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KATRIN_1 [288]
3 years ago
13

3/4(x+8)>1/2(2x+10)​

Mathematics
1 answer:
JulijaS [17]3 years ago
8 0

Answer:

x < 4

Step-by-step explanation:

3/4(x+8)>1/2(2x+10)​    (multiply both sides by 4)

(4)(3/4)(x+8)>(4)(1/2)(2x+10)​

3(x+8)>2 (2x+10)​  (expand parentheses by distribution property)

x(3)+8(3)>(2x)(2)+10(2)​

3x+24>4x+20   (subtract 3x from both sides)

​3x+24 - 3x > 4x+20- 3x

​24 > x+20 (subtract 20 from both sides)

24 - 20 > x + 20 - 20

4 > x  (rearrange)

x < 4

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How do i solve integrals?
Cerrena [4.2K]

Answer:

Step-by-step explanation:

So the integral of 2 is 2x + c, where c is a constant. An "S" shaped symbol is used to mean the integral of, and dx is written at the end of the terms to be integrated, meaning "with respect to x". This is the same "dx" that appears in dy/dx . To integrate a term, increase its power by 1 and divide by this figure.

8 0
3 years ago
find the angle between the vectors. (first find the exact expression and then approximate to the nearest degree. ) a=[-8,6]. B=[
just olya [345]

Answer:

49°

Step-by-step explanation:

Given vectors:

a = [-8, 6]

B = [√7, 3]

θ = ?

To find the angle between the two vectors, we will be using the formula,

                                   a.B = |a||B|cosθ

For simplicity, it is good to first calculate the dot product, and the magnitudes. Then we will substitute the values of the dot product, and the magnitudes of the vectors to solve for the angle.

Calculating the dot product

         a.B = (-8, 6) . (√7, 3)

               = (-8 × √7) + (6 × 3)

               = -8√7 + 18

               = 18 - 8√7

               = 10√7

Calculating the magnitude the vectors

1. The magnitude of vector (-8, 6)    

                |a| = \sqrt{(-8)^{2} + 6^{2}}

                |a| = \sqrt{64 + 36}

                |a| = \sqrt{100}

                |a| = 10

2. The magnitude of vector (√7, 3)

                |B| = \sqrt{(\sqrt{7})^{2} + 3^{2}}

                |B| = \sqrt{7 + 9}

                |B| = \sqrt{16}

                |B| = 4

Calculating the angle between the vectors,

                cosθ = \frac{a.B}{|a||B|}

                cosθ = \frac{10\sqrt{7}}{10 * 4}

                cosθ = 0.6614

                θ = cos⁻¹0.6614

                θ = 48.59°

                θ = 49°

5 0
4 years ago
Find two solutions to the equation (x^3 − 64)(x^5 − 1) = 0.
Rashid [163]

Answer:

the two roots are x = 1 and x = 4

Step-by-step explanation:

Data provided in the question:

(x³ − 64) (x⁵ − 1) = 0.

Now,

for the above relation to be true the  following condition must be followed:

Either  (x³ − 64) = 0 ............(1)

or

(x⁵ − 1) = 0 ..........(2)

Therefore,

considering the first equation, we have

(x³ − 64) = 0

adding 64 both sides, we get

x³ − 64 + 64 = 0 + 64

or

x³ = 64

taking the cube root both the sides, we have

∛x³ = ∛64

or

x = ∛(4 × 4 × 4)

or

x = 4

similarly considering the equation (2) , we have

(x⁵ − 1) = 0

adding the number 1 both the sides, we get

x⁵ − 1 + 1 = 0 + 1

or

x⁵ = 1

taking the fifth root both the sides, we get

\sqrt[5]{x^5}=\sqrt[5]{1}

also,

1 can be written as 1⁵

therefore,

\sqrt[5]{x^5}=\sqrt[5]{1^5}

or

x = 1

Hence,

the two roots are x = 1 and x = 4

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Answer:

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<span>A number t divided by 82 
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<span>t divided by 82 = t / 8 
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3 0
3 years ago
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