

x^2 + 14x + 48

7 0

2 years ago

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Lindsey was making got chocolate. She used 3 cups of water to 2 tsp of coco mix. What is the ratio of coco mix to the water?

sineoko [7]

Answer:

Ratio of coco mix to water is 2 : 3.

Step-by-step explanation:

Given that:

Lindsey used 3 cups of water to 2 tsp of coco mix to make chocolate.

Ratio is the quantitative relationship between two amounts.

Ratio of coco mix to water.

Tsp of coco mix : Cups of water

2 : 3

Hence,

Ratio of coco mix to water is 2 : 3.

3 0

2 years ago

Find the taylor polynomial t3(x) for the function f centered at the number

inysia [295]

Answer:

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

Step-by-step explanation:

We are given that

$f(x)=7tan^{-1}(x)$

a=1

$T_n(x)=\sum_{r=0}^{n}\frac{f^r(a)(x-a)^r}{r!}$

Substitute n=3 and a=1

$t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}$

$f(x)=7tan^{-1}(x)$

$f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}$

Where $tan^{-1}(1)=\frac{\pi}{4}$

$f'(x)=\frac{7}{1+x^2}$

Using the formula

$\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}$

$f'(1)=\frac{7}{2}$

$f''(x)=\frac{-14x}{(1+x^2)^2}$

$f''(1)=-\frac{7}{2}$

$f''(x)=-14x(x^2+1)^{-2}$

$f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})$

By using the formula

$(uv)'=u'v+v'u$

$f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}$

$f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}$

$f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}$

Substitute the values

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3$

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

7 0

3 years ago

Write and solve an inequality to find the values of x for which the perimeter of the rectangle is less than 140.

Mashcka [7]

Given:

Length of the rectangle = $x+4$

Width of the rectangle = $x$

The perimeter of the rectangle is less than 140.

To find:

The inequality for the given situation and solve it.

Solution:

We have,

$l=x+4$

$w=x$

We know that, the perimeter of a rectangle is

$P=2(l+w)$

$P=2(x+4+x)$

$P=2(2x+4)$

$P=4x+8$

It is given that the perimeter of the rectangle is less than 140.

$P$

$4x+8$

$4x$

$4x$

Divide both sides by 4.

$x$

$x$

Therefore, the correct option is A because the sign of inequality is < and the solution of inequality is $x$ .

3 0

2 years ago